text:
Mission, Vision & History
C4RJ Case Stories
Resources for Current Volunteers
Involve Your Community
Your Gift Matters
What Is Restorative Justice?
The Restorative Justice Approach
How the Circle Process Works
Success Data
I'm Making a Referral
Why Restorative Justice
The Officer's Role
Ready to Refer
Sample Forms & Policies
I'm Affected by a Crime
You've Been Affected by a Crime: Now What?
I'm Facing Charges
Facing Criminal Charges
C4RJ member police chiefs and other police representatives discuss C4RJ policies and practices (left); C4RJ member police chiefs and other police representatives in front of the Bedford Police Department following a quarterly Police Council in May, 2018 (center); Arlington Police Chief Frederick Ryan and C4RJ Executive Director Erin Freeborn training new volunteers (right).
Considering a referral
What are the basic criteria for referral?
The victim is willing to have the case come to us instead of pressing charges or letting the responsible party (RP) be summonsed.
The RP is admitting the wrongdoing. (They may not understand the full ripple effects of the harm. That's OK. Such understanding will emerge in the process.)
We can be reasonably sure of a safe process. If there are physical or emotional risks, we may return the case to you.
Do victims have to participate directly in the "circle" process?
No. Victims can allow the case to come to us and also elect NOT to participate or to participate indirectly (e.g., stating preferences for a community service site or asking for a letter of apology). We can represent their views and needs if they cannot or do not wish to be present.
What should I say to the victim before referring a case?
Your department may set guidelines for this conversation but we recommend that you say, "This case may be referred to a nonprofit organization with whom we partner called C4RJ. You'll have a chance to say how you've been affected and ask for direct repair. You'll be hearing from C4RJ in the days ahead." We also encourage you to give them the C4RJ leave-behind piece.
What should I say to the Responsible party (RP) before referring a case?
Your department may set guidelines on this conversation but we recommend that you conduct the investigation as if the case were proceeding to court. At the time of the incident, if you think the case might be a good fit, you can say, "This case may be referred to a nonprofit organization with whom we partner called C4RJ. You'll be asked to take responsibility and make repair to those who've been affected. You'll be hearing from C4RJ in the days ahead." We suggest that you not promise a referral since other issues could arise that would make referral inappropriate. We also encourage you to give them the C4RJ leave-behind piece.
It may also be useful to mention that C4RJ requires a $250 participation fee of RP's accepting referral. We have a sliding scale for those who cannot afford the payment, and we also offer payment plans.
The RP admits s/he did it, but doesn't seem remorseful. Is that OK?
Yes. It's very common for RPs to "admit" they did something but not own up fully or even show regret. They may feel as though they are the victims by virtue of being caught. Most of the time, we see these attitudes dissipate during the process, especially once the individual has been able to see: (a) how others were affected; and (b) that they are going to have a chance to be heard and to make amends—and avoid being stigmatized as a bad person. As you know, most of us, when we do something wrong, make excuses. Getting past that is the work of the circle process and the period after.
What if the Responsible Party has a "bad attitude"?
Restorative justice is not just for RPs with "good attitudes" or who've been exemplary citizens. RJ isn't a means for doling out favors – in fact, offering restorative justice to one RP but referring another to court under the same circumstances may prompt accusations of favoritism or bias.
Further, the victim may still want the opportunity to speak with the RP even if the RP has "mouthed off" to you or been unpleasant. We will do our best to ensure that the victim is not revictimized by a bad attitude. We deal with this by meeting ahead of time – perhaps several times – with the RP and supporters.
What if the Responsible party has been in trouble before?
Each department will decide what its referral criteria are, but re-offense or a "frequent flyer" status is not a barrier from C4RJ's point of view. In some instances, the supports we offer, an encounter with those affected by harm, and close supervision, may help produce an outcome different from that of the court, and a change in the RP's outlook. We can't guarantee that, but prosecution is always an option if our process doesn't work.
In some instances, RPs who have appeared in court for previous offenses have been excellent participants in restorative justice, fulfilling all obligations.
The Responsible Party is such a nice person and is already upset. Won't restorative justice just make it worse for him/her?
The victim and others affected may need the opportunity to speak with the RP, regardless of how bad the RP feels. Beyond that, making amends may be just what this RP needs in order to feel s/he can reclaim some self-worth. Further, s/he will likely benefit from the support of the circle and from working with our trained volunteers.
What if the Responsible Party isn't owning up?
If he or she isn't admitting his or her role, you wouldn't refer to us. But if they have engaged in a circle and aren't living up to commitments or if they just aren't really taking personal responsibility (they're blaming someone else or making excuses), we can reconvene the circle or return the case to you. This happens, but rarely.
How does C4RJ try to ensure a safe process?
We hold most circles in police departments. And we always include an officer in the circle. It's up to the department or individual officer whether to attend in uniform and/or armed. Part of our intake process is to help us gauge whether an RP will be able to engage appropriately in the circle. In many cases, we seek an evaluation beforehand with a qualified therapist.
We've already made an arrest or requested a summons. Is it too late to consider restorative justice?
No. RJ can be offered at various stages, including post-complaint. C4RJ has hosted successful circles:
As a condition of a plea agreement
As part of juvenile diversion contracts
While a clerk or judge held a charge pending completion of a restorative agreement
Restorative justice has also been used post-conviction to allow dialogue between incarcerated individuals (including those convicted of murder) and victims.
I'm the school resource officer in my town, and I have an incident that occurred in the school. Is restorative justice an option?
Yes, we often work with schools, with both criminal incidents and other wrongdoing. We hold circles in schools if that makes it easier for the parties involved.
Who will my contact be during the process?
C4RJ is organized in teams, led by one or two case coordinators. For each case, your contact(s) will be the case coordinators who have been assigned the case, and they will provide you with their contact information. You are also welcome to be in touch with C4RJ staff at 978.318.3447.
The circle process
Who is included in a circle?
Typically:
RP(s)
RPs supporters (family members, close friends or colleagues)
Victim(s)
Victim supporters
2-4 C4RJ personnel, including one who will lead the circle and those who will work with the RP and victim after the circle while the restorative agreement is being fulfilled
Concerned community members
We prefer smaller (6 to 10 people) to larger circles, but in instances in which multiple people have been involved and a victim wants to participate in just one circle, we may include more people.
Where are circles typically convened and how long do they last?
We prefer to meet in the police department of the jurisdiction where the offense occurred because it underscores the seriousness of the process and our partnership with you. But there may be times in which a meeting at the PD is inappropriate, such as these rare circumstances:
The victim prefers a different location for convenience or comfort
The number of persons involved would make it difficult to accommodate a circle in the PD's available space. In this case, we look for facilities in the town in which we can meet and still protect privacy.
Does the responding officer in a case have to be the officer who sits in the circle?
No. It's great to have the reporting officer there, in part to carry forward the thread of the relationships that may have started with the incident. But in many instances, a department's scheduling makes that impossible. If that's the case, we ask that the officer who will participate seek a briefing ahead of time by the responding officer.
Restorative justice data and results
What evidence is there that restorative justice works?
C4RJ has a recidivism rate of just 16%. In addition to this measure, we also track participant satisfaction rates. Victim satisfaction rates hover at 90-92%. Some 80% to 85% of RPs report they are satisfied. Hundreds of similar programs around the country and the world have documented similar success.
Does restorative justice ever fail?
Sometimes those responsible aren't ready to be held accountable, or they may not be ready to take concrete steps toward repair. That said, it is seldom that we have to return a case to the referring partner because RPs aren't fulfilling their obligations.
I hear restorative justice is soft on crime. Won't this encourage RPs and would-be RPs to think they're going to "get off easy"?
Recently, a judge observed that we "work" RPs a lot harder in our program than most judges and the District Attorney's office do in their sentences and diversion agreements. Once a young RP remarked, "If I'd known I had to do this much, I might have gone to court!" (We don't think his parents would have agreed!)
Whereas, generally, an RP goes to court once or twice and may see a probation officer periodically, RPs in our program have contact with one of our trained facilitators weekly, usually in person, over two to three months. Meetings focus on decision-making, habits of thinking, and developing empathy. Facilitators are also checking to ensure the RP is pursuing other obligations made in the restorative agreement, sometimes including community service, apologies, and any therapeutic or educational programs specified.
Does restorative justice always lead to forgiveness?
No. We stress that forgiveness is not a requirement of this process. Victims need not offer it. RPs should not ask for it since it might well place an additional obligation on the victim. If forgiveness comes about as part of the process, great. If not, that's OK too. The process we offer is about accountability and repair. If you speak with a victim who is not in a posture of forgiveness in the wake of harm suffered, that's OK. The process may yield important information or answers for them.
What oversight is there for C4RJ?
C4RJ is governed by a Board of Directors, and advised by our Police Council, a quarterly gathering of liaisons from each of our partnering departments. All C4RJ policies are vetted by the Police Council and approved by the Board. As a 501(c)(3), C4RJ also reports to state and federal bodies charged with regulation of nonprofit organizations.


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6.2.14. – Twizzlers are Made From Soy?!??
So today we found out our little man is allergic to MILK & SOY. They say it's common, but it's still hard to deal with. You wish health and happiness for your child for the 9 months he's cooking up in the womb and then he comes, you count 10 toes and 10 fingers and you hear that cry and you feel amazing. You never realize how many hurdles and milestones there are throughout the first few years. The love you feel for your child is so extreme that helplessness sets in… I hate feeling helpless.
I hate even more knowing that Jax will miss out on some of the world'a greatest joys. What an awful feeling to look at you son, so small and so innocent knowing that he'll never be able to enjoy a bowl of frosted flakes or a slightly melted cup of rocky road ice cream.
PS: Mom is struggling with the fact that she can't eat Twizzlers anymore… Who would have thought there was anything more than sugar and nastiness in those disgusting red sticks.
This entry was posted in Uncategorized and tagged allergies, dairy, soy, twizzlers. Bookmark the permalink.


text:
\section{Introduction}

For most of  realistic models of quantum many-particle systems an exact solution cannot be achieved. This is a typical situation condensed matter or  nuclear physics. As a consequence, a number of approximate methods have been developed. Among them, widely used are various types of the mean-field approaches.

By mean-field (MF) description of the problem we understand  making use of the Hamiltonian which depends on  extra parameters, having the meaning of expectation values of well-defined operators. Values of those parameters  are not   \textit{a priori} known and are to be determined.  The  Hartree-Fock method is a good  example of such approach.

MF methods provide us with a valuable insight into the physical properties of a system under the consideration. They  are capable of describing many interesting phenomena, especially those involving phase transitions,  such as that from normal metal to  superconductor transition (e.g. in the form of the Bardeen-Cooper-Schrieffer theory), or those from a paramagnet to either antiferro- or ferromagnet.

In implementing such a procedure, it is the  physics of the problem, not the mathematics, that tells us when the application of MF approach is justifiable.  Nonetheless,  we  may also be faced with the problem of considering the internal mathematical consistency of the method, as well as its underlying general principles. The aim of this paper is to clarify this issue to some extent, in a simple manner

The paper is organized as follows. Sec.  \ref{statement of the} contains  formulation of the problem and invokes basic relevant notions from statistical mechanics. In Sec. \ref{meanfield}, by means of the maximum entropy principle we construct the statistical-mechanical description of the system in the  mean-field case.  Explicitly, in \ref{consturction} we derive the grand-canonical (GC) MF density operator, in \ref{Thermodynamics} we comment on the MF thermodynamics and its consistence with the MF statistical mechanics, whereas in \ref{Hartree} we discuss the important case of the Hartree-Fock (HF) type MF Hamiltonians. Finally, in \ref{Bogoliubov} the connection of present method with the variational principle based on the Bogoliubov inequality is elaborated. In Sec. \ref{examples} two specific  examples of mean-field Hamiltonians are provided, for which our approach leads to significant corrections. Section \ref{summary} contains a summary.

\section{Statement of the problem\label{statement of the}}
We assume that the Hilbert-Fock state space is of finite dimension $D$. The MF Hamiltonians are denoted as $\hat{H}$, whereas those  which are not of MF type, will be called \textit{exact} and labelled with the subscript '\textit{e}'.

As already mentioned in the Introduction, by  \textit{mean-field approach} we understand the situation in which Hamiltonian depends on certain parameters, that are to be determined when  solving the problem at hand.
Such  MF Hamiltonian is usually derived from the exact Hamiltonian $\hat{H}_{e}$, by replacing some of the operators, $\{ \hat{A}_s \} = \hat{A}_1, \ldots, \hat{A}_M$,   appearing in $\hat{H}_{e}$,   by  $ \mathbb C$-numbers $\{A_s \} = A_1, A_2, \ldots,  A_M$ $(\equiv \vec{A})$.
Those numbers have a natural interpretation of expectation values of the corresponding operators $\{\hat{A}_s\}$. However, the mean-field Hamiltonian  $\hat{H}$ may be not be related in an obvious manner to any exact Hamiltonian.

We assume that the system is in thermal equilibrium with the particle reservoir characterized by the temperature
 $T = \beta^{-1}$  and the chemical potential $\mu$.  The  question then arises, how to determine the values of  parameters $A_s$  appearing in $\hat{H}$.
If the Hamiltonian of the system were exact, i.e. not expectation-value dependent, then the treatment of such situation is straightforward, in the sense, that the  grand canonical (GC) ensemble is well defined, and all information about the system is contained in the  density operator \cite{statistical mechanics, Zubariev} defined as
\begin{equation}
\hat{\rho}_{e} = Z_{e}^{-1} \exp \big(-\beta ( \hat{H}_{e} - \mu  \hat{N})\big),
\label{exact density matrix GCE}
\end{equation}
where $Z_{e} =  \text{Tr}[  \exp \big(-\beta ( \hat{H}_{e} - \mu  \hat{N})\big) ] $ is the grand-canonical partition function. The expectation value of any observable $\hat{A}$ is then given by
\begin{equation}
\langle \hat{A}\rangle_{e}  \equiv  \text{Tr}[\hat{A}  \hat{\rho}_{e}].
\label{mean value of A exact}
\end{equation}
Moreover, we assume, that in the mean-field case  this basic definition  still holds, i.e.
\begin{equation}
\langle \hat{A}\rangle  =  \text{Tr}[\hat{A}  \hat{\rho}(\vec{A})].
\label{mean value of A}
\end{equation}
Regardless of the  form of  MF  density operator $\hat{\rho}$,  we should be able to obtain any expectation value through (\ref{mean value of A}). In particular, the last definition allows us to determine the values of $A_1, A_2, \ldots,  A_M$ appearing in $\hat{\rho}$,
by solving  a system of  $M$ implicit  equations for $M$ variables (the so-called self-consistency equations), which take the form
\begin{equation}
\langle \hat{A}_{s}\rangle  = {A}_{s}  = \text{Tr}[\hat{A}_{s}  \hat{\rho}(A_1, A_2, \ldots,  A_M)], ~~~~ \forall_{s}.
\label{mean value of A set of eq}
\end{equation}
We may hope that this system should  have at least one  solution.

One would expect, that MF grand-canonical density operator should be given by (\ref{exact density matrix GCE}) with $\hat{H}_{e}$ replaced with $\hat{H}$, so
\begin{equation}
\hat{\tilde{\rho}} = \tilde{Z}^{-1} \exp \big(-\beta ( \hat{H} - \mu  \hat{N})\big), ~~~~\tilde{Z} =  \text{Tr}[  \exp \big(-\beta ( \hat{H} - \mu  \hat{N})\big) ].
\label{MF naive DO}
\end{equation}
This is, however, not necessarily the case, as is discussed in detail below.
In order to find the correct mean-field GC density operator for a given MF Hamiltonian $\hat{H}$, we  proceed  analogously to the exact (non-MF) case and \textit{derive} the form of such density operator from the basic principles of statistical mechanics.
Thus, before concentrating in the remaining part of the paper on the mean-field case, we first  would like to summarize briefly the non-MF situation.
Namely, for the exact Hamiltonian $\hat{H}_{e}$, the GC density operator (\ref{exact density matrix GCE})  may be derived  from the  requirement that it  maximizes the  von Neumann entropy, $S \equiv - \text{Tr} \hat{\rho}_{e} \ln \hat{\rho}_{e}$, under the conditions that the expectation values of $\hat{H}_{e}$ and  of the particle number operator $\hat{N}$, are fixed. This means, that we maximize the following functional (see \cite{statistical mechanics, Zubariev})
\begin{equation}
\mathcal{S}_{e} =   \text{Tr} \big( -\hat{\rho}_{e} \ln \hat{\rho}_{e} -\beta  \hat{\rho}_{e} \hat{H}_{e}
+ \beta \mu  \hat{\rho}_{e} \hat{N} - \omega \hat{\rho}_{e} \big ).
\label{entropy exact GCE}
\end{equation}
The first two conditions are enforced with the help of the Lagrange multipliers $\beta$ and $\mu$, whereas the parameter $\omega$ is introduced to ensure the normalization, $\text{Tr} [\hat{\rho}_{e}]=1$.
Furthermore,  $\hat{\rho}_{e}$ must be a  constant of motion, so from the quantum Liouville equation \cite{statistical mechanics, Zubariev}
\begin{equation}
i \hbar  \frac{\partial}{\partial t}  \hat{\rho}_{e}(t) = [ \hat{H}_{e}, \hat{\rho}_{e}(t)],
\label{quantum Liouville}
\end{equation}
it follows, that $ [ \hat{H}_{e},  \hat{\rho}_{e}] = 0$. If, additionally, $ [ \hat{H}_{e},  \hat{N}] = 0$, which is usually the case in non-relativistic quantum mechanics, we may simultaneously diagonalize each of the operators appearing in (\ref{entropy exact GCE}).

Parenthetically, the maximization procedure is most conveniently carried out in the basis, in which $\hat{\rho}_{e}$ is  diagonal, i.e. $\hat{\rho}_{e} = \sum_{i}^{D} p_i |i \rangle \langle i|$.
The resulting $\hat{\rho}_{e}$ is then of the form given by (\ref{exact density matrix GCE}), \cite{Zubariev}. Density operators for the canonical and the micro-canonical ensembles may be obtained in a similar manner.
The fact, that those density operators maximize the entropy functional (\ref{entropy exact GCE}) or analogous expressions, may be considered as their basic property  \cite{Jaynes}.

\section{Maximum entropy principle in the mean-field case \label{meanfield}}
\subsection{Construction of the mean-field density operator \label{consturction}}
We turn now to the  construction  of the proper form of MF density operator. We concentrate on the case, for which the entire system description is based \textit{solely} on  MF Hamiltonian.  In the previous Section it was stated, that maximum entropy condition  is basic for canonical density operators. Moreover, the principle of maximum entropy is one of basic postulates of statistics, it ensures uniqueness and consistency of probability assignments, \cite{Jaynes}.  Consequently, it should be applied also to the particular case of mean-field description.
Now, the difference with the standard, i.e. non-MF situation above is that the Hamiltonian depends on $M$ additional variables $\{A_s\}\equiv \vec{A}$, interpreted as the expectation values of corresponding operators $\{\hat{A}_{s}\}$.
Those $M$ expectation values, together with $D$ diagonal matrix elements of $ \hat{\rho}$, are now the variational parameters, with respect to which the functiona
\begin{equation}
\mathcal{S} =  \text{Tr} \big( - \hat{\rho} \ln  \hat{\rho} -\beta  \hat{\rho} \hat{H}(\vec{A})
+ \beta \mu \hat{\rho} \hat{N} - \omega (\hat{\rho} - \frac{1}{D}) \big),
\label{entropy MF no lambdas}
\end{equation}
is maximized. As earlier, $ - \omega (\hat{\rho}  - 1/D)$ term\footnote{Here taken in the form which slightly differs from that of (\ref{entropy exact GCE}), but this technical alteration is inessential.} is added to ensure normalization of the density operator. Next, if those  $M+D$  parameters  were treated as  independent, there is no certainty that $\{A_s\}$ obtained in that manner coincide with those resulting form the self-consistency conditions  (\ref{mean value of A set of eq}).
In principle, one may use (\ref{mean value of A set of eq})
to eliminate the $A_s$ variables from $\hat{H}$.  Nonetheless, the resulting dependence of $\mathcal{S}$ on $D$ diagonal elements of $\hat{\rho}$ is usually  very  complicated, and the MF density operator cannot be obtained in an explicit form.

To resolve the situation, we use the  Lagrange-multiplier method. Explicitly, the $M$  conditions (\ref{mean value of A set of eq}) are  enforced with the help of $M$ Lagrange multipliers $ \{\lambda_{s} \}\equiv \vec{\lambda} $. To  account for that, the terms of the form $\lambda_{s} (\text{Tr}[ \hat{\rho}\hat{A}_{s}] - A_{s} ) $ must be added to the functional (\ref{entropy MF no lambdas}). Under these circumstances, we have in total $D + 2M $ variables, $\{p_1, \ldots, p_D, A_1, \ldots, A_M,  \lambda_1, \ldots, \lambda_M \}$, treated as independent (apart from the normalization condition). In effect, the constrained entropy functional reads
\begin{equation}
\mathcal{S}_{\lambda} =  \text{Tr} \Big[ - \hat{\rho}_{\lambda} \ln  \hat{\rho}_{\lambda} -\beta \big( \hat{\rho}_{\lambda} \hat{H}
-  \mu \hat{\rho}_{\lambda} \hat{N} - \sum_{s=1}^{M}  \lambda_{s}\hat{\rho}_{\lambda} (\hat{A}_{s} -  A_{s})\big) - \omega (\hat{\rho}_{\lambda} - \frac{1}{D})\Big].
\label{entropy MF}
\end{equation}
The incorporation of the self-consistency constraints may be regarded as a redefinition of the MF Hamiltonian
\begin{equation}
\hat{H} ~~\to ~~ \hat{H}_{\lambda} ~= ~  \hat{H}-\sum_{s=1}^{M}   \lambda_{s}(\hat{A}_{s} - A_{s} ).
\label{GMF Hamiltonian with lambdas}
\end{equation}
The last step  is the basic novel ingredient of our method  and is discussed in detail in the remaining part of this paper.

Few  remarks are in place  here. First, we are mainly interested in a situation when  each operator $\hat{A}_{s}$ is bilinear in creation and/or  annihilation operators\footnote{In principle, the presented formalism applies also to the  general situation of the arbitrary mean-field Hamiltonian.}, as then $\hat{H}_{\lambda}$ is easily tractable. Second, some  combination of  the  operators  $\{ \hat{A}_{s}\}$ appearing in (\ref{GMF Hamiltonian with lambdas}) may be proportional to the number operator $\hat{N}$. Nonetheless, the presence of the chemical potential $\mu$ fixes the  average number of particles $N$.

Third, the   operators $\hat{H}$ and $\hat{N}$ in (\ref{entropy MF})  have a different  status than the operators $\{ \hat{A}_{s}\}$.  The expectation values of the former are assumed to be fixed \textit{a priori}, enforced by the values of  temperature and chemical potential. In contrast, the expectation values of the latter are obtained from a variational procedure, and are not \textit{a priori} given.
Consequently, the same distinction holds for parameters  $\beta$, $\mu $ from one side and  $\{ \lambda_{s}\}$ from the other, even though all of them play the role of Lagrange multipliers. Nonetheless, some of  $\{ \lambda_{s}\}$ may have a physical interpretation. Depending on the corresponding  operator $ \hat{A}_{s}$, they may be termed as local fugacities, molecular fields, etc. Note, that for $ \hat{A}_{s} =   \hat{A}^{\dag}_{s}$,  $ \lambda_{s}$ is real, but for non-Hermitian $ \hat{A}_{s}$, it is a complex number in general.

Apart from $ \hat{H}$ and  $\hat{N}$, there may be more observables with expectation values fixed \textit{a priori}. In such a situation the  terms $(- \xi_s \hat{A}_s)$, analogous to $- \mu \hat{N}$ term, should be also added, to   ensure that the quantity $\langle \hat{A}_s  \rangle$ is consistent with the prior information about the system.  If we ignore some available information about the physical system under consideration, we obtain a broader probability distribution.

Fourth, the operators appearing in  (\ref{GMF Hamiltonian with lambdas}) do not necessarily  commute with each other, and thus cannot be, in general, simultaneously diagonalized. However, in analogy to the canonical case, we require that $\hat{\rho}_{\lambda}$ thus obtained is stationary, i.e.  $  \partial   \hat{\rho}_{\lambda}(t)/\partial t =0$, from which, using the quantum Liouville equation (\ref{quantum Liouville}) for the MF case the condition $[\hat{H}_{\lambda}, \hat{\rho}_{\lambda}] = 0$ follows.
In particular, in the case  $[\hat{N}, \hat{H}_{\lambda}] \neq 0$ (e.g. in the BCS theory  of superconductivity), we can also redefine the Hamiltonian according to $\hat{H}_{\lambda} \to \hat{K}_{\lambda} = \hat{H}_{\lambda} - \mu \hat{N}$, and then demand, that the corresponding grand Hamiltonian commutes with the density operator, $[\hat{K}_{\lambda}, \hat{\rho}_{\lambda}] = 0$. In result,  diagonalization of $\hat{K}_{\lambda}$  provides us again with  the set  of $\vec{A}$  and $\vec{\lambda}$- dependent eigenvectors $\{ | i(\vec{A}, \vec{\lambda})\rangle \}_{i=1}^D$ and corresponding eigenvalues of the density operator $ \hat{\rho}_{\lambda}$.

Fifth, the present formalism may be easily modified and applied to the case of canonical ensemble or discrete classical models (e.g. lattice gas, or spin systems).

Finally, even if, strictly speaking, the $T=0$ situation cannot be investigated directly within the framework of the present formalism, (must be analyzed separately), it is fully legitimate as well as convenient to analyze the situation in question  as the $T\to 0$ limit, e.g. for finite, but sufficiently low temperature. A more detailed analysis of this point will be presented elsewhere \cite{Jedrak}.

Let us go back to the problem of finding the maximum of the entropy subject to the constraints.
The functional (\ref{entropy MF}), written in the common eigenbasis of $\hat{K}_{\lambda}$ and $\hat{\rho}_{\lambda}$ reads
\begin{equation}
\mathcal{S}_{\lambda} =  \sum_{i=1}^{D} \Big\{ - q_{i} \ln  q_{i} -\beta  q_{i}\Big( (\hat{H})_{ii}
-\mu  (\hat{N})_{ii}  - \sum_{s=1}^{M}    \lambda_{s}  \big( (\hat{A}_{s})_{ii} -  A_{s}\big)\Big)  - \omega (q_{i} - \frac{1}{D}) \Big\},
\label{entropy MF eigen}
\end{equation}
(with $ (\hat{N})_{ii} = \langle i | \hat{N} | i \rangle $ etc.) We assume, that the maximum of (\ref{entropy MF eigen}) corresponds to stationary point. Then, the necessary conditions for an extremum of $\mathcal{S}_{\lambda}$ with the normalization preserved, are
\begin{equation}
\frac{\partial \mathcal{S}_{\lambda}}{\partial  \omega}  =  1 - \sum_{j=1}^{D}  q_j = 0,
\label{derivative omega}
\end{equation}

\begin{equation}
\forall_{j}:  ~~~ \frac{\partial \mathcal{S}_{\lambda}}{\partial  q_{j}}  = - (1 + \omega) - \ln q_j -\beta  (\hat{H})_{jj}  + \beta \mu  (\hat{N})_{jj}  + \sum_{s=1}^{M}   \beta\lambda_{s}  \big( (\hat{A}_{s})_{jj} -  A_{s} \big)  =   0,
\label{derivative q i}
\end{equation}
\begin{equation}
\forall_{w}:~~~   \frac{\partial \mathcal{S}_{\lambda} }{\partial  A_{w}}  =  -\beta \sum_{i=1}^{D}
q_{i}   \Big(\frac{\partial \hat{H}_{ii} }{\partial  A_{w}}   + \lambda_{w} \Big) = 0,
\label{derivative A s}
\end{equation}
and
\begin{equation}
\forall_{w}:~~~ \frac{\partial \mathcal{S}_{\lambda}}{\partial  \lambda_{w}} =  \beta \sum_{i=1}^{D} q_i  \big( (\hat{A}_{w})_{ii} -  A_{w})  \big)  = 0.
\label{derivative lambda s}
\end{equation}
In (\ref{derivative A s}) and (\ref{derivative lambda s}), we ignored the possible explicit $\vec{A}$- and $\vec{\lambda}$- dependence of the (normalized) eigenvectors $|i \rangle$. This is justified due to the  Hellmann-Feynman theorem, \cite{Feynman theorem}. Using (\ref{derivative omega}) and (\ref{derivative q i})   we obtain the explicit, basis-independent form of  density operator $\hat{\rho}_{\lambda}$,
\begin{equation}
 \hat{\rho}_{\lambda} =  \mathcal{Z}_{\lambda}^{-1} \exp \big(-\beta ( \hat{H}_{\lambda} - \mu  \hat{N})\big), ~~~  \mathcal{Z}_{\lambda}= \text{Tr}[\exp \big(-\beta ( \hat{H}_{\lambda} - \mu  \hat{N})\big)].
\label{correct MF density matrix}
\end{equation}
We see that (\ref{correct MF density matrix}) differs from (\ref{MF naive DO})  by the presence of additional terms in the Hamiltonian, in accordance with (\ref{GMF Hamiltonian with lambdas}). Additionally, Eqs. (\ref{derivative A s}) and (\ref{derivative lambda s}) may be rewritten respectively as
\begin{equation}
 -\frac{1}{\beta} \frac{\partial \mathcal{S}_{\lambda} }{\partial  A_{w}}  = \text{Tr}[ \hat{\rho}_{\lambda}\Big(  \frac{\partial  \hat{H} }{\partial A_w}  + \lambda_w  \Big)] = \Big \langle  \frac{\partial  \hat{H} }{\partial A_w} \Big \rangle_{\lambda}  + \lambda_w   =      \Big  \langle   \frac{\partial  \hat{H}_ {\lambda} }{\partial A_w} \Big  \rangle_{\lambda} = 0,
\label{mean value of H GMF with lambda, d/dA}
\end{equation}
and
\begin{equation}
\frac{1}{\beta} \frac{\partial \mathcal{S}_{\lambda} }{\partial  \lambda_{w}}  = \text{Tr}[ \hat{\rho}_{\lambda}(\hat{A}_w -  A_{w})]=
\big \langle  \hat{A}_{w} \big \rangle_{\lambda}  -  A_{w} = \Big \langle   \frac{\partial  \hat{H}_ {\lambda} }{\partial \lambda_w} \Big \rangle_{\lambda} = 0.
\label{mean value of H GMF with lambda, d/dlambda}
\end{equation}
Next, we define the generalized  grand-canonical Landau functional
\begin{equation}
\mathcal{F}(\vec{A}, \vec{\lambda}) \equiv -\beta^{-1}\ln \mathcal{Z}_{\lambda} = -\beta^{-1} \mathcal{S}_{\lambda}( \hat{\rho}_{\lambda}(\vec{A}, \vec{\lambda}), \vec{A}, \vec{\lambda}),
\label{mathcal F functional}
\end{equation}
with $\mathcal{Z}_{\lambda}$ given by (\ref{correct MF density matrix}). That last relation may be easily verified by inserting $\hat{\rho}_{\lambda}$ given by
(\ref{correct MF density matrix}) into (\ref{entropy MF}). Thus, the constrained minimum of $\mathcal{F}(\vec{A}, \vec{\lambda})$ corresponds to the constrained maximum of $\mathcal{S}_{\lambda}( \hat{\rho}_{\lambda}(\vec{A}, \vec{\lambda}), \vec{A}, \vec{\lambda})$, here understood as a function of $\vec{A}$ and $ \vec{\lambda}$ only,  once Eqs. (\ref{derivative omega}) and (\ref{derivative q i}) are solved. From now on we will use $\mathcal{F}$ instead of $\mathcal{S}_{\lambda}$.
 With the help of (\ref{mathcal F functional}), the Eqs. (\ref{mean value of H GMF with lambda, d/dA}) and (\ref{mean value of H GMF with lambda, d/dlambda}) are equivalent to
\begin{equation}
\nabla_{A} \mathcal{F} =0, ~~~~~ \nabla_{\lambda} \mathcal{F} =0,
\label{derivative of mathcalF A, lambda}
\end{equation}
where $\nabla_{A} \equiv \big( \frac{\partial }{\partial  A_{1}}, \frac{\partial }{\partial  A_{2}}, \ldots, \frac{\partial }{\partial  A_{M}}  \big)$, and analogously for $\nabla_{\lambda}$.
When deriving the last condition we made use of the identity (c.f. \cite{Zubariev})
\begin{equation}
\frac{\partial }{\partial x} e^{C(x)} = \int_{0}^{1} d\tau e^{  \tau C(x)} \Big (\frac{\partial }{\partial x} C(x) \Big)  e^{ (1- \tau) C(x)}
\label{general derivative of exp op},
\end{equation}
in combination with the linearity of the trace operation, as well as  its invariance with respect to cyclic permutation of the operators.

For each solution of (\ref{derivative omega})-(\ref{derivative lambda s}) or, equivalently, of (\ref{derivative of mathcalF A, lambda}), it should be verified  separately  if  also the sufficient  conditions for the existence of the constrained  minimum of $\mathcal{F}$ are fulfilled. In practice, this will be most conveniently  carried out through numerical analysis.

We may expect that one of the solutions  of  (\ref{derivative of mathcalF A, lambda}), labeled  ($\vec{A}_0$,  $\vec{\lambda}_0$), yields such desired  global  minimum of $\mathcal{F}$   subject to constraints,  and this defines the equilibrium state. If  more than one solution with the minimal  value of  $\mathcal{F}$ exist, we select one of them, breaking the remaining  symmetry.

However,  $\mathcal{F}$ is well defined also for the values of $\vec{A}$ which differ from $\vec{A}_0$, and  may possess physical interpretation provided that the consistency conditions (\ref{mean value of H GMF with lambda, d/dlambda}) are met. Namely,  $\mathcal{F}(\vec{A}, \vec{\lambda}(\vec{A}))$ may be interpreted as a Landau functional corresponding to the given MF model.  The explicit analytical form of $\mathcal{F}(\vec{A}, \vec{\lambda}(\vec{A}))$ or  its series expansion in the powers of order parameter(s) may be difficult to obtain, and then again we have to resort to the  numerical analysis. Nonetheless,  the concepts of Landau theory of phase transitions may be utilized. E.g., the quantity $\mathcal{N}^{-1}\exp\big(-\beta  \mathcal{F}(\vec{A}, \vec{\lambda}(\vec{A}))\big)$, where $\mathcal{N}$ is normalization constant, may be interpreted as the probability distribution  of $\vec{A}$. Also, plots of $\mathcal{F}(\vec{A}, \vec{\lambda}(\vec{A}))$ may be used for qualitative analysis of the character of the emerging phase transition.

\subsection{Mean-field equilibrium thermodynamics\label{Thermodynamics}}
The equilibrium description is justified, provided that at the point $\vec{A} = \vec{A}_0$ the function ~$\mathcal{N}^{-1}\exp\big(-\beta  \mathcal{F}(\vec{A}, \vec{\lambda}(\vec{A}))\big)$ possesses a sharp peak, and consequently, the fluctuations of order parameter can be neglected. This condition is expected to hold for a wide class of  MF Hamiltonians defined on lattice, in the $\Lambda \to \infty$ limit, where $\Lambda$ is the number of lattice sites. Obviously, for such models, this condition is equivalent to the thermodynamic limit $ V \to \infty$, $N \to \infty$ with finite $N/V = n$. Then we  can define the thermodynamic grand potential
\begin{equation}
\Omega(T, V, \mu) = \mathcal{F}(T, V, \mu; \vec{A}_0(T, V, \mu), \vec{\lambda}_0(T, V, \mu)).
\label{Omega thermo MF}
\end{equation}
With this definition, the MF thermodynamics can be constructed, but we still have to demonstrate its consistency with MF statistical mechanical description based on (\ref{correct MF density matrix}), cf. \cite{Argyres Kaplan}. Such consistency is certainly present for an exact description, making use of non-mean-field Hamiltonians \cite{statistical mechanics, Zubariev, Jaynes, Argyres Kaplan}; in that case we have for example
\begin{equation}
S = -\Big( \frac{\partial \Omega}{\partial T}\Big)_{V, \mu}, ~~~ p = -\Big( \frac{\partial \Omega}{\partial V}\Big)_{T, \mu} , ~~~ N = -\Big( \frac{\partial \Omega}{\partial \mu}\Big)_{T, V}.
\label{S deriv Omega}
\end{equation}
One may ask if in the MF case the r.h.s. of (\ref{S deriv Omega}) correspond to their statistical-mechanical definitions, as the equilibrium MF Hamiltonian $\hat{H}_{\lambda}$  obviously depends on temperature and chemical potential through its dependence on  the equilibrium values of mean fields and Lagrange multipliers\footnote{
We exclude the possibility of explicit dependence of $\hat{H}_{\lambda}$ on $\beta = 1/T$ or $\mu$.}, $(\vec{A}_{0},\vec{\lambda}_{0})$.
It can  be shown, that (\ref{S deriv Omega}) indeed holds, due to the underlying variational principle. Explicitly,
\begin{eqnarray}
\Big( \frac{\partial \Omega}{\partial T}\Big)_{V, \mu} &=&  \beta^{2} \partial_{ \beta}\big( \beta^{-1} \ln \mathcal{Z}_{\lambda}\big) = \beta \mathcal{Z}_{\lambda}^{-1}  \partial_{ \beta} \mathcal{Z}_{\lambda} - \ln \mathcal{Z}_{\lambda} \nonumber \\
&=& - \beta \mathcal{Z}_{\lambda}^{-1} \text{Tr}[ (\beta \partial_{ \beta} \hat{K}_{\lambda} + \hat{K}_{\lambda}) e^{-\beta \hat{K}_{\lambda}} ] - \ln \mathcal{Z}_{\lambda} \nonumber \\
&=& - \beta^{2}  \langle \partial_{ \beta} \hat{K}_{\lambda}\rangle_{\lambda} - \beta \langle \hat{K}_{\lambda} \rangle_{\lambda}  - \ln \mathcal{Z}_{\lambda} \nonumber \\
&=& - \beta^{2}  \langle \partial_{ \beta} \hat{K}_{\lambda}\rangle_{\lambda}  + \langle \ln \hat{\rho}_{\lambda}\rangle_{\lambda} = - \beta^{2}  \langle \partial_{ \beta} \hat{K}_{\lambda}\rangle_{\lambda} -S,
\end{eqnarray}
where $\hat{K}_{\lambda} = \hat{H}_{\lambda} -\mu \hat{N}$ and $\mathcal{Z}_{\lambda}$ are evaluated for $\vec{A}= \vec{A}_{0}$, $\vec{\lambda}= \vec{\lambda}_{0}$. Also, $\ln \mathcal{Z}_{\lambda}(\vec{A}_{0},\vec{\lambda}_{0}) = -\beta \Omega $, and $\langle \hat{K}_{\lambda}\rangle_{\lambda} = U - \mu N$. In effect, we obtain the desired result (\ref{S deriv Omega}), provided that the term  $\langle \partial_{ \beta} \hat{K}_{\lambda}\rangle_{\lambda}$ vanishes. Indeed, in equilibrium we have
\begin{eqnarray}
\langle \partial_{ \beta} \hat{K}_{\lambda}\rangle_{\lambda}&=& \Big \langle \sum_{s = 1}^{M}\Big( \frac{\partial  \hat{K}_{\lambda}}{\partial A_s} \frac{\partial A_{s0}}{\partial \beta} + \frac{\partial  \hat{K}_{\lambda}}{\partial \lambda_s} \frac{\partial \lambda_{s0}}{\partial \beta}\Big)\Big \rangle_{\lambda} \nonumber \\
&=&  \sum_{s = 1}^{M}\Big( \Big \langle \frac{\partial  \hat{H}_{\lambda}}{\partial A_s}\Big \rangle_{\lambda} \frac{\partial A_{s0}}{\partial \beta} + \Big \langle \frac{\partial  \hat{H}_{\lambda}}{\partial \lambda_s} \Big \rangle_{\lambda} \frac{\partial \lambda_{s0}}{\partial \beta}\Big) = 0, \nonumber \\
\end{eqnarray}
due to relations (\ref{mean value of H GMF with lambda, d/dA}) and (\ref{mean value of H GMF with lambda, d/dlambda}). The same reasoning applies to the derivatives of $\Omega$ with respect to  $\mu$ and $V$. In the latter case, Hamiltonian depends explicitly on $V$ (e.g. through the volume dependence of its parameters). In effect, we recover proper expressions for average number of particles $N$ and pressure $p$. Other thermodynamic potentials may be constructed in the standard manner, e.g. $F = \Omega + \mu N$.

Note that as the $\beta$- and $\mu$-  dependences of equilibrium values of mean-fields $\vec{A}_{0}$ obtained through self-consistent Eqs. (\ref{mean value of A set of eq}) are nontrivial in general, and also because (\ref{mean value of H GMF with lambda, d/dA}) with $\vec{\lambda} = \vec{0}$ usually does not  hold in such case, there is no reason for $\langle \partial_{ \beta} \hat{K}\rangle$ (as well as for analogous expressions for the derivatives with respect to $\mu$, $V$ etc.) to vanish. This feature results in the lack of consistency between MF thermodynamics and MF statistical mechanics even on the level of the first-order derivatives of the corresponding thermodynamic potential and constitutes an obvious and serious drawback of  non-variational self-consistent approach\footnote{A notable exception, the case of pure Hartree or Hartree-Fock type of MF Hamiltonians, is discussed in the next paragraph.}.
The problem of such consistency is also analyzed in Ref. \cite{Argyres Kaplan}, however, from a slightly different perspective.
The analysis of the second-order derivatives of  $\Omega$ is slightly more complicated and is carried out in \cite{Jedrak}.

\subsection{Redundancy conditions  \label{Hartree}}
The natural question is, when the presence of Lagrange multipliers in the Hamiltonian (\ref{GMF Hamiltonian with lambdas}) is unnecessary. We will not provide the general answer, but instead present an important example. Suppose we have the exact  Hamiltonian  of the following form
\begin{equation}
\hat{H}_{e} = \hat{H}_{0e} + \sum_{\{\kappa, \gamma \}} V_{\kappa \gamma} \hat{\mathcal{O}}_{\kappa}\hat{\mathcal{O}}_{\gamma},
\label{no lambda H parent}
\end{equation}
where each  $\hat{\mathcal{O}}_{\kappa}$ is assumed to be bilinear in the creation and/or
 annihilation operators, and the summation is taken over  all  pairs $\{\kappa, \gamma\}$ of multi indices.
If the interaction term in  (\ref{no lambda H parent}) is decoupled  (i.e. replaced by its MF counterpart)  according to\footnote{Usually, the $\hat{\mathcal{O}}_{\kappa}\hat{\mathcal{O}}_{\gamma}$ term may be decoupled in more then one way, but this does not change the arguments presented.}
\begin{equation}
 \hat{\mathcal{O}}_{\kappa}\hat{\mathcal{O}}_{\gamma} \to     \hat{A}_{s} A_{t}  + A_{s} \hat{A}_{t} - A_{s} A_{t},
\label{AB BF}
\end{equation}
with $A_s$, $A_t$ given by (\ref{mean value of A set of eq}) for the appropriate density matrix, then the standard Hartree or the Hartree-Fock (HF) type mean-field Hamiltonian is obtained. With  the constraint terms it reads
\begin{eqnarray}
\hat{H}_{\lambda} &=&  \hat{H}_{0e} +  \sum_{\{s,t\}} \tilde{V}_{st} \big( \hat{A}_{s} A_{t}  + A_{s} \hat{A}_{t} - A_{s} A_{t}\big)  - \sum_{s}  \lambda_s (\hat{A}_{s}  - A_{s}) \nonumber \\
&\equiv&  \hat{H}  - \sum_{s}  \lambda_s (\hat{A}_{s}  - A_{s}).
\label{no lambda H}
\end{eqnarray}
In the above,  $\hat{H}_{0e}$ does not depend on $\vec{A}$, and  $\tilde{V}_{st}$ are the  combinations of matrix elements $V_{\kappa \gamma}$. Then, one can find that
\begin{equation}
 \Big \langle  \frac{\partial  \hat{H} }{\partial A_t} \Big \rangle_{\lambda}  =  \Big \langle \sum_{s}\tilde{V}_{st} \big( \hat{A}_{s} - A_{s}\big)  \Big \rangle_{\lambda}=  0,
\label{mean value of H GMF with lambda, d/dA HF type}
\end{equation}
which together with (\ref{mean value of H GMF with lambda, d/dA}) implies that $\lambda_t =0$, for any $t$. Also, for  Hamiltonian of such form, the solution of   $M$ self-consistent equations (\ref{mean value of A set of eq}) coincide with those of  $M$ equations $\nabla_{A} \mathcal{F} =0 $ of unconstrained variational approach. A related discussion of this problem  may be found in \cite{FB}.

Summarizing, for any MF Hamiltonian of the Hartree-Fock type,  application of our method is not necessary.
However, there are important examples of the MF Hamiltonians not being of HF form, in Section \ref{examples} we present two such  examples.

\subsection{Relation of the present method to the variational principle of Bogoliubov and Feynman \label{Bogoliubov}}

In the previous Sections we considered  the situation, when the description of the system has been based  \textit{solely} on  the MF Hamiltonian $\hat{H}$. In other words, we made no use of any exact Hamiltonian  $\hat{H}_{e}$ present in a problem at hand,  from which  $\hat{H}$ may be derived and for  which it  may play the role of  a simplified counterpart.

However, frequently the MF density operators are used  as a trial variational states in order  to determine the bounds on the grand potential $\Omega_e$ or free energy $F_e$ of the system described by  $\hat{H}_{e}$. Such bounds are provided by the well-known inequality due to Bogoliubov and Feynman \cite{Feynman, Decoster}
\begin{equation}
\Omega_{e} \leq \langle\hat{K}_{e} - \hat{K}_0 \rangle_0 +  \tilde{\Omega} = \text{Tr}[ \hat{\rho}_0 (\hat{K}_{e} - \hat{K}_0)] +   \tilde{\Omega}.
\label{Bogoliubov GC}
\end{equation}
In the above, $\hat{H}_0$ is a trial Hamiltonian, $\tilde{\Omega} = -\beta^{-1}\ln \text{Tr}[ \exp \big( - \beta( \hat{H}_0-\mu_0 \hat{N}) \big)]$, and $\Omega_e = -\beta^{-1}\ln \text{Tr}[ \exp \big( - \beta( \hat{H}_e-\mu_e \hat{N}) \big)]$. Also, $\hat{K}_{e}=\hat{H}_e-\mu_e \hat{N}$ and analogously for $\hat{K}_0$, with $\mu_e$ ($\mu_0$) being respective chemical potentials. There are several ways to prove (\ref{Bogoliubov GC}), probably the shortest of them  is that of making use of the Klein inequality, \cite{Nielsen}
\begin{equation}
\text{Tr}[ \hat{\rho} \ln  \hat{\rho}] \geq \text{Tr}[ \hat{\rho} \ln  \hat{\sigma}],
\label{Klein e}
\end{equation}
which holds for any normalized density operators $\hat{\rho}$ and $ \hat{\sigma}$. Inserting
\begin{equation}
\hat{\rho} = \hat{\rho}_0 = Z_0^{-1} \exp(-\beta(\hat{H}_0 -\mu_0 \hat{N})),  ~~~~~ \hat{\sigma} =\hat{\rho}_{e} = Z_{e}^{-1} \exp(-\beta(\hat{H}_{e} -\mu_e \hat{N})),
\label{}
\end{equation}
(with $Z_0 = \text{Tr}[\exp(-\beta(\hat{H}_0 -\mu_0 \hat{N}))]$ and $Z_e = \text{Tr}[\exp(-\beta(\hat{H}_e -\mu \hat{N}))]$) into (\ref{Klein e}),  we obtain (\ref{Bogoliubov GC}).
If one wants to work with the chemical potential as an independent variable, then $\mu_e = \mu_0 \equiv \mu$ and
(\ref{Bogoliubov GC}) reduces to
\begin{equation}
\Omega_{e} \leq \langle\hat{H}_{e} - \hat{H}_0 \rangle_0 +  \tilde{\Omega} = \text{Tr}[ \hat{\rho}_0 (\hat{H}_{e} - \hat{H}_0)] +   \tilde{\Omega}.
\label{Bogoliubov GC mu}
\end{equation}
However, usually it is more convenient to have particle number $N$ as an independent variable. In such a case $ \mu_0$  and (unknown) $ \mu_e$ are not equal in general, as they are determined from different conditions, i.e. $\langle \hat{N}\rangle_0 = N $ and $\langle \hat{N}\rangle_e = N $, respectively. However, (\ref{Bogoliubov GC}) may be then given the form
\begin{equation}
F_e = \Omega_{e} + \mu_e N   \leq \langle\hat{H}_{e} - \hat{H}_0 \rangle_0 +   \tilde{\Omega}+\mu_0 N.
\label{Bogoliubov GC N}
\end{equation}
Bogoliubov inequality is the basis of the variational principle, the best  trial state  $\hat{\rho}_0$  is the one  minimizing r.h.s.  of (\ref{Bogoliubov GC mu}), or (\ref{Bogoliubov GC N}), depending on the choice of independent variables, $\mu$ or  $N$, respectively.

If one wants to use the Bogoliubov inequality in order to determine upper bounds for $\Omega_e$ or $F_e$ of some  $\hat{H}_{e}$, but with the   self-consistency of the approach being preserved, then the density operators (\ref{MF naive DO}) are not good candidates for the trial states. Clearly, they have no free variational parameters\footnote{We neglect the situation, in which there appear some extra variational parameters, not being expectation values of operators present in a problem, and thus not requiring the corresponding Lagrange multipliers.}
 if only the values of mean-fields, $\vec{A}=\vec{A}_0$, are obtained in a self-consistent fashion through (\ref{mean value of A set of eq}). On the other hand, if  $\vec{A}_0$  were obtained from unwary minimization of the r.h.s. of (\ref{Bogoliubov GC mu}), (\ref{Bogoliubov GC N})  with  $\hat{\rho}$ given by (\ref{MF naive DO}), then  the self-consistency of the approach would  be violated in general (again with the notable exception of the Hartree-Fock MF Hamiltonians).

In contrast, consider the trial state chosen in the form (\ref{correct MF density matrix}), with the trial Hamiltonian  $\hat{H}_{0} = \hat{H}_{\lambda}$, yet unrelated to the $\hat{H}_{e}$ in question. Then the r.h.s. of (\ref{Bogoliubov GC mu}) reads
\begin{equation}
\text{Tr}[ \hat{\rho}_{\lambda} (\hat{H}_{e} - \hat{H}_{\lambda})] + \mathcal{F}(\vec{A}, \vec{\lambda}) \equiv  \mathcal{B} (\vec{A}, \vec{\lambda}),
\label{Bogoliubov GC for rho lambda mu}
\end{equation}
whereas that of (\ref{Bogoliubov GC N}), respectively (with $\mu_0 \equiv \mu$)
\begin{equation}
\text{Tr}[ \hat{\rho}_{\lambda} (\hat{H}_{e} - \hat{H}_{\lambda})] + \mathcal{F}(\vec{A}, \vec{\lambda}) + \mu N =  \mathcal{B} (\vec{A}, \vec{\lambda}) + \mu N \equiv \mathcal{B}_N (\vec{A}, \vec{\lambda}).
\label{Bogoliubov GC for rho lambda N}
\end{equation}
Depending on the choice of independent variable, the necessary\footnote{Similarly as previously, here we assume, that the desired minimum corresponds to  a stationary point.} $2M$ conditions for r.h.s of the Bogoliubov inequality, i.e $\mathcal{B}$ or $\mathcal{B}_N $, to have a  minimum subject to constrains of self-consistency acquire slightly different form. For (\ref{Bogoliubov GC for rho lambda mu}) we have
\begin{equation}
\nabla_{A} \mathcal{B}(\vec{A}, \vec{\lambda}) =0, ~~~~~~~   \nabla_{\lambda} \mathcal{F}(\vec{A}, \vec{\lambda})=0.
\label{Bogoliubov set of eq}
\end{equation}
On the other hand, for (\ref{Bogoliubov GC for rho lambda N}) we have
\begin{equation}
\nabla^{\prime}_{A} \mathcal{B}_N(\vec{A}, \vec{\lambda}) =0, ~~~~~~~   \nabla_{\lambda}(\mathcal{F}(\vec{A}, \vec{\lambda}) + \mu N)=0, ~~~~ \frac{\partial}{\partial N}(\mathcal{F}(\vec{A}, \vec{\lambda}) + \mu N)= \mu.
\label{Bogoliubov set of eq N}
\end{equation}
In the above, $\nabla^{\prime}_{A}$ labels the gradient taken with respect to all mean-fields except $A_1 = N$. Also
\begin{equation}
\nabla^{\prime}_{A}\mathcal{B}_N(\vec{A}, \vec{\lambda})=\nabla^{\prime}_{A}\mathcal{B}(\vec{A}, \vec{\lambda}), ~~~ ~~ \nabla_{\lambda}(\mathcal{F}(\vec{A}, \vec{\lambda}) + \mu N)=\nabla_{\lambda}\mathcal{F}(\vec{A}, \vec{\lambda}).
\label{prime is not a prime}
\end{equation}

If $\mathcal{B}(\vec{A}, \vec{\lambda}) \neq \mathcal{F}(\vec{A}, \vec{\lambda})$, the solution of (\ref{Bogoliubov set of eq}) or (\ref{Bogoliubov set of eq N}) leads  to different values of $\vec{A}_{0}$ and $\vec{\lambda}_{0}$, than those obtained from (\ref{derivative of mathcalF A, lambda}). Then,  $\hat{\rho}_{\lambda}$ is   no longer a true grand-canonical MF density operator for $\hat{H}_{\lambda}$. The above  remarks hold  for other inequalities (i.e. various generalizations of (\ref{Bogoliubov GC})), and the corresponding variational principles built on them, \cite{Decoster}. One may use  the $\hat{\rho}_{\lambda}$  (\ref{correct MF density matrix}) as useful ans\"{a}tze in such cases as well. Nonetheless,  the entropy functional (\ref{entropy MF}) with the corresponding variational principle plays a special role, and it  determines the form (\ref{correct MF density matrix})  of GC MF density operator. Also, the self-consistency conditions may be conveniently formulated with the help of $\mathcal{F}(\vec{A}, \vec{\lambda})$, as in (\ref{Bogoliubov set of eq}) and (\ref{Bogoliubov set of eq N}).
However, clearly, for any  pair of $\hat{H}_{e}, \hat{H}_{\lambda}$, for which the condition
\begin{equation}
\langle \hat{H}_{e}-\hat{H}_{\lambda} \rangle_{\lambda} =0
\label{}
\end{equation}
 holds, so consequently also $\tilde{\Omega}(\vec{A}, \vec{\lambda}) = \mathcal{F}(\vec{A}, \vec{\lambda}) = \mathcal{B}(\vec{A}, \vec{\lambda})$, the Bogoliubov - Feynman and Maximum Entropy  variational principles coincide. In such case the optimal, from the point of view of maximum entropy inference, i.e. the most uninformative MF density operator $\hat{\rho}_{\lambda}$ provides us also with the upper bound for exact grand potential $\Omega_e$  or free energy $F_e$ of $\hat{H}_{e}$. For (\ref{Bogoliubov set of eq}) this is obvious. For (\ref{Bogoliubov set of eq N}), this may be shown using $\partial\mathcal{F}/ \partial\mu = -N$  and replacing $\mathcal{B} \to \mathcal{F}$, then we obtain the identity
\begin{eqnarray}
\mu &= & \frac{\partial}{\partial N}(\mathcal{F}(\vec{A}, \vec{\lambda}) + \mu N)= \frac{\partial\mathcal{F}}{\partial N} + \frac{\partial\mathcal{F}}{\partial \mu} \frac{\partial\mu}{\partial N} + N \frac{\partial\mu}{\partial N} = \frac{\partial\mathcal{F}}{\partial N} + \mu.
\label{prove F}
\end{eqnarray}
Consequently, $\partial\mathcal{F}/ \partial N=0$ for the solutions of Eqs. (\ref{Bogoliubov set of eq N}).
From Eqs. (\ref{prime is not a prime}) and (\ref{prove F}), it follows that any solution of  (\ref{Bogoliubov set of eq N}), i.e. $(\vec{A}_0, \vec{\lambda}_0, \mu)$ is  also a solution of (\ref{Bogoliubov set of eq}), if  $N$ and $\mu$ related by the condition $\langle \hat{N} \rangle = N = -\partial \mathcal{F}/\partial \mu$. Also, conversly, Eqs. (\ref{prime is not a prime}), (\ref{prove F}) imply that  the solutions of (\ref{Bogoliubov set of eq}) are also solutions of  (\ref{Bogoliubov set of eq N}),  thus both schemes  are equivalent.
\section{Two nontrivial examples\label{examples}}
We now discuss briefly two MF models, for which the proposed method leads to  nontrivial corrections. To show this, we make use of Eq. (\ref{mean value of H GMF with lambda, d/dA}), and  argue, that the condition $\vec{\lambda}=\vec{0}$ cannot be, in general, satisfied.

\subsection{Renormalized  t-J model \label{tJ}}

The first example we analyze is the MF Hamiltonian  of the so-called renormalized t-J model
\cite{ZGRS}-\cite{LiZhouWang}. It originates from the standard t-J model \cite{Spalek Oles},  believed to describe correctly the essential physics of the cuprate high-temperature superconductors, that is expressed by the Hamiltonian
\begin{equation}
 \hat{H}_{e} = \sum_{i, j(i), \sigma} t_{ij} \tilde{c}_{i \sigma}^{\dag} \tilde{c}_{j \sigma} + J \sum_{\langle i j \rangle}  ( \mathbf{S}_{i}\cdot \mathbf{S}_{j} - \frac{1}{4}\hat{\nu}_i \hat{\nu}_{j}) -\mu \sum_{i,  \sigma} \hat{\nu}_{i \sigma}.
\label{t-J exact}
\end{equation}
Here $\tilde{c}_{i \sigma}= (1- n_{i -\sigma} )c_{i \sigma}$ and $ \hat{\nu}_i = \sum_{\sigma}= \tilde{c}^{\dag}_{i \sigma}\tilde{c}_{i \sigma}$ are the operators with double occupancies on site $i$ projected out. The corresponding renormalized MF Hamiltonian may be taken in the form \cite{Didier, Marcin 2}
\begin{eqnarray}
\hat{H} &=&  \sum_{\langle i j \rangle \sigma}\Big( \big (t_{ij} g^{t}_{ij} c_{i \sigma}^{\dag} c_{j \sigma} +  \text{H.c.} \big)  - \frac{3}{4} J g^{J}_{ij} (\chi_{ji}  c_{i \sigma}^{\dag} c_{j \sigma} + \text{H.c.} - |\chi_{ij}|^{2}) \nonumber \\ &+& - \frac{3}{4}J g^{J}_{ij} (\Delta_{ij}   c^{\dag}_{j \sigma}c_{i -\sigma}^{\dag} + \text{H.c.}   - |\Delta_{ij}|^{2})\Big)  -\mu  \sum_{i \sigma}   c_{i \sigma}^{\dag} c_{i \sigma},
\label{ren tJ Ham}
\end{eqnarray}
with  $n_i = \sum_{\sigma} \langle  c_{i \sigma}^{\dag}  c_{i \sigma} \rangle $,   $\chi_{ij} = \langle  c_{i \sigma}^{\dag}  c_{j \sigma} \rangle$, and $\Delta_{ij} = \langle  c_{i -\sigma}  c_{j \sigma} \rangle = \langle  c_{j -\sigma}  c_{i \sigma} \rangle$.
In this form the projections have been abolished at the price of introducing the expectation-value dependent renormalization factors $g^{t}_{ij} $ and $g^{J}_{ij} $ resulting from the Gutzwiller approximation \cite{Vollhardt}.
At first, we consider  the renormalization factors  depending solely on  $x_i = 1- n_i$, i.e.
\begin{equation}
 g^{t}_{ij} = \sqrt{\frac{4x_i x_j}{(x_i + 1)(x_j + 1)}} ~~~~\text{and} ~~~ g^{J}_{ij} = \frac{4}{(x_i + 1)(x_j + 1)}.
\label{g ren fac x only}
\end{equation}
The task is  to  determine the values of the mean-field parameters
$\chi_{ij}$, $\Delta_{ij}$ and $x_{i}$. This could be  achieved with the help of self-consistent Eqs. (\ref{mean value of A set of eq}) for $\hat{H}$ given by (\ref{ren tJ Ham}) (cf. \cite{ZGRS}- \cite{Marcin 2})

On the other hand, we may proceed in the framework of  our method, and redefine the Hamiltonian (\ref{ren tJ Ham}) according to
\begin{eqnarray}
\hat{H}_{\lambda} = \hat{H} &-& \sum_{i} \lambda^{n}_{i}( \sum_{\sigma} c_{i \sigma}^{\dag}  c_{i \sigma} - n_{i}) - \sum_{\langle i j \rangle \sigma}  \big( \lambda^{\chi}_{ij}  (c_{i \sigma}^{\dag}  c_{j \sigma} - \chi_{ij}) + \text{H.c.} \big) \nonumber \\ &- &  \sum_{\langle i j \rangle \sigma} \big( \lambda^{\Delta}_{ij}  (c_{i - \sigma}  c_{j \sigma} - \Delta_{ij})  + \text{H.c.} \big).
\label{MF tJ lambda term}
\end{eqnarray}
Assuming no additional symmetries, using $\partial / \partial n_i = - \partial / \partial x_i$,  converting the sums over bonds to that over sites in  (\ref{ren tJ Ham}) according to  $2 \sum_{\langle i, j \rangle} = \sum_{i, j(i)} $ and finally taking $A_w \equiv n_i $ in Eqs.  (\ref{mean value of H GMF with lambda, d/dA}) we obtain

\begin{eqnarray}
2 \lambda^{n}_{i} &=& \Big \langle \sum_{j(i), \sigma} t_{ij}  \frac{\partial g^{t}_{ij}}{\partial x_i} \big( c_{i \sigma}^{\dag} c_{j \sigma} + \text{H.c.} \big) \Big \rangle_{\lambda} \nonumber \\
&- &
\Big \langle\frac{3}{4}J     \sum_{j(i), \sigma} \frac{\partial g^{J}_{ij}}{\partial x_i} (\chi_{ji}  c_{i \sigma}^{\dag} c_{j \sigma} + \text{H.c.} - |\chi_{ij}|^{2})\Big \rangle_{\lambda} \nonumber \\
&- &
\Big \langle   \frac{3}{4}J     \sum_{j(i), \sigma} \frac{\partial g^{J}_{ij}}{\partial x_i} (\Delta_{ij}   c^{\dag}_{j \sigma}c_{i -\sigma}^{\dag} + \text{H.c.}   - |\Delta_{ij}|^{2})\Big \rangle_{\lambda} \nonumber \\
&=& \sum_{j(i)} \Big( 4 t_{ij} \frac{\partial g^{t}_{ij}}{\partial x_i} \text{Re}\chi_{ij} - \frac{3}{2}J  \frac{\partial g^{J}_{ij}}{\partial x_i} \big( |\chi_{ij}|^{2} + |\Delta_{ij}|^{2}  \big) \Big).
\label{lambda x i zero}
\end{eqnarray}
Below it will be shown that in this particular case, $\lambda^{\chi}_{ij}  = \lambda^{\Delta}_{ij}= 0$, for each bond $\langle ij \rangle$.
Consequently, if  for all $i$, we put $ \lambda^{n}_{i} = 0$, then the density operator, and all averages coincide with those of the standard self-consistent MF treatment. However, in such situation  equations (\ref{lambda x i zero}) and  the self-consistent equations (\ref{mean value of A set of eq}) for $\hat{H}$ given by (\ref{ren tJ Ham}) cannot  in general  be simultaneous satisfied. For example, if one sets   $J=0$ and $  x_{i}\in (0, 1)$, then from  (\ref{lambda x i zero}) it follows tha
\begin{equation}
\Big \langle  \sum_{j(i) \sigma} \big( \frac{\partial g^{t}_{ij}}{\partial x_i}  c_{i \sigma}^{\dag} c_{j \sigma} +  \text{H.c.} \big)  \Big  \rangle  = 0.
\label{}
\end{equation}
This is a senseless result, as in such case our Hamiltonian reduces essentially to that of free fermions, at least  for  the  case of homogeneous solution, $x_i \equiv x$. Similarly, for $t_{ij} = 0$, one obtains  $ |\chi_{ij}|^{2} = 0$, $ |\Delta_{ij}|^{2} = 0$ for each bond $ij$, provided that each $t_{ij}$ is of the same sign,  as  $\partial g^{t}_{ij}/\partial x_i > 0$, for any $x_i$ and $\langle ij \rangle$. This is in contradiction  with the numerical results of Raczkowski \textit{et al.}, \cite{Marcin 1}. In other words, in this particular case, we can infer about the predictions of our method using the previous, non-variational  results. Consequently, it follows that $\lambda^{n}_{i}\neq 0$, for every $i$.

From the point of view of our findings, a related situation can be found in Ref. \cite{LiZhouWang}. Namely, the inhomogeneous solutions of the model (\ref{ren tJ Ham}) (with slightly different definitions of  $\chi_{ij}$ and $\Delta_{ij}$), and with the renormalization factors given by (\ref{g ren fac x only}) are analyzed. However, the additional terms of the form $\sum_{i}\epsilon_i (\sum_{\sigma}  c_{i \sigma}^{\dag} c_{i \sigma} + x_i - 1)$, are included, where $\epsilon_i$ is the local fugacity. The quantity $\epsilon_i$  plays a role analogous to $ \lambda^{n}_{i}$ in our method. Inclusion of that additional constraint allows to treat each  ${x}_{i}$ as a variational parameter, and leads to the equations analogous to (\ref{lambda x i zero}). This  should be clear in the light of the discussion in the preceding Sections.

Next, we pose the question of redundancy of the Lagrange multipliers $\lambda^{\chi}_{ij}$ and $\lambda^{\Delta}_{ij}$. We may expect, according to discussion carried out in Section \ref{Hartree},  that these quantities  vanish, as the Hamiltonian (\ref{ren tJ Ham}) with  $g^{t}_{ij}$ and $ g^{J}_{ij}$  given by (\ref{g ren fac x only}) is of the Hartree-Fock type with respect to both $\chi_{ij}$ and $\Delta_{ij}$. This is indeed the case, we may use once more  Eq. (\ref{mean value of H GMF with lambda, d/dA}) with $A_w = \chi_{ij} $ or $A_w = \Delta_{ij} $, respectively, to obtain
\begin{equation}
 \lambda^{\chi}_{ij} = \Big \langle   \frac{3}{4}J \sum_{\sigma}g^{J}_{ij} ( c_{j \sigma}^{\dag} c_{i \sigma} - \chi_{ij}^{\ast})\Big \rangle_{\lambda} \equiv 0,
\label{HF like for chi}
\end{equation}

\begin{equation}
 \lambda^{\Delta}_{ij} = \Big \langle  \frac{3}{4}J \sum_{ \sigma}g^{J}_{ij} ( c_{j \sigma}^{\dag} c_{i -\sigma}^{\dag}  - \Delta_{ij}^{\ast})\Big \rangle_{\lambda}  \equiv 0.
\label{HF like for Delta}
\end{equation}
Obviously, those self-consistency equations must be fulfilled both in the previous treatments ($\vec{\lambda} = \vec{0}$) as well as in the present method.

Next, we consider  the renormalization factors of the form \cite{Marcin 2},
\begin{equation}
 g^{t}_{ij} = \sqrt{\frac{4x_i x_j(1-x_i)(1-x_j)}{(1-x_i^{2})(1-x_j^{2}) + 8(1-x_i x_j)|\chi_{ij}|^{2} + 16|\chi_{ij}|^{4}}},
\label{gt ren fac x chi delta}
\end{equation}
\begin{equation}
 g^{J}_{ij} = \frac{4(1-x_i)(1-x_j)  }{(1-x_i^{2})(1-x_j^{2}) + 8x_i x_j(|\Delta_{ij}|^{2}-|\chi_{ij}|^{2}) + 16(|\Delta_{ij}|^{4}+|\chi_{ij}|^{4})}.
\label{gJ ren fac x chi delta}
\end{equation}
 In such more general situation, the equation (\ref{lambda x i zero}) is still valid, and Eqs. (\ref{HF like for chi}) and  (\ref{HF like for Delta}) acquire  the form, respectively

\begin{eqnarray}
\lambda^{\chi}_{ij} &=& - \sum_{ \sigma} \Big \langle   t_{ij}  \frac{\partial g^{t}_{ij}}{\partial \chi_{ij}} \big( c_{i \sigma}^{\dag} c_{j \sigma} + \text{H.c.} \big)  -  \frac{3}{4}J  \frac{\partial g^{J}_{ij}}{\partial \chi_{ij}}  (\chi_{ji}  c_{i \sigma}^{\dag} c_{j \sigma} + \text{H.c.} - |\chi_{ij}|^{2})  \nonumber \\ &+& - \frac{3}{4}J g^{J}_{ij} ( c_{j \sigma}^{\dag} c_{i \sigma} - \chi_{ij}^{\ast}) - \frac{3}{4}J  \frac{\partial g^{J}_{ij}}{\partial \chi_{ij}}(\Delta_{ij} c^{\dag}_{j \sigma} c_{i -\sigma}^{\dag}  + \text{H.c.}  - |\Delta_{ij}|^{2}) \Big \rangle_{\lambda} \nonumber \\
&=& -  4 t_{ij} \frac{\partial g^{t}_{ij}}{\partial  \chi_{ij}} \text{Re}\chi_{ij} + \frac{3}{2}J  \frac{\partial g^{J}_{ij}}{\partial  \chi_{ij}} ( |\Delta_{ij}|^{2} + |\chi_{ij}|^{2}),
\label{lambda chi ij zero most general}
\end{eqnarray}

\begin{eqnarray}
\lambda^{\Delta}_{ij} &=&  \sum_{\sigma} \Big   \langle   \frac{3}{4}J  \frac{\partial g^{J}_{ij}}{\partial \Delta_{ij}}  (\chi_{ji}  c_{i \sigma}^{\dag} c_{j \sigma} + \text{H.c.} - |\chi_{ij}|^{2}) +   \frac{3}{4}J g^{J}_{ij} (  c_{i -\sigma}^{\dag} c^{\dag}_{j \sigma} - \Delta_{ij}^{\ast}) \nonumber \\
&+&   \frac{3}{4}J  \frac{\partial g^{J}_{ij}}{\partial \Delta_{ij}} (\Delta_{ij} c^{\dag}_{j \sigma} c_{i -\sigma}^{\dag}  + \text{H.c.}  - |\Delta_{ij}|^{2})    \Big \rangle_{\lambda} \nonumber \\ &=& \frac{3}{2}J  \frac{\partial g^{J}_{ij}}{\partial  \Delta_{ij}} \big( |\Delta_{ij}|^{2} + |\chi_{ij}|^{2}\big).
\label{lambda Delta ij zero most general}
\end{eqnarray}
Again, if we put $\vec{\lambda}=\vec{0}$ in order to reduce our approach  to the standard treatment, that Eqs. (\ref{lambda chi ij zero most general}) and (\ref{lambda Delta ij zero most general})
cannot be satisfied. For Eq. (\ref{lambda chi ij zero most general}) this can be  easily shown for the case  $t_{ij}=-t$ by adding it to its complex conjugate part. Then, at least for the class of solutions with  for $\text{Re}\chi_{ij}> 0$, one can convince oneself by explicitly computing respective derivatives of $g^{t}_{ij}$ and $ g^{J}_{ij}$, that the resulting expression should have nonzero value, for a wide range of values $x_i$. For Eq. (\ref{lambda Delta ij zero most general})  or (\ref{lambda chi ij zero most general}) in the case of $t_{ij}=0$,  no additional explanation is required.

Similar arguments apply if, instead of $\vec{\lambda}=\vec{0}$, a weaker conditions, $\lambda^{\chi}_{ij}=0$ or $\lambda^{\Delta}_{ij}=0$, are imposed. In such situation we can no longer make use of the numerical or analytical results  of the purely self-consistent approach. Nonetheless, it is unlikely that such conditions do not lead to contradictions. E.g. from  Eqs. (\ref{lambda chi ij zero most general}) with $t_{ij}=0$ or from  (\ref{lambda Delta ij zero most general}) in a general case,  we infer, that $ \chi_{ij} = \Delta_{ij}=0$.

In that slightly hand-waving manner we tried to argue, that for the Hamiltonian (\ref{ren tJ Ham}) our variational approach does not  reduce to that based solely on the self-consistency equations.

This model has been recently analyzed  and solved numerically in a separate publication \cite{Jedrak Spalek}.

\subsection{Electrons with spin-dependent effective masses}
In Ref.  \cite{Spalek Gopalan}, an approximate one-particle description of a strongly-correlated electron system was introduced, based on the observation,  that the effective masses of quasiparticles may be spin-dependent. The MF Hamiltonian of this model depends on $ A_{1} = N$ and $A_2 = M$, where
\begin{equation}
 \hat{N} =  \hat{A}_1 = \sum_{\mathbf{k}} \sum_{\sigma} c^{\dag}_{\mathbf{k} \sigma}c_{\mathbf{k} \sigma}, ~~~~~  \hat{M} = \hat{A}_2 =\sum_{\mathbf{k}} \sum_{\sigma} \sigma c^{\dag}_{\mathbf{k} \sigma}c_{\mathbf{k} \sigma}.
\label{}
\end{equation}
With additional constraint terms, the Hamiltonian reads
\begin{equation}
  \hat{H}_{\lambda}^{SDM} =  \Big( \sum_{\mathbf{k}} \sum_{\sigma}  (\Phi_{\sigma}(N, M)  \epsilon_{\mathbf{k} \sigma}   -\sigma h_a  - \mu )\hat{n}_{\mathbf{k} \sigma} \Big) - \lambda_{1}(\hat{N} - N) - \lambda_{2}(\hat{M} - M),
\label{sdm Hamiltonian}
\end{equation}
where $(-\sum_{\mathbf{k}} \sigma h_a \hat{n}_{\mathbf{k} \sigma})$ is the Zeeman term, and $h_a$ is the reduced  applied magnetic field.
The present case is, in fact, quite similar to the previous one, the band-narrowing factors $\Phi_{\sigma}(N, M)$ are  also derived using the Gutzwiller approach. Then analogously, by making use of Eq. (\ref{mean value of H GMF with lambda, d/dA}), either with $A_w = N $ or $A_w = M $,  we  obtain respectively
\begin{eqnarray}
\lambda_{1} &=& -\Big \langle   \sum_{\mathbf{k}} \sum_{\sigma} \frac{\partial \Phi_{\sigma}(N, M)}{\partial N}  \epsilon_{\mathbf{k} \sigma}  \hat{n}_{\mathbf{k}\sigma} \Big \rangle_{\lambda},
\label{lambda N zero}
\end{eqnarray}
\begin{eqnarray}
\lambda_{2} &=& -\Big \langle   \sum_{\mathbf{k}} \sum_{\sigma} \frac{\partial \Phi_{\sigma}(N, M)}{\partial M}  \epsilon_{\mathbf{k} \sigma}  \hat{n}_{\mathbf{k}\sigma} \Big \rangle_{\lambda}.
\label{lambda M zero}
\end{eqnarray}
Regardless of the detailed  analytical form of $\Phi_{\sigma}(N, M)$, we  multiply (\ref{lambda M zero}) by
\begin{equation}
-\frac{\partial}{\partial N} \Phi_{\downarrow}(N, M) \Big(\frac{\partial }{\partial M} \Phi_{\downarrow}(N, M)\Big)^{-1} \nonumber
\end{equation}
and add it to (\ref{lambda N zero}). The resulting r.h.s is then proportional to $ \big  \langle E_{kin, \uparrow}  \big \rangle_{\lambda} =  \big  \langle \sum_{\mathbf{k}}  \epsilon_{\mathbf{k} \uparrow}  \hat{n}_{\mathbf{k}\uparrow}  \big \rangle_{\lambda} $, the average band energy of particles with $z$-spin component $\sigma = \uparrow$. If we put $\lambda_{1}= \lambda_{2}=0$, all the averages become identical to those of the standard treatment. Consequently, for (\ref{lambda N zero}) and (\ref{lambda M zero}) to hold in such a case we have to require that $ \big  \langle E_{kin, \uparrow}  \big \rangle= 0$, for any admissible value of $N, M$, and functional dependence of $\Phi_{\sigma}(N, M)$ on its variables,  which is a trivial and inconsistent result. Obviously, the same  holds  for $\sigma = \downarrow$.

The numerical studies of this model and its extensions are carried out recently, \cite{Kaczmarczyk Spalek}.

\section{Summary and conclusions\label{summary}}
In this paper we have presented the method,  based on the maximum entropy principle,  of constructing the grand-canonical density operator for a wide class of  mean-field Hamiltonians. We have shown, that such a density operator is not, in general, obtained by replacing the exact Hamiltonian by its mean-field counterpart in the grand canonical density operator. Usually, some extra terms, which may be interpreted as a kind of source terms or molecular fields, must be added to the mean-field Hamiltonian. This modification also ensures consistency of basic thermodynamic relations, which is not guaranteed for self-consistent methods which are not based on the variational principle.

It is also shown, that although for the Hartree-Fock type of mean-field Hamiltonians application of our method is not necessary, there are also some important examples, for which it does not reduce to the standard approach. The renormalized  mean-field   t-J Hamiltonian is  such case.

If there is $M$ expectation  values appearing in a mean-field Hamiltonian that are to be determined, the method presented here results in  a system of $2M$ equations. This is in contrast to the standard treatment, where only $M$ such equations appear

The problem of the application of maximum entropy inference in the mean-field case was studied also in a series of the papers \cite{Argentynczycy}. However, the presented point of view and the results obtained  differ essentially from ours.

Also, the problem partly related to that discussed in the present paper, namely the construction of  variational principles suited for optimization of physical quantities of interest, is examined in an interesting article \cite{Balian}. This paper contains  a detailed  analysis of  the problem, as well as covers a large area of the subject (for example the authors discuss time-dependent formalism).

\subsection*{Acknowledgments}
The authors are grateful to Krzysztof Ro\'{s}ciszewski for  his valuable comments and insightful remarks.
Discussions with Marcin Raczkowski and Andrzej Kapanowski are also warmly acknowledged.
The authors acknowledge the Grant No. N N 202 128 736 from the Ministry of Science and Higher Education.

\vspace{0.3cm}

$\ast$e-mail: jedrak@th.if.uj.edu.pl

$\dag$e-mail: ufspalek@if.uj.edu.pl\\

redpajama_set_name: RedPajamaArXiv


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Here are some common questions you should be ready to answer when interviewing. Take time to rehearse you answers or work with a friend. Make sure it feels natural and not forced and remember to smile.
Answer this question with your "elevator speech" – keep it professional (no talking about your pets or kids), targeted to how your experience fits the position, and what makes you unique.
2. Share with me why you are looking to leave your current position and why you left each position you've held in the last 5 years.
The interviewer is looking for red flags as to why you left previous jobs (job hopping, disgruntled, conflict) and how you feel about your current/previous employer. Be sure your answer does not come across as though you are angry, bitter or have a chip on your shoulder. Even in bad situations there are things you've learned. Share a professionally honest response as to why you left and keep your answer positive.
3. Describe for me the best & your worst bosses you've worked for and why they were the best/worst.
DON'T be negative! The interviewer wants to understand the kind of management that motivates you and de-motivates you, how you respond to leadership, and how you've worked with various styles of management.
4. Share with me an example of a professional conflict you were involved in – what was the situation and how did you respond?
The interviewer wants to know how you react to conflict (do you resolve or give in, are you a problem solver) and that you can handle yourself in a difficult situation.
5. If I were to call your current/most recent supervisor and ask which of your performance areas need the most improvement, how would he/she respond?
The interviewer is looking for your weaknesses. Remember, every weakness is an opportunity to learn & grow professionally, so when answering this be sure to address what you do to mitigate your weaknesses and how you are working to improve in these areas.
6. What success/accomplishment in your career are you most proud of and why?
The interviewer is listening for what energizes you, and the kind of work that makes you feel really good. This is a good chance to share your strengths and how your strengths helped you achieve this success.
7. Hindsight being 20/20, if you could go back and re-do a decision or a failed project, what decision/project would you chose and what would you do differently?
The interviewer is seeking to understand if you are accountable for your mistakes/decisions and if you learn from your experiences.
8. From everything you've learned about this role, the company, tell me why you are the best candidate for the position.
This is your chance to sell yourself. Speak to what you know the company needs in this role and how you can be a valuable player on their team.
9. Describe for me the type of work environment in which you thrive.
Be honest, but also know that if you describe an environment opposite of what you see around you in the interview, you've lost the job. Then again, if it is not the right environment for you, it's not the right job for you!
10. What is your current salary? What are your salary expectations?
Do not avoid this question! If you are asked your current salary, give the base amount plus any bonuses/commissions/perks you receive. When asked your salary expectations, give a range (typically a $10 – $15K range) and be sure that your minimum amount in the range is a salary you are comfortable earning. Employers will typically offer you between your minimum and midpoint.
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