Question: In quantum mechanics it is well known that if you have a Lagrangian $\mathcal{L}$ and you want to quantise it, there is no unique way of doing this. This is because when you construct the Hamiltonian $\mathcal{H}$ and try to promote observables to operators, canonical momenta and position don't commute.

However the other way of doing quantum mechanics is in terms of path integrals. If one has a classical Lagrangian $\mathcal{L}$ then you can write down the propagator as $\int \mathcal{D}[q] e^{i S[q]}$ where $S[q] = \int \mathcal{L} dt$ is the action. It would therefore seem like $\mathcal{L}$ *uniquely* specifies the appropriate quantum theory, at least for any measurement that only involves propagators.

So my question is: how can one have non-uniqueness for canonical quantisation but apparent uniqueness for path integrals if the two formulations are supposed to be equivalent? Is it that the propagators do not fully determine the theory? Does this then mean that there are some quantities that cannot, even in principle, be calculated with path integrals?

Answer: 
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> In quantum mechanics it is well known that if you have a Lagrangian L and you want to quantise it, there is no unique way of doing this.
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This is correct, however note that you also have the constraint that H needs to be Hermitian. 

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> However the other way of doing quantum mechanics is in terms of path integrals. If one has a classical Lagrangian L then you can write down the propagator as $∫D[q]e^{iS[q]}$ where $S[q]=∫Ldt$ is the action. It would therefore seem like L uniquely specifies the appropriate quantum theory, at least for any measurement that only involves propagators.
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This procedure is also not unique, but it is subtle to see why. It is true that the Lagrangian is uniquely defined, however path integrals can only be understood in discrete time.

The correct way to understand path integral is to write it in discrete time. The continuous notation is only a shorthand and a practical way to write path integrals. Therefore one needs to define the Lagrangian in discrete time, where the definition is not unique. To see this, consider the discrete time form of the path integral where the path is divided into infinitesimal segment starting fromtime $t$ and finishing $t+\Delta t$. The question is then how should one define the Lagrangian? Is it with respect to time $t$, $t+\Delta$, or may be an average. This arbitrary choice is directly connected to how you choose to quantize your Hamiltonian. In particular, one can show that in order to recover the Schrodinger equation, the Lagrangian has to be defined as $L(\dot{q}(t),\bar{q}(t))$.

Suppose you start with a lagragian that contains $q(t)\dot q(t)$. When you discretize this term there will be an ambiguity:

$$
q(t)\dot q(t)dt\qquad\rightarrow\qquad q\_k (q\_k-q\_{k-1}) \qquad or \qquad (q\_k-q\_{k-1}) q\_{k-1}
$$ 

or any superpostion of that. This is related with $[q,p]\neq 0$.
The conventional wisdom in physics is that that the operator formulation with Hilbert space and [$S$-matrix](https://en.wikipedia.org/wiki/S-matrix) is the first principles, which connect to physics and experiments; whereas the [path/functional integral formulation](https://en.wikipedia.org/wiki/Path_integral_formulation) is usually derived from the operator formulation (using an operator ordering and time slicing prescription).

Ignoring all the well-known open problems in the definition of the path/functional integral, one may possibly claim (after future mathematicians have solved this) that there exists a unique most natural definition of a path/functional integral by itself.

However, one would still have to translate this path/functional integral into the operator formalism in order to connect to physics. And this translation is non-unique because of different operator ordering prescriptions, different choices of the vacuum state, etc, cf. e.g. [this](https://physics.stackexchange.com/q/56070/2451) & [this](https://physics.stackexchange.com/q/252213/2451) Phys.SE posts.