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.

  • 1
    See e.g. Stone - von Neumann theorem on Wikipedia. – Qmechanic Nov 27 '12 at 23:10
  • 2
    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 $[{\hat x}, {\hat P}] = [{\hat x}, {\hat p_{x}}] = i\hbar$ – Jerry Schirmer Nov 28 '12 at 0:40
up vote 12 down vote accepted

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 $\hat x$.

You have already got "practical" answers, so I intend to answer form another point of view.

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 $U : L^2(\mathbb R, dx)\to H$ 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 verify (1) on a common invariant dense subspace $S\subset H$, (b) $X^2+P^2$ is essentially self-adjoint on $S$ and (c) all vectors in $S$ are cyclic for $X$ and $P$. Then there exists a unitary operator $U : L^2(\mathbb R, dx)\to H$ such that (2) are valid for $\psi \in C_0^{\infty}(\mathbb R)$.

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 $C_0^\infty(\mathbb R)$ for the Schwartz space ${\cal S}(\mathbb R)$ 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 $\mathbb R^n$, i.e., concerning particles in space for $n=3$. Dropping the irreducibility requirement 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 $L^2(\mathbb R,dx)$ 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.

  • +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? – Jia Yiyang Dec 8 '13 at 14:56
  • 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. – Valter Moretti Dec 8 '13 at 15:04

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?

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