# What Are the Methods to Solve Commutators in Quantum Mechanics?

• fifthrapiers
It should read= i\hbar \frac{p}{2m} \psi + \frac{p}{2m}i\hbar \psi + p[x,\frac{1}{2m}] \psi = i\hbar \frac{p}{2m} \psi + \frac{p}{2m}i\hbar \psi .In summary, the problem involves using commutator identities and formulas to solve for [x,H] in two different ways. The first method involves using [x,p]=i\hbar and other identities, while the second method uses the "direct way." The correct steps for both methods are provided, along with a reminder of the commut
fifthrapiers
I'm having trouble with commutators. I have to solve them 2 ways. First, using $$[x,p]=i\hbar$$ and other identities/formulas, and the the second method the "direct way".

1.) $$x,\hat{H}$$

My work:

$$[x,\hat{H}]\psi &= x\hat{H}\psi - \hat{H}x\psi$$
$$= x \left ( \frac{p^2}{2m} + V(x) \right )\psi - \left ( \frac{p^2}{2m} + V(x) \right )x\psi$$
$$= \frac{xp^2\psi}{2m} + x V(x)\psi - \frac{p^2x\psi}{2m} - V(x) x\psi$$

2.) $$[\hat{p}, \hat{H} + x]$$

$$[\hat{p}, \hat{H} + x]\psi &= \hat{p}(\hat{H}+x)\psi + (\hat{H}+x)\hat{p}\psi$$
$$= i\hbar\frac{\partial}{\partial p} \left(\left( \frac{p^2}{2m} + V(x) \right ) + x\right)\psi + \left(\left( \frac{p^2}{2m} + V(x) \right ) + x\right)i\hbar\frac{\partial}{\partial p}\psi$$

Yikes.

What is the original question? i.e the problem stated.

Use the [x,p] and other commutator idendites, you have not done that,
for example:

[x,H] = [x,pp + V] = {i skip the constant factors} = [x,pp] + [x,V] = ? continue

malawi_glenn said:
What is the original question? i.e the problem stated.

Use the [x,p] and other commutator idendites, you have not done that,
for example:

[x,H] = [x,pp + V] = {i skip the constant factors} = [x,pp] + [x,V] = ? continue

The orig question was [x,h] for # 1, I just forgot to put in the brackets...

Yes, that's the whole point, I'm not sure how. I showed my work.

do you know the commutator relation for

[A,B*C] = ??

find that, and use it.

malawi_glenn said:
do you know the commutator relation for

[A,B*C] = ??

find that, and use it.

Ah yes! That was quite helpful. How does this look:

$$[x,\hat{H}]\psi = [x,\frac{p^2}{2m}]\psi$$
$$= [x,p\cdot\frac{p}{2m}]\psi$$
$$= [x,p]\frac{p}{2m}\psi + p[x,\frac{p}{2m}]\psi$$
$$= i\hbar\frac{p}{2m}\psi + p[x,p\cdot\frac{1}{2m}]\psi$$
$$= i\hbar\frac{p}{2m}\psi + p[x,p]\frac{1}{2m}\psi + p[x,\frac{1}{2m}]\psi$$
$$= i\hbar\frac{p}{2m}\psi + \frac{p}{2m}i\hbar + p\frac{x}{2m}\psi + \frac{x}{2m}\psi$$
$$= \boxed{2i\hbar\frac{p}{2m}\psi + 2\frac{px}{2m}\psi}$$

Not sure if it's right. The direct way, which I tried earlier, is still causing probs. I'm guessing for the second one, I should use [A,B+C] = [A,B] + [A,C]?

fifthrapiers said:
Ah yes! That was quite helpful. How does this look:

$$[x,\hat{H}]\psi = [x,\frac{p^2}{2m}]\psi$$
$$= [x,p\cdot\frac{p}{2m}]\psi$$
$$= [x,p]\frac{p}{2m}\psi + p[x,\frac{p}{2m}]\psi$$
$$= i\hbar\frac{p}{2m}\psi + p[x,p\cdot\frac{1}{2m}]\psi$$
So far so good but you don't need to stick in a wavefunction when you do it this way. One has to put a test function only when one uses explicit representations in terms of differential operators.

$$= i\hbar\frac{p}{2m}\psi + p[x,p]\frac{1}{2m}\psi + p[x,\frac{1}{2m}]\psi$$
No!
$$[x,\frac{p}{2m}] = \frac{1}{2m} [x,p]$$
simply (the 1/2m is an overall constant which youcan pull out of the commutator.

kdv said:
No!
$$[x,\frac{p}{2m}] = \frac{1}{2m} [x,p]$$
simply (the 1/2m is an overall constant which youcan pull out of the commutator.
For the opening poster... technically, you weren't wrong; you were merely encumbered. ($[x, a] = 0$ for any constant a)

Hurkyl said:
For the opening poster... technically, you weren't wrong; you were merely encumbered. ($[x, a] = 0$ for any constant a)

Yes, that is true. My apologies if I gave the wrong impression.

Technically, it is correct to write

$$[x,P/2m] = \frac{1}{2m} [x,P] + [x, \frac{1}{2m}] P$$

Nobody woul dnormally write it this way but it is indeed correct.
The next step is to use the fact that [x,1/2m] = 0.

However, the next line in the demonstration of the OP is incorrect.

## 1. What is a commutator and why is it important?

A commutator is a mathematical tool used in group theory to describe how elements of a group interact with each other. It is important because it helps us understand the structure and properties of a group and its elements.

## 2. Why do I keep getting errors when trying to use commutators?

There could be several reasons for this, such as incorrect syntax, using incompatible elements, or not specifying the correct group. Double-check your code and make sure to consult a reference if needed.

## 3. How can I simplify a commutator?

To simplify a commutator, you can use known identities and properties of the group to rearrange the elements. You can also use software or a calculator to calculate the commutator for you.

## 4. Can commutators be used in other areas of science?

Yes, commutators have applications in various fields of science, such as quantum mechanics, electromagnetism, and signal processing. They are also used in computer science and cryptography.

## 5. How can I improve my understanding of commutators?

To improve your understanding of commutators, it is important to have a strong foundation in group theory and abstract algebra. You can also practice solving problems and working with different types of commutators to gain more experience.

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