Can we correctly define momentum operator only by means of position operator and their commutation relation?

In "J.M. Ziman. Electrons and Phonons: The Theory of Transport Phenomena in Solids" the author formally introduces the position (displacement) operator and then defines the momentum operator with the correct commutation relation $$[\hat{u}_{l}, \hat{p}_{l'}] = i \hbar \delta_{l,l'}$$ between these two. Is such approach formally correct?

Edit: This suggestion is wrong: Can this lead to some degenerate form of "momentum operator"? First that comes to my mind is \begin{aligned}\hat{p}_{l} = -i\hbar\frac{\partial}{\partial u_{l}} \quad \text{if} \quad u < u_{0} \\ \hat{p}_{l} = -i\hbar\frac{1}{u_{l}} \quad \text{if} \quad u \geq u_{0}.\end{aligned} Putting this theoretical construction aside, the core of the question is that if generally such definition of an operator can be done?

(Disclaimer: The more rigourously inclined individual may be better suited by looking at the Stone-von Neumann theorem, as Qmechanic notes)

One can deduce that the momentum operator takes the form $\hat p = -\mathrm{i}\hbar\partial_x$ in the position representation from the fact that the momentum operator generates the infinitesimal translations as $T(\epsilon) = \boldsymbol{1} - \frac{\mathrm{i}}{\hbar}\epsilon \hat p$ alone:

Observe that

$$T(\epsilon)\lvert \psi \rangle = \int \mathrm{d}x \lvert x \rangle \langle x \rvert T(\epsilon)\lvert \psi \rangle \tag{1}$$

Now, $\langle x \rvert T(\epsilon) \lvert \psi \rangle= \langle x - \epsilon\vert\psi\rangle = \psi(x - \epsilon)$ . A Taylor expansion of this wavefunction yields

$$\psi(x - \epsilon) = \psi(x) - \epsilon(\partial_x\psi)(x) + \mathcal{O}(\epsilon^2) \tag{2}$$

Put $(2)$ into $(1)$, forget about $\mathcal{O}(\epsilon^2)$ and get

$$(\boldsymbol{1} - \frac{\mathrm{i}}{\hbar}\epsilon\hat p)\lvert \psi \rangle = \int \mathrm{d}x \psi(x) \lvert x \rangle - \int\mathrm{d}x \lvert x \rangle\epsilon \partial_x \psi(x) \implies \hat p\ = \int\mathrm{d}x \lvert x \rangle (-\mathrm{i}\hbar\partial_x)\langle x \rvert$$

Therefore, $\hat p = -\mathrm{i}\hbar\partial_x$ in the position basis. This already suggests that you have essentially no freedom in how to choose the momentum operator.

• I had actually started typing something very much like this up (except I included the derivation of the form of the infinitesimal translation operator) but gave up because it was getting too long (probably because of the included derivation). – Kyle Kanos Jul 30 '14 at 16:01
• Precisely what I was looking for!! – Sjorszini Sep 27 '16 at 6:39

I) Comment to the question (v1):

The Schrödinger position representation

$$\hat{p}_k ~=~ \frac{\hbar}{i} \frac{\partial }{\partial x^k}, \qquad \hat{x}^j ~=~x^j,$$

correctly reproduces the canonical commutation relations

$$[\hat{x}^j,\hat{p}_k ]~=~i\hbar ~\delta^j_k ~{\bf 1},$$

while the proposal

$$\hat{p}_k ~=~ \frac{\hbar}{i} \frac{1}{x^k}, \qquad \hat{x}^j ~=~x^j, \qquad\leftarrow \text{(Wrong!)}$$

simply commutes

$$[\hat{x}^j,\hat{p}_k ]~=~0.$$

II) Comment to the question (v2): It seems OP is essentially asking:

What is the most general expression for the position representation of the momentum operator?

This was e.g. asked in this Phys.SE post. See also Stone von Neumann theorem.

• You're right, was too hasty with my calculation from top of my head. – Jan Hirschner Jul 30 '14 at 15:29