# Eigenvalues of Hermitian operators

1. Oct 16, 2006

### 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

2. Oct 16, 2006

### AKG

A2v = A(Av).

3. Oct 16, 2006

### 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...

4. Oct 16, 2006

### 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.

5. Oct 16, 2006

### 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.

6. Oct 16, 2006

### 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.

7. Oct 16, 2006

### 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.