# Compatible Observables

## Homework Statement

So-called compatible observables correspond to operators which commute, i.e. [A, B] = 0, where [A, B] stands for AB − BA.

a) [3pt] Suppose Hermitian operators A and B represent two compatible observables, and all eigenvalues of A are different. Show that eigenstates of A are also eigenstates of B. Thus we can label the common eigenstates of A and B by the corresponding eigenvalues as |a',b'>

## Homework Equations

[A,B] = AB-BA = 0

## The Attempt at a Solution

A|a'> = a'|a'>
B|b'> = b'|b'>

Multiply A|a'> = a'|a'> by B

BA = B(a'|a'>) (1)

Since AB-BA = 0
AB = BA

I assume I can multiply by |a'> so

Multiply by |a'>

AB|a'> = BA|a'> = Ba'|a'>

I'm not sure if what I'm on the correct path but this is my attempt so far. I'm not sure how I can equate prove the question so I just tried something.

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TSny
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AB|a'> = BA|a'> = a'|a'>
What can you conclude about the state B|a'> from this result?

What can you conclude about the state B|a'> from this result?

I guess I could say that if B operates on |a'>, the eigenvector would be a' with eigenstate |a'>.

Is that correct?

Edit: Sorry nevermind about that. Give me a couple of minutes

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TSny
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I guess I could say that if B operates on |a'>, the eigenvector would be a' with eigenstate |a'>.

Is that correct?
Note that the phrase the eigenvector would be a' doesn't make sense because a' is a number, not a vector.

At the end of your first post you have
AB|a'> = BA|a'> = a'|a'>
There's a mistake in the last expression on the right which I overlooked. Can you see the mistake and fix it?

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Note that the phrase the eigenvector would be a' doesn't make sense because a' is number, not a vector.

At the end of your first post you have

There's a mistake in the last expression on the right which I overlooked. Can you see the mistake and fix it?

Yes it should be AB|a'> = BA|a'> = Ba'|a'>

So a' is the eigenvalue and the eigenstate would be |a'>

TSny
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Yes it should be AB|a'> = BA|a'> = Ba'|a'>
Yes. It might be helpful to write this as A(B|a'>) = a'(B|a'>). Be sure you see how this follows from what you wrote.

Does this tell you something about the state B|a'>?

Yes it should be AB|a'> = BA|a'> = Ba'|a'>

So a' is the eigenvalue and the eigenstate would be |a'>

Or I could say let χ = B|a'> then

Aχ = a'χ

So χ is the eigenstate of A with eigenvalue a'?

Yes. It might be helpful to write this as A(B|a'>) = a'(B|a'>). Be sure you see how this follows from what you wrote.

Does this tell you something about the state B|a'>?

Am I also allowed to say that B|a'> = a'|a'>

I would think not as I haven't proved any of it

TSny
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Or I could say let χ = B|a'> then

Aχ = a'χ

So χ is the eigenstate of A with eigenvalue a'?
Yes. Good. But be careful with the notation. If χ is an eigenstate of A, then we should use ket notation and write the eigenstate as |χ>. Then you have A|χ> = a'|χ>.

Think about how |χ> must be related to |a'>. Here, you will need to use the information given in the problem that all eigenvalues of A are different.

TSny
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Am I also allowed to say that B|a'> = a'|a'>
no

I would think not as I haven't proved any of it
Right.

Yes. Good. But be careful with the notation. If χ is an eigenstate of A, then we should use ket notation and write the eigenstate as |χ>. Then you have A|χ> = a'|χ>.

Think about how |χ> must be related to |a'>. Here, you will need to use the information given in the problem that all eigenvalues of A are different.

Am I allowed to say that

|χ> must also be equal to b'|b'>
since |χ> = B|a'>

I'm not sure how to use the fact that the information of all eigenvalues of A are different. Does that mean the eigenvalues of A must have the eigenvalues of B?

Or could I say that B|a'> = |χ> then the eigenvalue of |χ> is just 1? but I'm not sure where this would lead me.

TSny
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Am I allowed to say that

|χ> must also be equal to b'|b'>
since |χ> = B|a'>
I don't see how you are arriving at "|χ> must also be equal to b'|b'>" or what that really means.

I'm not sure how to use the fact that the information of all eigenvalues of A are different. Does that mean the eigenvalues of A must have the eigenvalues of B?
No, A and B will generally have different eigenvalues.

You know that A|a'> = a'|a> and also that A|χ> = a'|χ> (from post #7). So, both the vectors |a'> and |χ> are eigenvectors of A with the same eigenvalue a'. The problem assumes that all eigenvalues of A are different. This means that there cannot be two different states of the system which are eigenstates of A with the same eigenvalue. So, the vectors |a> and |χ> must represent the same state. However, this does not mean that |a'> and |χ> are the same vectors. Recall that you can multiply a state vector by any nonzero constant and it will still represent the same state of the system. So, although |a'> and |χ> represent the same quantum state, you cannot conclude that |a'> = |χ>. What relation must there be between the vectors |a'> and |χ>?

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TSny
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Or could I say that B|a'> = |χ> then the eigenvalue of |χ> is just 1?
No.

I don't see how you are arriving at "|χ> must also be equal to b'|b'>" or what that really means.

No, A and B will generally have different eigenvalues.

You know that A|a'> = a'|a> and also that A|χ> = a'|χ> (from post #7). So, both the vectors |a'> and |χ> are eigenvectors of A with the same eigenvalue a'. The problem assumes that all eigenvalues of A are different. This means that there cannot be two different eigenstates of the system which have the same eigenvalue. That is, the vectors |a> and |χ> must represent the same state. However, this does not mean that |a'> and |χ> are the same vectors. Recall that you can multiply a state vector by any nonzero constant and it will still represent the same state of the system. So, although you cannot conclude that |a'> = |χ>, what relation must there be between |a'> and |χ>?

|χ> must be a multiple of |a'> since they must represent the same system but they are not necessarily the same vectors?

I'm sorry, in some parts of the response above you use |a'> and sometimes you use |a>. Do you mean |a'> for all of them?

TSny
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|χ> must be a multiple of |a'> since they must represent the same system but they are not necessarily the same vectors?
Yes. |χ> and |a'> represent the same quantum state of the system, but they are not necessarily the same vectors.

I'm sorry, in some parts of the response above you use |a'> and sometimes you use |a>. Do you mean |a'> for all of them?

So, you have come to the conclusion that |χ> must be some (nonzero) multiple of |a'>.
That is, there exists a nonzero number λ such that |χ> = λ|a'>. Now, remember how you defined |χ>.

right, so λ is the eigenvalues of... B in the eigenstate |a'>?

|χ> = B|a'> = λ|a'>

Since λ is the eigenvalues of B then...I have to solve for λ?

TSny
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right, so λ is the eigenvalues of... B in the eigenstate |a'>?

|χ> = B|a'> = λ|a'>
Yes. You have sown that the state vector |a'> must be an eigenvector of B.

Since λ is the eigenvalues of B then...I have to solve for λ?
No, there is no way to deduce the value of λ. But the problem doesn't ask you to do that.

Yes. You have sown that the state vector |a'> must be an eigenvector of B.

No, there is no way to deduce the value of λ. But the problem doesn't ask you to do that.

Right, there's no way to deduce λ as I'm not given B. However, I'm not sure what to do next.

Should I say that Since A|a'> = B|a'> then A and B must share a common eigenbasis? So the eigenstates of A must be the eigenstates of B?

edit: and since they only differ by a constant λ then they must be non-degenerate?

TSny
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Since A|a'> = B|a'>
Does this have to be true? Note that A|a'> = a'|a'>. But B|a'> = λ|a'>. Since there is no reason why λ must equal a', you cannot say that A|a'> = B|a'>.

You are asked to show that any eigenstate |a'> of A is also an eigenstate of B. What does it mean for |a'> to be an eigenstate of B?

Does this have to be true? Note that A|a'> = a'|a'>. But B|a'> = λ|a'>. Since there is no reason why λ must equal a', you cannot say that A|a'> = B|a'>.

You are asked to show that any eigenstate |a'> of A is also an eigenstate of B. What does it mean for |a'> to be an eigenstate of B?

Right... so in this case, I can say that λ = b' since it asks me to label the common eigenstates as |a',b'>

So for an eigenstate of A to be an eigenstate of B, they must have |a'> on the RHS of the equation

Edit: I was confused because I thought it told me to show that

A|a'> = a'|a'> = B|b'> = b'|b'>

but I guess that doesn't really make sense.

PeroK
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Right... so in this case, I can say that λ = b' since it asks me to label the common eigenstates as |a',b'>

So for an eigenstate of A to be an eigenstate of B, they must have |a'> on the RHS of the equation

Edit: I was confused because I thought it told me to show that

A|a'> = a'|a'> = B|b'> = b'|b'>

but I guess that doesn't really make sense.

First, you need to focus on what you need to prove. You are asked to show that an eigenvector of ##A## is also an eigenvector of ##B##. Mathematically, this means that you need to do something like:

Let ##|a\rangle ## be an eigenvector of ##A## ... then (some mathematics) ... ##|a\rangle## is an eigenvector of ##B##.

Introducing an eigenvector of ##B## such as ##|b\rangle ## is not going to help.

The next thing to do is to write down what it means for ##|a\rangle ## to be an eigenvector of ##A##. So, therefore, we can say:

##A|a\rangle = a|a\rangle##, where ##a## is the corresponding eigenvalue.

The next thing is to hit that equation with ##B##, as you did in your OP. So, therefore, we can say that:

##BA|a\rangle = Ba|a\rangle##

Now, can you take it up from there? Hint: what do you know about ##BA##?

(Note that formal proofs have a logical structure, which is a skill to learn. In this case, the key is to use what you know and try to recognise the overall logic structure of your proof must take.)

TSny
TSny
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Right... so in this case, I can say that λ = b' since it asks me to label the common eigenstates as |a',b'>
Yes, I think you now have all the pieces. So, you should try to formulate a proof in a series of steps that form a clear, logical structure. @PeroK has provided some nice suggestions.