How Do Commutators of Functions of Observables Work in Quantum Mechanics?

AI Thread Summary
The discussion focuses on the commutation relations in quantum mechanics, specifically how to derive the commutators of functions of observables. It establishes that for any functions F(x) and G(p), the relations [x_i, G(p)] = iħ(∂G/∂p_i) and [p_i, F(x)] = -iħ(∂F/∂x_i) hold true when expanded in power series. The conversation highlights confusion regarding the treatment of operators as scalars during differentiation and the implications of higher-order terms in Taylor expansions. The resolution suggests that using the general form of series expansion simplifies the derivation process. Understanding these concepts is crucial for solving problems related to quantum mechanics and operator algebra.
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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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WisheDeom said:
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.
 
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