For the scalar field$\phi(\vec{x}=0,t=0)$, we can use translation operator$U=e^{iP\cdot x}=e^{i(Ht-\vec{P}\cdot\vec{x})}$ to move it to$\phi(\vec{x},t)$:
$$\phi(\vec{x},t)=e^{iP\cdot x}\phi(0,0)e^{-iP\cdot x}.$$
For the space-like part:
$$\phi(\vec{x})=e^{-i\vec{P}\cdot\vec{x}}\phi(\vec{x}=0)e^{i\vec{P}\cdot\vec{x}}$$
which$P^i\equiv\int d^3x\ \pi\partial^i\phi$ is the total momentum operator. One can derive this part by solving the equation:
$$i\partial_i\phi(\vec{x})=[\phi(\vec{x}),P_i].$$
A common way to derive this equaiton is to compute$[\phi,P_i]=[\phi,\int d^3x \ \pi\partial^i\phi]$. I find the relation in NRQM:
$$[F(x),p_x]\psi=\left(-\mathrm{i}F\frac{\partial \psi}{\partial x}+\mathrm{i}\frac{\partial F}{\partial x}\psi+\mathrm{i}F\frac{\partial \psi}{\partial x}\right)=\mathrm{i}\frac{\partial F}{\partial x}\psi\Rightarrow \mathrm{i}\frac{\partial F(\mathbf{x})}{\partial x^i}=[F(\mathbf{x}),p^i]$$i.e.$i\partial_iF(x)=[F,p^i]$. It appears that the two formulas differ by a negative sign, i.e.
$$i\partial_iF(x)=[F,p^i]\quad in \ NRQM\\i\partial_i\phi(\vec{x})=[\phi(\vec{x}),P_i]=-[\phi(\vec{x}),P^i]\quad in\ QFT$$
My question is how to realize this sign. Does it mean that we can NOT simply replace the relation in QM to QFT?
2 Answers2
If one looks at your calculations carefully...
I) Let us look on what$[F(x), p_x]$ means in NRQM. It is actually$[F(\hat{x}), \hat{p}_x]$.
$\hat{x}$ is operator here.
$F(x)$ - it is implied that this is some power series. We know what$\hat{x}^n$ is; we know what power series of$\hat{x}$ is; we know what$F(\hat{x})$ is.$F(\hat{x})$ is operator since it is power series of operator$\hat{x}$.
Various power series produce different$F(\hat{x})$ operators.
The relation$[F(\hat{x}),\hat{p}_x]=i\frac{\partial F}{\partial x}(\hat{x})$ actually comes from commutator betweenoperator$\hat{x}$ and operator$\hat{p}_x$:$[\hat{x},\hat{p}_x]=i$.
II) Let us look on what$\hat{\phi}(\mathbf{x})$ means in QFT.
$\mathbf{x}$ is not an operator here. It is simply a point in 3D space, a vector of three real numbers.
$\hat{\phi}(\mathbf{x})$ is operator. But the fact$\hat{\phi}(\mathbf{x})$ being operator is not related to$\mathbf{x}$.$\hat{\phi}(\mathbf{x})$ is operator because it is a linear combination of annihilation operators$\hat{a}_\mathbf{p}$ (and creation operators$\hat{a}_\mathbf{p}^+$, as well) acting on the Fock space. This linear combination is different for various$\mathbf{x}$, so we have operator depending on$\mathbf{x}$:$\hat{\phi}(\mathbf{x})$.
The expression for$\hat{\phi}(\mathbf{x})$ is unique in the whole theory of quantum scalar field. There is no room for choice here.
III) So,$[F(\hat{x}), \hat{p}_x]$ and$[\hat{\phi}(\mathbf{x}), \hat{P}_x]$ may look similar, but the underlying meaning is vastly different.
I think the pictures described in parts I and II are so different that they can hardly be combined. I would not consider analogies between$[F(\hat{x}), \hat{p}_x]$ and$[\hat{\phi}(\mathbf{x}), \hat{P}_x]$, since the underlying meaning is very different.
tl;dr: The sign change in your last line follows from a metric convention that doesn't fundamentally change the relationship between observables and their spatial evolution as defined by the commutation relation.
From your final line, the contradiction you're citing seems to be referring to the way, in QFT, 3D-vectors with upper space indices can be converted into vectors with lower space indices through a minus sign if you use the "west-coast"$(+, -, -, -)$ metric conventionhttps://en.wikipedia.org/wiki/Sign_convention#Relativity. In your NRQM expression you have a spatial derivative with a lower index and a momentum operator with an upper index thus seeming to contradict the analogous QFT expression with lower and upper indices in the same respective places.
Your contradiction follows from two sources: The sign convention you follow and your proper interpretation of it.
Sign Conventions: Both the West Coast and East Coast metric conventions (again,https://en.wikipedia.org/wiki/Sign_convention#Relativity) provide permissable formulations of relativity. If you follow the east coast$(-, +, +, +)$ then$P^i = P_i$ and there is no apparent contradiction.
Intepreting West Coast Convention: But even if you follow the west coast convention, there is no contradiction between the NRQM and QFT expressions as long as you have the appropriate interpretation of the upper and lower operators. In either convention, either the upper or lower index is perceived as the true operator and the other is defined in terms of it. So taking$P_\mu$ to be the true operator,$P^{\mu}$ is defined in terms of it and the metric as$P^{\mu}= \eta^{\mu \nu} P_{\nu}.$ In this case, it is the "true operator"$P_\mu$ that has a spatial part which always corresponds to the momentum operator we find in NRQM. The upper index operator (in the west coast convention) is the negative of said operator.
However, this upper and lower sign convention from QFT can't be trasported to NRQM. There are no upper and lower indices for vectors in NRQM since within NRQM, Lorentz invariant quantities like$x^\mu x_\mu = \eta_{\mu\nu}x_\mu x_\nu$ are not relevant. Put another way, NRQM always follows the East Coast convention in which spatial vectors don't change sign in moving from upper to lower index representation. So for the NRQM expression, it isn't quite right to say that the spatial deriative has a lower index, and the momentum operator has an upper spatial index in your second to last line. All spatial indices are defined on the same "level" in NRQM, and that level is always one which is reproduced by the results of the "true operator" in the QFT extension.
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