# Compatible Observables

• jasonchiang97
In summary: Yes I see that the eigenvalues of A and B are different and that A|a> = a'|a> and also that A|χ> = a'|χ>. So I could say that the eigenvalue of |χ> is just a' and that because all eigenvalues must be different, |a'> and |χ> must be different eigenvectors of A and therefore different vectors. So the only relation between |a'> and |χ> is that they are both eigenvectors of A with the same eigenvalue a'?Yes, that's the key point. So, what can you say about B|a'>?B|a'> is not necessarily the same as |a'> as

## 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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jasonchiang97 said:
AB|a'> = BA|a'> = a'|a'>
What can you conclude about the state B|a'> from this result?

TSny said:
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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jasonchiang97 said:
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
jasonchiang97 said:
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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TSny said:
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'>

jasonchiang97 said:
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'>?

jasonchiang97 said:
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'?

TSny said:
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

jasonchiang97 said:
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.

jasonchiang97 said:
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.

TSny said:
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.

jasonchiang97 said:
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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jasonchiang97 said:
Or could I say that B|a'> = |χ> then the eigenvalue of |χ> is just 1?
No.

TSny said:
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?

jasonchiang97 said:
|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?

jasonchiang97 said:
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.

TSny said:
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?

jasonchiang97 said:
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?

TSny said:
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.

jasonchiang97 said:
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
jasonchiang97 said:
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.

## What are "Compatible Observables"?

"Compatible Observables" refer to two or more physical quantities that can be measured simultaneously without affecting each other. In other words, compatible observables can be measured together without altering the outcome of the measurements.

## Why is it important to know if observables are compatible?

It is important to know if observables are compatible because it allows us to make accurate and precise measurements in experiments. If observables are not compatible, then measuring one observable could affect the outcome of measuring another, leading to inaccurate results.

## What types of observables are typically considered compatible?

Observable quantities that are considered compatible are usually quantities that do not share a common physical property. For example, position and momentum are considered compatible observables, while position and velocity are not.

## How can you determine if observables are compatible?

The commutator of two observables is used to determine if they are compatible. If the commutator of two observables is equal to zero, then they are considered compatible. If the commutator is non-zero, then the observables are not compatible.

## Can incompatible observables ever be measured together?

No, incompatible observables cannot be measured simultaneously without affecting each other. This is a fundamental principle of quantum mechanics known as the Heisenberg uncertainty principle. It states that the more precisely one observable is measured, the less precisely the other observable can be measured.