Special Cases of the BCH Identity

In summary, the problem is to find the momentum of a particle in a system of particles following the equation of motion
  • #1
WWCY
479
12

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!
 

Attachments

  • Screen Shot 2018-09-16 at 11.54.42 PM.png
    Screen Shot 2018-09-16 at 11.54.42 PM.png
    15 KB · Views: 885
Physics news on Phys.org
  • #2
WWCY said:

Homework Statement


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

View attachment 230761Here, 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
Hi @nrqed , I have actually managed to work out the problem, many thanks for your pointers!
 

Related to Special Cases of the BCH Identity

1. What is the BCH identity and how does it relate to special cases?

The BCH identity, also known as the Baker-Campbell-Hausdorff formula, is a mathematical identity that expresses the exponential of a sum of operators in terms of a product of exponentials. Special cases of the BCH identity refer to specific situations where the formula can be simplified or expanded to make calculations easier.

2. What are some examples of special cases of the BCH identity?

Some common examples of special cases of the BCH identity include the commutator identity, the Jacobi identity, and the Zassenhaus formula. These special cases arise when certain operators in the exponential sum commute or have specific properties.

3. How is the BCH identity used in physics and engineering?

The BCH identity is frequently used in physics and engineering to simplify calculations involving exponentials of operators. It is particularly useful in quantum mechanics, where operators do not necessarily commute, and in control theory, where it can be used to derive analytical solutions to differential equations.

4. Can the BCH identity be extended to non-commuting operators?

Yes, the BCH identity can be extended to non-commuting operators using the Zassenhaus formula. This formula involves an infinite series and is used to calculate the exponential of a sum of non-commuting operators.

5. Are there any limitations to the use of the BCH identity in special cases?

While the BCH identity is a powerful tool for simplifying calculations, it does have some limitations. In some cases, the formula may not converge or may produce inaccurate results. Additionally, the use of special cases can sometimes lead to oversimplification, which may not accurately reflect the true behavior of the system being studied.

Similar threads

  • Advanced Physics Homework Help
Replies
14
Views
1K
  • Advanced Physics Homework Help
Replies
3
Views
1K
  • Advanced Physics Homework Help
Replies
3
Views
950
  • Advanced Physics Homework Help
Replies
1
Views
669
  • Advanced Physics Homework Help
Replies
9
Views
921
  • Advanced Physics Homework Help
Replies
3
Views
1K
  • Advanced Physics Homework Help
Replies
24
Views
976
Replies
4
Views
1K
  • Advanced Physics Homework Help
Replies
8
Views
1K
  • Advanced Physics Homework Help
Replies
3
Views
1K
Back
Top