Special Cases of the BCH Identity

  • #1
479
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Homework Statement


Hi all, I'm having trouble working on the following problem. Any assistance will be greatly appreciated.

Screen Shot 2018-09-16 at 11.54.42 PM.png



Here, the capital letters stand for Position and Momentum operators while the ##x', p'## stand for eigenvalues.


Homework Equations




The Attempt at a Solution



a) and b)

It seems that a) and b) can be written in the form of
$$e^{\frac{i}{\hbar}x'P}e^{\frac{i}{\hbar}p'X}$$
which I can then express in the form of the BCH identity
$$e^{\frac{i}{\hbar}x'P}e^{\frac{i}{\hbar}p'X} = e^{\frac{i}{\hbar}x'P + \frac{i}{\hbar}p'X + \frac{i}{2\hbar}x'p' + 0 + O'}$$
where I denote the other terms by ##O'##. If ##O' = 0##, I obtain b), if ##O' = \frac{i}{2\hbar}x'p' ##, I get a). Is this what the question was getting at? It does seem too convenient.

c)

I'm not too sure where to start with this one.

Thanks in advance for any help!
 

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Answers and Replies

  • #2
nrqed
Science Advisor
Homework Helper
Gold Member
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279

Homework Statement


Hi all, I'm having trouble working on the following problem. Any assistance will be greatly appreciated.

View attachment 230761


Here, the capital letters stand for Position and Momentum operators while the ##x', p'## stand for eigenvalues.


Homework Equations




The Attempt at a Solution



a) and b)

It seems that a) and b) can be written in the form of
$$e^{\frac{i}{\hbar}x'P}e^{\frac{i}{\hbar}p'X}$$
which I can then express in the form of the BCH identity
$$e^{\frac{i}{\hbar}x'P}e^{\frac{i}{\hbar}p'X} = e^{\frac{i}{\hbar}x'P + \frac{i}{\hbar}p'X + \frac{i}{2\hbar}x'p' + 0 + O'}$$
where I denote the other terms by ##O'##. If ##O' = 0##, I obtain b), if ##O' = \frac{i}{2\hbar}x'p' ##, I get a). Is this what the question was getting at? It does seem too convenient.
It is not the best terminology to say that you express something "in the form of the BCH identity". People don't usually put it this way. What you have to do is to prove in each case that the left hand side is equal to the rhs, and the BCH identity can help you demonstrate this. So you should start from the expression on one side and end up on the expression on the other side, using any identity or algebraic manipulation you can do.

Focus on (a). Start from the left side and show that you get to the right side. This is almost what you did, but you did not complete your proof. You wrote

$$e^{\frac{i}{\hbar}x'P}e^{\frac{i}{\hbar}p'X} = e^{\frac{i}{\hbar}x'P + \frac{i}{\hbar}p'X + \frac{i}{2\hbar}x'p' + 0 + O'}$$

Now you must determine ##O'##. You cannot assume or choose the form it has. It has a unique expression and you must determine it. Once you will have shown what ##O'## is in this expression, you will have proven (a). Then we can discuss (b).
 
  • #3
479
12
Hi @nrqed , I have actually managed to work out the problem, many thanks for your pointers!
 

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