Orthogonality in QM

  • #1
referframe
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Main Question or Discussion Point

In non-relativistic QM, given a Hilbert Space with a Hermitian operator A and a generic wave function
Ψ. The operator A has an orthogonal eigenbasis, {ai}.

I have often read that the orthogonality of such eigenfunctions is an indication of the separateness or distinctiveness of the associated eigenvalues, i.e. that orthogonality in QM means separate and independent.

But what if the probability distribution for Ψ is peaked at one value causing most of the eigenvalues to be clustered in a very narrow range? How could one then say, from a practical point-of-view, that the eigenvalues are separate or distinct, even though the eigenfunctions themselves remain orthogonal?

Thanks in advance
 
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  • #3
jfizzix
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...what if the probability distribution for Ψ is peaked at one value causing most of the eigenvalues to be clustered in a very narrow range? How could one then say, from a practical point-of-view, that the eigenvalues are separate or distinct, even though the eigenfunctions themselves remain orthogonal?

Thanks in advance
The eigenvalues of an observable define the set of outcomes that such a function can have. For example, the position observable [itex]\hat{x}[/itex] has a continuous spectrum of eigenvalues spanning all the real numbers.

The eigenvalues and eigenstates of an observable depend on the observable, and are completely indpendent of the quantum state [itex]|\psi\rangle[/itex] or the quantum wavefunction [itex]\langle x|\psi\rangle = \psi(x)[/itex].

The eigenstates associated to different eigenvalues of an observable are orthogonal. What this means from a practical standpoint is that if a particle is prepared in an eigenstate [itex]|a_{i}\rangle[/itex] of observable [itex]\hat{A}=\sum_{i}a_{i} |a_{i}\rangle\langle a_{i}|[/itex], the probability that measuring [itex]\hat{A}[/itex] will give any other eigenvalue is zero, since the inner product [itex]\langle a_{i}|a_{j}\rangle = 0[/itex] if [itex]i\neq j[/itex], and the probability goes as the magnitude square of this inner product.
 
  • #4
PeroK
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I have often read that the orthogonality of such eigenfunctions is an indication of the separateness or distinctiveness of the associated eigenvalues, i.e. that orthogonality in QM means separate and independent.
I have never read that.

But what if the probability distribution for Ψ is peaked at one value causing most of the eigenvalues to be clustered in a very narrow range?
The eigenvalues and eigenvectors of an operator have nothing to do with the particular state of a system - or any particular vector - they are properties of the operator itself.
 
  • #5
stevendaryl
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In non-relativistic QM, given a Hilbert Space with a Hermitian operator A and a generic wave function
Ψ. The operator A has an orthogonal eigenbasis, {ai}.

I have often read that the orthogonality of such eigenfunctions is an indication of the separateness or distinctiveness of the associated eigenvalues, i.e. that orthogonality in QM means separate and independent.
For Hermitian operators, eigenfunctions for distinct eigenvalues have to be orthogonal.

Easy proof:Suppose ##A|\psi_1\rangle = \lambda_1 |\psi_1\rangle## and ##A|\psi_2\rangle = \lambda_2|\psi_2\rangle##. Then on the one hand,

##\langle \psi_2|A|\psi_1 \rangle = \langle \psi_2|(A|\psi_1\rangle) = \lambda_1 \langle \psi_2|\psi_1 \rangle##

On the other hand,

##\langle \psi_2|A|\psi_1 \rangle = (\langle \psi_2|A)|\psi_1\rangle = (A |\psi_2\rangle)^\dagger |\psi_1\rangle = (\lambda_2 |\psi_2\rangle)^\dagger |\psi_1\rangle = \lambda_2 \langle \psi_2|\psi_1\rangle##

So those two have to be equal: ##\lambda_1 \langle \psi_2|\psi_1\rangle = \lambda_2 \langle \psi_2|\psi_1\rangle##. That's only possible if ##\lambda_1 = \lambda_2## or ##\langle \psi_2|\psi_1\rangle = 0##
 
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  • #6
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Suppose A|ψ1⟩=λ1A|ψ1⟩=λ1A|\psi_1\rangle = \lambda_1 and A|ψ2⟩=λ2A|ψ2⟩=λ2A|\psi_2\rangle = \lambda_2.
I think this was meant to read ##A|\psi_1\rangle=\lambda_1 |\psi_1\rangle## and ##A|\psi_2\rangle=\lambda_2 |\psi_2\rangle## (may be a problem with my viewer).

Translating to plain English, I concur that orthogonality of states with different eigenvalues for the same Hermitian operators does mean "separate and distinctive"; if the state is ##|\psi_1\rangle##, then the state is not ##|\psi_2\rangle##; if the eigenvalue is ##\lambda_1##, then it is not ##\lambda_2##.

This crucially depends on the operator being Hermitian. A famous non-Hermitian operator is the annihilation operator a operating on a coherent state of the harmonic oscillator (or formally equivalent the field operator a operating on coherent states of the electromagnetic field). Coherent states are the closest approach to classical analog of these systems and satisfy the eigenvalue equation ##a|\alpha\rangle=\alpha|\alpha\rangle##. But a is not Hermitian, the eigenvalues ##\alpha## are complex rather than real, and the states ##|\alpha\rangle## are not orthogonal - rather they form an over-complete basis. A good article can be found here https://en.wikipedia.org/wiki/Coherent_states.
 
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  • #8
referframe
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Can you give some specific references?
I listed 3 references in my reply to PeroK above.
 
  • #9
PeterDonis
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I listed 3 references in my reply to PeroK below.
Yes, I see them, but I don't see how they support what you said in the OP. The references say that eigenvectors are mutually orthogonal. They don't say the eigenvalues are "separate and distinct" in the way you appear to be using that concept in the OP--for example, they don't say the eigenvalues have to have any minimum "spacing" (difference) between them. In fact, your third reference specifically talks about the case of degeneracy, where multiple orthogonal eigenvectors have the same eigenvalue.

So it looks to me like your OP is based on a misconception.
 
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  • #10
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It is a misconception to assume that in general orthogonal eigenstates of an observable have distinct eigenvalues. Frequently they do but there may be degeneracies. Even the one-dimensional free particle has a degenerate energy spectrum, since eigenstates of momentum ##\hat p## with respective eigenvalues p and -p have the same energy p2/2m, but are nevertheless orthogonal eigenstates of the energy observable ##\hat p^2/2m##. Here, however, the spectrum of ##\hat p## is non-degenerate. Due to the density of the real number line, momentum eigenstates ##|p_1\rangle## and ##|p_2\rangle## can have eigenvalues p1 and p2 that are arbitrarily close together, but as long as p1p2, the eigenstates are orthogonal, i.e., ##\langle p_1|p_2\rangle = 0##.
 

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