Fields in QFT: What are the Initial Conditions, $\phi(t=0,x)$?
Solving a differential equation generally goes something like this:
* One is given a set of initial conditions. There is a large freedom of choice in doing so. For example, in Quantum Mechanics any normalizable $\psi(x,0)$ will do
* The differential equation then determines how the state evolves, giving $\psi(x,t)$ for all times $t$.
Consider now QFT, in particular the free real scalar field as a simple example. The (<PERSON>-Gordon) equation of motion for fields in the <PERSON> picture is solved by
$$\phi(\vec{x},t)=e^{itH}\phi(\vec{x},0)e^{-itH}$$
For the real scalar field, the explicit solution is written $$\phi(\vec{x},t)=\int (a(\vec{p})e^{-ip\cdot x}+a^{\dagger}(\vec{p})e^{ip\cdot x}) \frac{d^3\vec{p}}{\sqrt{(2\pi)^3 2E_0}}$$
In QFT lecture, it is often stated that this is the most general $\phi(x,t)$ which solves the equation of motion. Let's investigate that claim. In particular, plugging in $t=0$, we have assumed the initial condition
$$\phi(\vec{x},0)=\int (a(\vec{p})e^{i\vec{p}\cdot \vec{x}}+a^{\dagger}(\vec{p})e^{-i\vec{p}\cdot \vec{x}}) \frac{d^3\vec{p}}{\sqrt{(2\pi)^3 2E_0}}$$
But as far as I can tell, the only constraint on $\phi(x,0)$ introduced in lecture is that it should obey the commutation relations. 281
$\phi(x,0)$ is a very important quantity, because in the <PERSON> picture it is the field operator, just as analogously $x$ and $p=i\nabla$ were the position and momentum operators in Quantum Mechanics.
Therefore I'd like to ask for one of two possible solutions:
* Can someone provide a proof that we must have this particular initial condition for the fields? My feeling is that the commutation relations alone are not enough to give this specific form, as operators have a huge amount of degrees of freedom.
* Just as interesting would be to provide a counterexample: an initial condition $\phi(\vec{x})$ which satisfies the commutation relations but is not equal to $\phi(\vec{x},0)$ above.
Either of these answers would be very elucidating for me. If not a proof, then a reference to a proof would of course also be just as helpful.
EDIT: After more thought, the answer to this question is sort of a mixture of <PERSON> and <PERSON> answers. As <PERSON> remarked, the form $\phi(x,0)$ is not uniquely defined by the commutation relations or KG equation, because you could just redefine $\phi(x,0)$ to be the field with $t=1$ and all equations would still hold. This would be equivalent to redefining the operators $a(\vec{p})$ by adding a phase.
However, there is a strong restriction on the $a(\vec{p})$ operators, namely* they can only be redefined up to a momentum-dependent phase $a(\vec{p}) \to e^{i \alpha(\vec{p})}a(\vec{p})$. The proof is along the lines of what Cosmos Zachos' was mentioning and the treatment is done in many QFT textbooks. The essential reason this is possible is that redefining $a(\vec{p}) \to e^{i \alpha(\vec{p})}a(\vec{p})$ does not change the commutation relations.
*The momentum-dependence is my own realization and not taken from a book, but seeing as the commutation relations remain unchanged under the redefinition, it seems clear that they cannot constrain the phase of $a(\vec{p})$. In addition, redefining $\phi(\vec{x},0) \to \phi(\vec{x},1)$ is a momentum-dependent change in the $a$ operators: $a(\vec{p}) \to e^{i\sqrt{\vec{p}^2+m^2} t} a(\vec{p})$ 281