# Does the canonical commutation relation fix the form of the momentum operator?

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For one dimensional quantum mechanics $$[\hat{x},\hat{p}]=i\hbar$$

Does this fix univocally the form of the $\hat{p}$ operator? My bet is no because $\hat{p}$ actually depends if we are on coordinate or momentum representation, but I don't know if that statement constitutes a proof. Moreover if we choose $\hat{x}\psi=x\psi$ is the answer of the following question yes?

For the second one

$$(\hat{x}\hat{p}-\hat{p}\hat{x})\psi=x\hat{p}\psi-\hat{p}x\psi=i\hbar\psi$$

but I don't see how can I say that $\hat{p}$ must be proportional to $\frac{\partial}{\partial x}$. I don't know if trying to see that $\hat{p}$ must satisfy the Leibniz rule and thus it should be proportional to the $x$ derivative could help. Or using the fact that $\hat{x}$ and $\hat{p}$ must be hermitian

Any hint will be appreciated.

This post imported from StackExchange Physics at 2014-04-12 19:04 (UCT), posted by SE-user Jorge
edited Jun 5, 2014
See e.g. Stone - von Neumann theorem on Wikipedia.

This post imported from StackExchange Physics at 2014-04-12 19:04 (UCT), posted by SE-user Qmechanic
It can't, because, amongst other things, we have $[{\hat x}, {\hat y}] = [{\hat y}, {\hat p_{x}}] = 0$, so if we define ${\hat P} \equiv {\hat p} + {\hat y}$, then we also have $[{\hat x}, {\hat P}] = [{\hat x}, {\hat p_{x}}] = i\hbar$

This post imported from StackExchange Physics at 2014-04-12 19:04 (UCT), posted by SE-user Jerry Schirmer

@JerrySchirmer: The construction you give is forbidden by the canonical commutation relations for $p_x$ and $p_y$, $[p_x,p_y]=0$. If you omit the p-p commutation relation, leaving only the x-x commutation relation and the x-p commutation relation, you end up describing the situation of a particle in an external magnetic field. The phases are not integrable, and can't be removed by a redefinition of the x basis.

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There is a quite famous theorem due to Stone and von Neumann, later improved by Mackay, and finally by Dixmier and Nelson, roughly speaking establishing the following result within the most elementary version. (Another version of the theorem focuses on the unitary groups generated by $X$ and $P$ avoiding problems with domains, however I stick here to the self-adjoint operator version.)

THEOREM. (rough statement "for physicists") If you have a couple of self-adjoint operators $X$ and $P$ defined on a Hilbert space $H$ such that are conjugated to each other:

\begin{equation} [X,P] = i \hbar I \quad\quad\quad (1) \end{equation} and there is a cyclic vector for $X$ and $P$, then there exists a unitary operator $U : L^2(R, dx)\to H$ such that:

$$(U^{-1} X U )\psi (x)= x\psi(x)\quad \mbox{and}\quad (U^{-1} P U )\psi (x)= -i\hbar \frac{d\psi(x)}{dx}\:.\quad (2)$$

(The rigorous statement, in this Nelson-like version is reads as follows

THEOREM. Let $X$ and $P$ be a pair of self-adjoint operators on a complex Hilbert space $H$ such that (a) they veryfy (1) on a common invariant dense subspace $S\subset H$, (b) $X^2+P^2$ is essentially self-adjoint on $S$ and (c) there is a cyclic vector in $S$ for $X$ and $P$. Then there exists a unitary operator $U : L^2(R, dx)\to H$ such that (2) are valid for $\psi \in C_0^{\infty}(R)$.

Notice that the operators defined in the right-hand sides of (2) admits unique self-adjoint extensions so they completely fix the operators representing respective observables. We can equally replace $C_0^\infty(R)$ for the Schwartz space ${\cal S}(R)$ in the last statement.)

Barring technicalities, all that means that commutation relations actually fix position and momentum observables as well as the Hilbert space. For instance, referring to Murod Abdukhakimov's answer, if the addition of $\partial f$ to the standard expressions of $X$ and $P$ gives rise to truly self-adjoint operators, then a unitary transformation (just that connecting $\psi$ to $\psi'$ in Murod Abdukhakimov's answer) gets rid of the deformation restoring the standard expression. Remember that unitary transformations do not alter all physical objects of QM.

The result extends to $R^n$, i.e., concerning particles in space for $n=3$. Dropping the irreducibility requirement (there is a cyclic vector for $X$ and $P$) the thesis holds anyway but $H$ decompose into a direct sum (not direct integral!) of closed subspaces where the strong statement is valid.

There are important consequences of this fundamental theorem. First of all $H$ must be saparable as $L^2(R,dx)$ is. Moreover no time operator $T$ (conjugated with the Hamiltonian operator $H$) exists if the Hamiltonian operator id bounded below as physics requires. The latter statement is due to the fact that the theorem fixes the spectra of $X$ and $P$ as the whole real axes in both cases, so that the spectrum of $H$ would not be bounded below if $T,H$ were a conjugated pair of operators. A similar no-go theorem arises concerning quantization of a particle on a circle when one tries to define position and impulse self-adjoint operators. The attempt to solve these no-go results gave rise to more general formulation of quantum mechanics based on the notion of POVM and eventually turned out to be very useful in other contexts as quantum information theory.

An important observation is that Stone-von Neumann - MacKay - Dixmier -Nelson's result fails when dealing with infinite dimensional systems. That is, roughly speaking, passing from the (symplectic space) of a finite number of particle to the (symplectic space) of a field. In that case the canonical commutation relations of $X_i$ and $P_j$ are replaced by those of the quantum fields. E.g:,

$$[\phi(t, x), \pi(t, y)] = i \hbar \delta(x,y) I$$

or more sophisticated versions of them. In this juncture, there exist infinitely many representations of the algebra of observables that cannot be connected by unitary operators. This is a well-known phenomenon in QFT or quantum statistical mechanics (in the thermodynamic limit). For instance the free theory and the interacting theory of a given quantum field cannot be represented in the same Hilbert space once one assumes standard requirements on states and observables (the so called Haag's theorem and this is the deep reason why LSZ formalism uses the weak topology instead of the strong one as in standard quantum theory of the scattering).

If one includes superselections charges in the algebra of observables, non unitarily equivalent representations of the algebra arise automatically giving rise to sectors.

In QFT in curved spacetime the appearance of inequivalent representations of the algebra of observables is a quite common phenomenon due to the presence of curvature of the spacetime.

This post imported from StackExchange Physics at 2014-04-12 19:04 (UCT), posted by SE-user V. Moretti
answered Dec 8, 2013 by (2,085 points)
+1.I'm curious exactly how LSZ uses weak topology instead of strong one, you know, physics books don't usually care. Any good reference?

This post imported from StackExchange Physics at 2014-04-12 19:04 (UCT), posted by SE-user Jia Yiyang
Maybe you could find this comment on Haag's textbook. I think I found this remark many many years ago in a book by Hepp. However, the problem is the so called Haag's theorem, the formal Moller operators cannot define a unitary transformation from the Hilbert space of the free theory to that of the interacting one without violating Haag's theorem. So the limit have to be computed for matrix elements, as in fact is done in LSZ formalism.

This post imported from StackExchange Physics at 2014-04-12 19:04 (UCT), posted by SE-user V. Moretti
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No. You can add an arbitrary constant shift (or an arbitrary operator commuting with $x$) without affecting the CCR.

For 1-dimensional QM, the general solution of the CCR with $\hat x$ represented as multiplication by $x$ on wave functions with argument $x$ is $\hat p=\hat p_0-A(\hat x)~~$, where $\hat p_0$ is the canonical momentum operator , and $A(x)$ is an arbitrary function of $x$.
Proof. The difference $\hat A:=\hat p_0-\hat p~$ commutes with $\hat x$, hence is a function of $\hat x$.

This post imported from StackExchange Physics at 2014-04-12 19:04 (UCT), posted by SE-user Arnold Neumaier
answered Nov 28, 2012 by (15,787 points)

This ambiguity is resolved by redefining the x-basis with a phase rotation, then the p operator is brought to canonical form. This is discussed by Dirac.

Yes, the resulting representations are equivalent (also by von Neumann's theorem), but the question was whether $\hat p$ had to be proportional to $\partial_x$. which is not the case.

But then to say "it is not the case" is misleading, because it's only a basis rotation that makes it the case. A basis rotation doesn't change anything except the names of states, it's not a truly different thing. It's just choosing coordinates on Hilbert space. Usually people ask questions about the structure of the thing, not the structure of the coordinates you are choosing.

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Stone's Theorem says that the self-adjoint (unbounded) operators $H$ are in 1-1 correspondence with the corresponding strongly-continuous 1-parameter unitary groups $U(t)$ via the correspondence $U(t)=\exp(itH)$, so it is sometimes more convenient to work directly with the unitary groups instead of the unbounded operators.  The difficulties caused by the unboundedness of X and P may be circumvented by formulating commutation relations between the corresponding 1-parameter unitary groups of operators $\exp(itX)$ and $\exp(isP)$, where $s$ and $t$ are real parameters.

These (not-strictly equivalent) relations between unitary operators

$\exp(itX) \exp(isP) = \exp(-its) \exp(isP) \exp(itX)$

are so-called "Weyl-Relations"  and indeed their irreducible representations given generators X and P which are unitarily equivalent to the usual forms of X and P.  This is explained in Reed & Simon's Method of mathematical physics volume III (scattering theory).

Furthermore, working with the unitary groups instead of with X and P allows one to ponder finite-dimensional analogues.  In particular, the unitary groups $\{U(t) = \exp (itX) | t\in\bf{R}\}$ and $\{V(s) =\exp (isP) | s\in \bf {R}\}$ become finite cyclic groups of unitary operators $\{U^k | { k=0,...,n-1}\}$ and $\{ V^k|k=0,...,N-1\}$satisfying

$U^N=V^N=I$

$UV=\exp(-2\pi i/N)VU$.

The irreducible representations in this finite-dimensional case were first worked out in 1950 by Hermann Weyl in his book Theory of Groups and Quantum Mechanics.

There are difficulties and outright counter-examples when one works with the usual physics-book canonical commutation relations $[X,P]=i$ , although one of the other answers reports extra hypotheses to get around it the difficulties.  See, for example, Nelson's example in Reed & Simon's book Methods of mathematical physics  volume 1 (Functional analysis).  If I recall correctly it was in a section called "formal manipulation is a touchy business".  (Sorry, my books are on another continent at present.)  Furthermore, one cannot have a relationship of the form $[X,P]=i$ in finite dimensions, as one sees by taking the trace of both sides. answered May 28, 2014 by Jon Tyson
edited May 28, 2014
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In more general case it could be:

$p_x = -ih\frac{∂}{∂x}+\frac{∂f}{∂x}$

$p_y = -ih\frac{∂}{∂y}+\frac{∂f}{∂y}$

$p_z = -ih\frac{∂}{∂z}+\frac{∂f}{∂z}$

where $f(x,y,z)$ - arbitrary function.

But you can also replace wave function $\psi'=e^{-\frac{i}{h}f(x,y,z)}\psi$ which brings you back to traditional form.

Looks like gauge transformation, isn't it?

This post imported from StackExchange Physics at 2014-04-12 19:04 (UCT), posted by SE-user Murod Abdukhakimov
answered Nov 29, 2012 by (505 points)

It doesn't "look like a gauge transformation", it is the original gauge transformation. This transformation removes the ambiguity in the p operators as you say, and is the key to the Stone Von-Neumann theorem, which simply wants to make this argument rigorous.

The key point is that you also need the commutation relation [p_x,p_y] = 0 between different components of p. If you remove this condition, you can't conclude that the phase term is a gradient, and you end up defining a magnetic field. In the 1940s, Feynman would use this to "derive Maxwell's equations from QM", what he was doing was weakening the commutation relations to show that you can introduce a gauge field easily by a simple modification of the canonical commutation relation. Feynman knew this was not deep, he didn't publish it, but it was preserved for posterity in a pedagogical American Journal of Physics article by Dyson, after Feynman's death.

@Ron Maimon . Feynman didn't use the momentum. He just assumed a second order equations of motion, commutativity of position coordinates, and "canonical" commutation between position and mass times velocity. He didn't define a Lagrangian, nor the momentum. He didn't make use of relativistic invariance or of any particular form of the commutation relation among different components of velocity.

You are giving Feynman's stated assumptions. The only crucial property of these assumptions is that he assumes that different position operators commute, but he _leaves out_ the commutativity of different components of the momentum. That's the crucial thing--- he is omitting the condition that different components of p commute, all the rest is standard nonsense.

When you leave out the commutativity of the different components of momentum, you end up with a non-integrable phase-factor in the definition of the states, so you can't pass to the standard situation, where p is a partial derivative. This introduces a magnetic field, which automatically ends up closed. The gauge invariance is automatic, because this is the freedom to choose your local basis, so you get the magnetic Maxwell equations for free when you make the quantum equation of motion self-consistent for the charged particle.

He gets the electric equations by doing the same trick with the energy operator, it's the same thing. It's really not deep, it's just the observation that when you omit the p-p commutation relations, you get electromagnetic fields. It was weird to people in the 1940s, because the gauge field wasn't as intuitive back then, but Feynman knew all about it, so he didn't consider this stuff as anything more than a trick.

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At least the representation of the momentum operator follows from the definition of the position operator and the commutation relation:

Let $\hat{Q}$ be a position operator with action $\hat{Q}\psi(x)=x\psi(x)$. The matrix elements of $\hat{Q}$ in position space are $\langle x|\hat{Q}|x'\rangle=x' \delta(x-x')$. Let $\hat{P}$ denote the (unknown) momentum operator in position space. Then, we can apply the commutator:

$([\hat{Q},\hat{P}])\psi(x)=\int dx\,\langle x|[\hat{Q},\hat{P}]|x'\rangle \psi(x')=\int dx'\,(x-x')\langle x|\hat{P}|x'\rangle\psi(x').$

On the other hand, by the commutator relation we have

$([\hat{Q},\hat{P}])\psi(x)=\int dx'\,i\hbar \,\delta(x-x')\psi(x').$

Therefore, the matrix elements of $\hat{P}$ have to satisfy:

$(x-x')\langle x|\hat{P}|x'\rangle=i\hbar \,\delta(x-x').$

From analysis we know the general identity of the Dirac distribution:

$-x\frac{d}{dx}\delta(x)=\delta(x).$

By this identity, the matrix elements of the momentum operator can be found to be

$\langle x|\hat{P}|x'\rangle=\frac{\hbar}{i}\frac{d}{dx}\delta(x-x'),$

or applied to $\psi$:

$\hat{P}\psi(x)=\frac{\hbar}{i}\frac{d}{dx}\psi(x).$

answered Dec 29, 2016 by (50 points)
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Other answers have already mentioned the Stone-von-Neumann theorem and its generalizations, which essentially say that any representation of the commutation relation $[X,P]=i\hbar$ is equivalent to the standard representation where $X=x$ and $P=-i\hbar \frac{d}{dx}$.

However, I would like to add here that while the representations are equivalent, they are not necessarily equal, and I want to give an example of what this means. Here is a counterintuitive realization of the operators X and P:

$$P = k, \quad X = i\hbar \frac{d}{dk}$$

They obviously satisfy the commutation relation, so it's perfectly valid. However, now it is the operator $X$ which is a differentiation operator, and $P$ which is a multiplication operator.

This representation has everything backwards, but it is still equivalent to the other representation, though: The equivalence is given by the Fourier transform. This is actually quite useful in some calculations: Just like the momentum is the differentiation with respect to the position, the position is differentiation with respect to momentum.

answered Dec 29, 2016 by (775 points)
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This very interesting question has already been answered correctly and mathematically rigorously here many times using Stone-von Neumann theorem. However, it is possible, and would be very profitable to the majority of physics students (and even practitioners like experimental physicists which may be unfamiliar with functional analysis), to have an answer ("Yes, up to a phase") using only physical language, which I present below. This also have the advantage of showing how the momentum operator is related to the generator of translations.

We begin by recalling the basic postulate of QM that observables are to be represented by operators $\boldsymbol{A}$ whose eigenkets $|a\rangle$ describes states which are sure to return the associated eigenvalue $a$ as a measurement result. Now, consider the following assuptions:

1. For a quantum particle in the line, position is represented by the operator $\boldsymbol{x}$ with eigenkets $|x\rangle$.
2. We are given an operator $\boldsymbol{p}$ with the following property: $[\boldsymbol{x},\boldsymbol{p}]=i\hbar$.

From these two assumptions, we can show that $\boldsymbol{p}=i\hbar d/dx$ (up to a phase) and moreover, that $\boldsymbol{p}$ is the generator of the translation operator.

Indeed, define the operator $T(x_0)=e^{-ix_0\boldsymbol{p}/\hbar}$ for some $x_0\in \mathbb{R}$. Begin with a particle localized at x, so that its state is (normalized to) $|x\rangle$. Lets now compute the action of the position operator on the new state $T(x_0)|x\rangle$, namely, $\boldsymbol{x} \big(T(x_0)|x\rangle \big)$.

The first step, which is an exercise on mathematical induction (or combinatorics) is to show that $[\boldsymbol{x},\boldsymbol{p}^n]=i\hbar n\boldsymbol{p}^{n-1}$. Second, we expand the exponential operator as follows:

$\boldsymbol{x} T(x_0)|x\rangle =\sum_{n=0}^\infty\frac{1}{n!}\big(\frac{-ix_0}{\hbar}\big)^n \boldsymbol{x}\boldsymbol{p}^n|x\rangle \\ =\sum_{n=0}^\infty \frac{1}{n!}\big(\frac{-ix_0}{\hbar}\big)^n ([\boldsymbol{x},\boldsymbol{p}^n]+\boldsymbol{p}^n\boldsymbol{x}) |x\rangle\\ =\sum_{n=1}^\infty \frac{1}{n!} \big(\frac{-ix_0}{\hbar}\big)^n \times i\hbar n\boldsymbol{p}^{n-1}|x\rangle + \sum_{n=0}^\infty \frac{1}{n!}\big(\frac{-ix_0}{\hbar}\big)^n \boldsymbol{p}^n (\boldsymbol{x} |x\rangle)\\ =\sum_{n=1}^\infty \frac{1}{(n-1)!} \big(\frac{-ix_0}{\hbar}\big)^{(n-1)}\boldsymbol{p}^{n-1}(x_0|x\rangle)+xT(x_0)|x\rangle\\ =(x+x_0)T(x_0)|x\rangle.$

Hence, we know that $T(x_0)|x\rangle$ is an eigenstate of the position operator with eigenvalue $x+x_0$, so a particle on that state is described by

$T(x_0)|x\rangle = e^{i\theta}|x+x_0\rangle,$

where $e^{i\theta}$ is where the "up to a phase" part of the answer comes from. Ignore it and set $\theta = 0$. Expand the right-hand side above in a series in $x_0$:

$T(x_0)|x\rangle = |x\rangle + x_0\frac{d}{dx}|x\rangle+\mathcal{O}(x_0^2).$

Finally, using the power series definition of the exponential operator,

$T(x_0)|x\rangle=e^{-ix_0\boldsymbol{p}/\hbar}|x\rangle=|x\rangle+\big(\frac{-ix_0\boldsymbol{p}}{\hbar}\big)|x\rangle+\mathcal{O}(x_0^2).$

Comparing the first order terms of the above equations gives our result, namely, $\boldsymbol{p}=i\hbar d/dx$ (up to a phase).

answered Dec 29, 2016 by (550 points)

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