What Are the Eigenvalues of a Hermitian Operator if \(\hat{A}^2 = 2\)?

[kcirick]

Hi again,

Question: \hat{A} is an Hermitian Operator. If \hat{A}^{2}=2, find the eigenvalues of \hat{A}

So We have:
\hat{A}\left|\Psi\right\rangle=a\left|\Psi\right\rangle

But I actually don't know how to even begin. \hat{A} is a general Hermitian operator, and I don't know where even \hat{A}^{2} would fit in with the question asked.

Any help is appreciated! Thank you!
-Rick

 

[AKG]

A²v = A(Av).

 

[kcirick]

Umm...

\hat{A}\left(\hat{A}\left|\Psi\right\rangle\right) = a\left(a\left|\Psi\right\rangle\right)

\hat{A}^{2}\left|\Psi\right\rangle = a^{2}\left|\Psi\right\rangle

2\left|\Psi\right\rangle = a^{2}\left|\Psi\right\rangle

a = \sqrt{2} ?

I'm not sure...

 

[AKG]

How did you get the first equation? What's |\Psi \rangle? What's a? How do you get a = \sqrt{2}, and not, say, -\sqrt{2}? I mean, you have the right idea, but you haven't put that right idea into the form of a proper proof.

 

[kcirick]

Well, I'm using Dirac bracket notation to be consistent with the rest of my work, and ket space is just a real space, so it's same thing as \

\hat{A}^{2}\Psi = a^{2}\Psi

where a is the eigenvalue of the Hermitian operator.


You're right, a can be negative too, but is the final answer really a=\pm\sqrt{2}? It seems too simple to be true.

 

[kcirick]

Well, I'm using Dirac bracket notation to be consistent with the rest of my work, and ket space is just a real space, so it's same thing as \

\hat{A}^{2}\Psi = a^{2}\Psi

where a is the eigenvalue of the Hermitian operator.


You're right, a can be negative too, but is the final answer really a=\pm\sqrt{2}? It seems too simple to be true.

 

[AKG]

I understand the notation, but you just introduced a and |\Psi \rangle without saying a thing about them. Okay, so a is an eigenvalue of \hat{A} and |\Psi \rangle is a (non-zero) eigenvector corresponding to a.

2|\Psi \rangle = \hat{A}^2|\Psi \rangle = \hat{A}(\hat{A}|\Psi \rangle ) = \hat{A}(a|\Psi \rangle ) = a(\hat{A}|\Psi \rangle) = a(a|\Psi \rangle) = a^2|\Psi \rangle

Therefore a = \pm \sqrt{2}. Note there is some indeterminacy. \hat{A} could have all it's eigenvalues positive, or all negative, or some positive and some negative.