Two Questions in Quantum Theory

In summary: 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. \
  • #1
WisheDeom
12
0
Quantum Theory: Commutators of Functions of Observables

Homework Statement



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

Show that

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

and

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

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



Homework Equations



Definition of a Taylor series of a function [itex]f[/itex] of variable [itex]x[/itex] expanded around point [itex]a[/itex]:

[tex] f(x) = \sum^{\infty}_{n=0} \frac{f^{(n)}(a)}{n!}(x-a) [/tex]

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



The Attempt at a Solution



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

[tex]\frac{\partial}{\partial x}(x^2) = 2x[/tex]

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


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

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

Thank you!
 
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  • #2


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

[tex]\frac{\partial}{\partial x}(x^2) = 2x[/tex]

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

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

Thank you!

Here, [itex]\mathbf{x}[/itex] and [itex]\mathbf{p}[/itex] 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 [itex]G(\mathbf{p})[/itex] (and your commutation relation of course). In 3 dimensions, the series expansion has the general form [itex]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}[/itex]. That is, you expect a series that in general has has a term for each possible combination of (positive integer) powers of each [itex]p_j[/itex].

What is [itex][x_j, G(\mathbf{p})][/itex] according to this general expansion? What is [itex]\frac{\partial G}{\partial p_i}[/itex]?
 
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  • #3
Ah! Thank you, that was straightforward.
 

Related to Two Questions in Quantum Theory

1. What is the concept of superposition in quantum theory?

In quantum theory, superposition refers to the ability of a quantum system to exist in multiple states at the same time. This means that until a measurement is made, the system exists in a combination of all possible states, rather than a single definite state.

2. How does the observer affect the outcome of a quantum experiment?

In quantum theory, the observer plays a crucial role in determining the outcome of an experiment. This is because the act of observation causes the quantum system to collapse into a single state, known as wave function collapse. The specific outcome of the experiment is then determined by the probabilities associated with each possible state.

3. What is the uncertainty principle in quantum theory?

The uncertainty principle, also known as Heisenberg's uncertainty principle, states that it is impossible to simultaneously know the exact position and momentum of a particle. This is because the act of measuring one of these quantities causes the other to become more uncertain. This principle is a fundamental aspect of quantum mechanics.

4. Can quantum entanglement be used for communication?

No, quantum entanglement cannot be used for communication. While two entangled particles may have a strong correlation, there is no way to control which specific state each particle will be in when measured. Therefore, it is not possible to use entanglement to transmit information faster than the speed of light.

5. What is the role of probability in quantum theory?

In quantum theory, probability plays a fundamental role in describing the behavior of particles and systems. This is because the behavior of particles at the quantum level is inherently probabilistic, and can only be described in terms of probabilities rather than definite outcomes. Probability is used to calculate the likelihood of a particle being in a certain state or location, and is a central concept in understanding quantum phenomena.

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