# Hermitian operator

ehrenfest

## Homework Statement

Since the momentum operator is Hermitian why is this wrong:

<psi| (p-hat)^2 |psi> = <psi| p-hat p-hat |psi> = <p-hat psi| p-hat |psi> = (p)^2 where p is the expectation value of the momentum.

## The Attempt at a Solution

Homework Helper
Gold Member
<p-hat psi| p-hat |psi> this is not a allowed thing.

p-operator on a state gives you the state back + its eigenvalue, so you dont change the state.

since psi is a general state here, we dont know if psi is a eigenstate to p-hat, so one must insert some completness relations to take care of the the operator. Now there are many ways to do this, start with the simplets case and see if you can make progress.

ehrenfest
<p-hat psi| p-hat |psi> this is not a allowed thing.

p-operator on a state gives you the state back + its eigenvalue, so you dont change the state.

since psi is a general state here, we dont know if psi is a eigenstate to p-hat
I see. But when psi is an eigenstate of p-hat, then my equations hold, right?

Homework Helper
Gold Member
yes, but only then. And then you usally label them as: |p>

Do you use any perticular book in this course?

ehrenfest
Robinett (2nd Edition)

I think this quote from my book is downright wrong then:

$$<Y_{l,m}|(L_{x}^2+L_{y}^2+L_{z}^2)|Y_{l,m}> = <L_x Y_{l,m}|L_x Y_{l,m}> + <L_y Y_{l,m}|L_y Y_{l,m}> + <L_z Y_{l,m}|L_z Y_{l,m}>$$

where the L_i is the ith component of the angular momentum operator. The angular momenta components definitely do not commute so we know that Y_{l,m} is an eigenfunction of L_{z} only.

This is equation 16.49 on Robinett page 456.

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Homework Helper
Gold Member
I think the notation is very bad, <| is a state! You never take in the operator inside the bracket.

Then Y_lm is not the state!

$$Y_{lm} = <\vec{n}| lm>$$

Y_lm is a wavefunction in direction space (n is direction vector). It is a difference of state and wavefunction.

I have never seen this strange notation anyware in scientific litterature, it is very confusing.

Rather write:
$$<l,m| L_x^2 |l,m> = <l,m| L_xL_x |l,m>$$
Now L_x is not hermitian, so it MUST operate to the right, on the ket (if you operate with a non hermitian operator to the left, you get the comlex conjugate of its eigenvalue, see dual correspondence). So the book of yours do a fatal misstake there also. L_z is hermitian so it can operate both to left and to the right. In the case of L_x, you can rewrite it as a superposition of the ladder operators and see for yourself.

ehrenfest
Now L_x is not hermitian, so it MUST operate to the right, on the ket (if you operate with a non hermitian operator to the left, you get the comlex conjugate of its eigenvalue, see dual correspondence).

But L_x is Hermitian. It corresponds to an observable.

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Homework Helper
Gold Member
oh yes, sorry, my mistake :) Was confusing it with the ladder operators.

But my point was that the notation your book uses is quite confusing.

ehrenfest
But the equation in post #5 is still wrong, isn't it?

ehrenfest
Actually, my book does justify that:

$$<\psi_1|O|\psi_2> = <O ^\dagger \psi_1|\psi_2>$$

So, that leaves me still confused since isn't this true:

$$<(p-hat)^2> = <\psi|(p-hat)^2|\psi> = <(p-hat) \psi|(p-hat)\psi> = <p-hat>^2$$

which one of those equalities is wrong? Is it the last one?

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Staff Emeritus
Gold Member
Robinett's equation is fine,

Actually, my book does justify that:

$$<\psi_1|O|\psi_2> = <O ^\dagger \psi_1|\psi_2>$$

So, that leaves me still confused since isn't this true:

$$<(p-hat)^2> = <\psi|(p-hat)^2|\psi> = <(p-hat) \psi|(p-hat)\psi> = <p-hat>^2$$

which one of those equalities is wrong? Is it the last one?

The last one.

ehrenfest
I think I figured out what is wrong my thinking. It should be:

$$<\hat{p}^2> = <\psi|\hat{p}^2|\psi> = <\hat{p} \psi|\hat{p}\psi>$$

which is equal to
= p^2

only when psi is an eigenstate of of the momentum operator.

Staff Emeritus
Gold Member
Common notation:

$$\left< \psi |A| \phi \right> = \left< \psi |A \phi \right> = \left<A^\dagger \psi | \phi \right> .$$

ehrenfest
But was my statement in post #12 right?

Staff Emeritus
Gold Member
But was my statement in post #12 right?

Yes, if $\hat{p} \left| \psi \right> = p \left| \psi \right>$, then

$$\left< \psi | \hat{p}^2 | \psi \right> = p \left< \psi | \hat{p} | \psi \right> = p^2 \left< \psi | \psi \right> = p^2.$$

Homework Helper
Gold Member
Well yes it is full leagal to take in the operator in the ket, etc. But as I have learned, it is not a clear notation. I have seldom seen it, but George is more experienced than me, so :)

Staff Emeritus
Gold Member
Well yes it is full leagal to take in the operator in the ket, etc. But as I have learned, it is not a clear notation. I have seldom seen it, but George is more experienced than me, so :)

Well, what you find confusing, I might find clear, and what I find clear, you might find confusing. A third person might have completely different ideas than either of us on what's confusing and what's clear. We are all individuals, and we all look at things in different ways. This is part of what makes life interesting! The important thing is to have enough common ground that communication is possible.

For example, I find Dirac notation's handling of general operators and adjoints to be confusing. I always have to translate this into the notation I learned in math class, fool around bit until I (hopefully!) get the answer, and then translate back into Dirac notation. On the other hand, I find that Dirac notation deals with projection operators and insertions of complete sets of state in a manner that is much more transparent than standard math notation.

Cheers!

ehrenfest
Robinett (2nd Edition)

I think this quote from my book is downright wrong then:

$$<Y_{l,m}|(L_{x}^2+L_{y}^2+L_{z}^2)|Y_{l,m}> = <L_x Y_{l,m}|L_x Y_{l,m}> + <L_y Y_{l,m}|L_y Y_{l,m}> + <L_z Y_{l,m}|L_z Y_{l,m}>$$

where the L_i is the ith component of the angular momentum operator. The angular momenta components definitely do not commute so we know that Y_{l,m} is an eigenfunction of L_{z} only.

This is equation 16.49 on Robinett page 456.

OK. I now think that it is me that was wrong in writing this post, not my QM book. Can someone verify that? I mean verify that $$<\psi O^{\dagger}|\chi> = <\psi|O\chi>$$ is true regardless of whether psi and chi are eigenfunctions of O.

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