# 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?

• Perhaps my mathematics/QM is a little rusty, but I don't believe that the latter "degenerate form" satisfies the commutation relation. Jul 30 '14 at 15:17
• You might be interested in this Physics.SE post and this Wikipedia entry. Jul 30 '14 at 15:51

(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). Jul 30 '14 at 16:01
• Precisely what I was looking for!! 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?