# Two Questions in Quantum Theory

• WisheDeom
According to the general expansion, [x_j, G(\mathbf{p})] would be equal to \sum_{n_1=0}^\infty \sum_{n_2=0}^\infty \sum_{n_3=0}^\infty g(n_1, n_2, n_3 ) (p_1)^{n_1}(p_2)^{n_2}(p_3)^{n_3}. So [x_j, G(\mathbf{p})] would be a sum of terms, each of which would be a product of the position operator x_i and the momentum operator p_j. \f

#### WisheDeom

Quantum Theory: Commutators of Functions of Observables

## Homework Statement

First is a question from Sakurai Modern Quantum Mechanics, 2nd ed., 1.29a.

Show that

$$[x_i,G(\mathbf{p})] = i\hbar\frac{\partial G}{\partial p_i}$$

and

$$[p_i,F(\mathbf{x})] = - i\hbar\frac{\partial F}{\partial x_i}$$

for any functions $F(\mathbf{x})$ and $G(\mathbf{p})$ which can be expanded in power series of their arguments.

## Homework Equations

Definition of a Taylor series of a function $f$ of variable $x$ expanded around point $a$:

$$f(x) = \sum^{\infty}_{n=0} \frac{f^{(n)}(a)}{n!}(x-a)$$

Commutator of position and momentum operators: $[x_i,p_j]=i\hbar\delta_{ij}$.

## The Attempt at a Solution

I tried a general solution, i.e. looking at the commutator $[F(\mathbf{x}), G(\mathbf{p})]$. The first problem I'm having is with the concept of taking derivatives with respect to operators. Can I simply treat $x$ as a scalar while computing $\frac{\partial F}{\partial x}$, for example? Is

$$\frac{\partial}{\partial x}(x^2) = 2x$$

valid?

Assuming I have that right, I am really stuck on the Taylor expansion itself. I don't see why powers of higher than first order would disappear. I assume my solution will look something like

$$[F(\mathbf{x}),G(\mathbf{p})] = [\mathbf{x},\mathbf{p}]\frac{\partial F}{\partial x_i}\frac{\partial G}{\partial p_i}$$

which would satisfy the problem, but I don't know how to get there.

Thank you!

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I tried a general solution, i.e. looking at the commutator $[F(\mathbf{x}), G(\mathbf{p})]$. The first problem I'm having is with the concept of taking derivatives with respect to operators. Can I simply treat $x$ as a scalar while computing $\frac{\partial F}{\partial x}$, for example? Is

$$\frac{\partial}{\partial x}(x^2) = 2x$$

valid?

Assuming I have that right, I am really stuck on the Taylor expansion itself. I don't see why powers of higher than first order would disappear. I assume my solution will look something like

$$[F(\mathbf{x}),G(\mathbf{p})] = [\mathbf{x},\mathbf{p}]\frac{\partial F}{\partial x_i}\frac{\partial G}{\partial p_i}$$

which would satisfy the problem, but I don't know how to get there.

Thank you!

Here, $\mathbf{x}$ and $\mathbf{p}$ are vector operators, so if you use a Taylor expansion, you will need to use the multivariable form, given in equation (31) here.

Luckily, you can avoid that entirely, all you need for this problem is the general form of the series expansion of $G(\mathbf{p})$ (and your commutation relation of course). In 3 dimensions, the series expansion has the general form $G(\mathbf{p}) = \sum_{n_1=0}^\infty \sum_{n_2=0}^\infty \sum_{n_3=0}^\infty g(n_1, n_2, n_3 ) (p_1)^{n_1}(p_2)^{n_2}(p_3)^{n_3}$. That is, you expect a series that in general has has a term for each possible combination of (positive integer) powers of each $p_j$.

What is $[x_j, G(\mathbf{p})]$ according to this general expansion? What is $\frac{\partial G}{\partial p_i}$?

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Ah! Thank you, that was straightforward.