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ehrenfest
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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.
I see. But when psi is an eigenstate of p-hat, then my equations hold, right?malawi_glenn said:<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 don't change the state.
since psi is a general state here, we don't know if psi is a eigenstate to p-hat
malawi_glenn said: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).
ehrenfest said:Actually, my book does justify that:
[tex] <\psi_1|O|\psi_2> = <O ^\dagger \psi_1|\psi_2> [/tex]
So, that leaves me still confused since isn't this true:
[tex] <(p-hat)^2> = <\psi|(p-hat)^2|\psi> = <(p-hat) \psi|(p-hat)\psi> = <p-hat>^2 [/tex]
which one of those equalities is wrong? Is it the last one?
ehrenfest said:But was my statement in post #12 right?
malawi_glenn said: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 :)
ehrenfest said:Robinett (2nd Edition)
I think this quote from my book is downright wrong then:
[tex] <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}>[/tex]
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.
A Hermitian operator is a mathematical term used in quantum mechanics to describe an operator that is equal to its own adjoint. In other words, the operator remains unchanged when its values are complex conjugated.
(p-hat)^2 is the square of the momentum operator, while (p)^2 is the square of the momentum variable. This may seem like a small difference, but it has significant implications in quantum mechanics.
The momentum operator (p-hat) is a Hermitian operator, meaning it is equal to its adjoint. However, when we square the momentum operator, it is no longer equal to its adjoint, making it a non-Hermitian operator. This is why (p-hat)^2 is not equal to (p)^2.
The fact that (p-hat)^2 is not equal to (p)^2 has significant implications in quantum mechanics. It means that the square of the momentum operator does not have all the properties of a Hermitian operator, which can affect the outcomes of calculations and experiments.
One example is the calculation of the average momentum. When using the momentum operator (p-hat), we will get a different result than when using the momentum variable (p). This is because (p-hat)^2 and (p)^2 have different eigenvalues, leading to different outcomes for the average momentum.