Solving Operator Eigenstates: Constants & Normalization

[CalcYouLater]

Homework Statement

I have some questions about the eigenstates of an operator.

A state is an eigenstate of an operator if the application of the operator on the state results in a constant complex multiple of the state. The constant complex multiple will be the eigenvalue.

For instance, given a wave function:

\psi(r,\theta,\phi)=4{\left|\right\psi_1\rangle}+\sqrt{\frac{4}{3\pi}}{\left|\right\psi_2\rangle}

And an operator, A^hat, (with eigenvalue a), then psi is an eigenfunction of A^hat if:

\hat{A}\psi(r,\theta,\phi)=a\psi(r,\theta,\phi)

If the above is a correct understanding, then my question are:

How do the constants in front of the states (psi_1, and psi_2) affect (if at all) the eigenvalue a?

Does the wavefunction need to be properly normalized to determine whether or not it is an eigenstate of the operator?

Would the operator A^hat applied to psi(r,theta,phi) result in:


\hat{A}\psi(r,\theta,\phi)=4\hat{A}{\left|\right\psi_1\rangle}+\sqrt{\frac{4}{3\pi}}\hat{A}{\left|\right\psi_2\rangle}

yielding:

{a}\psi(r,\theta,\phi)=a(4{\left|\right\psi_1\rangle}+\sqrt{\frac{4}{3\pi}}{\left|\right\psi_2\rangle})

Homework Equations

The Attempt at a Solution

I have a specific example in mind involving the angular momentum operator, but I want to see if I understand the concepts of eigenstates, eigenvalues, and eigenfunctions.

The way I see it, the constants in front of the states, psi_1 and psi_2, simply multiply the eigenvalue "a" after the operator A^hat has been applied. If that is true, then it does not matter whether or not the wave function is normalized. However, to determine the probability of obtaining a specific eigenvalue, the wave function must be properly normalized.

Is this correct? If not, what is incorrect?

 

[vela]

Homework Statement

I have some questions about the eigenstates of an operator.

A state is an eigenstate of an operator if the application of the operator on the state results in a constant complex multiple of the state. The constant complex multiple will be the eigenvalue.

For instance, given a wave function:

\psi(r,\theta,\phi)=4{\left|\right\psi_1\rangle}+\sqrt{\frac{4}{3\pi}}{\left|\right\psi_2\rangle}

And an operator, A^hat, (with eigenvalue a), then psi is an eigenfunction of A^hat if:

\hat{A}\psi(r,\theta,\phi)=a\psi(r,\theta,\phi)

If the above is a correct understanding, then my question are:

How do the constants in front of the states (psi_1, and psi_2) affect (if at all) the eigenvalue a?

No. If \vert \psi \rangle is an eigenstate with eigenvalue \lambda, you always know, by definition, that

\hat{A}\vert \psi \rangle = \lambda \vert \psi \rangle

regardless of how \vert \psi \rangle is represented. You can't really say much about what A does to \vert \psi_1 \rangle and \vert \psi_2 \rangle individually, but you do know that A applied to that particular linear combination of the two kets yields a multiple of the combination.

Does the wavefunction need to be properly normalized to determine whether or not it is an eigenstate of the operator?

No. As you may recall from linear algebra, eigenvectors are unique only up to a multiplicative constant.

Would the operator A^hat applied to psi(r,theta,phi) result in:

\hat{A}\psi(r,\theta,\phi)=4\hat{A}{\left|\right\psi_1\rangle}+\sqrt{\frac{4}{3\pi}}\hat{A}{\left|\right\psi_2\rangle}

yielding:

{a}\psi(r,\theta,\phi)=a(4{\left|\right\psi_1\rangle}+\sqrt{\frac{4}{3\pi}}{\left|\right\psi_2\rangle})

Yup. And just to reiterate, you don't have

\begin{align*}<br /> \hat{A}\vert \psi_1 \rangle &amp;= a\vert \psi_1 \rangle \\<br /> \hat{A}\vert \psi_2 \rangle &amp;= a\vert \psi_2 \rangle<br /> \end{align*}

You just know that A applied to that particular combination of the two kets yields a times the ket.

The Attempt at a Solution

I have a specific example in mind involving the angular momentum operator, but I want to see if I understand the concepts of eigenstates, eigenvalues, and eigenfunctions.

The way I see it, the constants in front of the states, psi_1 and psi_2, simply multiply the eigenvalue "a" after the operator A^hat has been applied. If that is true, then it does not matter whether or not the wave function is normalized. However, to determine the probability of obtaining a specific eigenvalue, the wave function must be properly normalized.

Is this correct? If not, what is incorrect?

Yes, you've got it.

On a side note, you've messed up the notation. If you have a ket on one side of an equation, you have to have a ket on the other side, so writing

\psi(r,\theta,\phi) = 4 \vert \psi_1 \rangle + \sqrt{\frac{4}{3\pi}} \vert \psi_2 \rangle

isn't correct. It should be

\vert \psi \rangle = 4 \vert \psi_1 \rangle + \sqrt{\frac{4}{3\pi}} \vert \psi_2 \rangle

If you now apply the bra \langle r, \theta, \phi \vert on the left to that equation, you get

\langle r, \theta, \phi \vert \psi \rangle = 4 \langle r, \theta, \phi \vert \psi_1 \rangle + \sqrt{\frac{4}{3\pi}} \langle r, \theta, \phi \vert \psi_2 \rangle

The products yield wave functions, e.g. \langle r, \theta, \phi \vert \psi \rangle = \psi(r,\theta,\phi), and you end up with

\psi(r,\theta,\phi) = 4 \psi_1(r,\theta,\phi) + \sqrt{\frac{4}{3\pi}}\psi_2(r,\theta,\phi)

 

[CalcYouLater]

Thanks very much vela.

In a related issue, I just came across an expression that I do not understand involving the angular momentum operator Lz. I hope it is a typo in the book. Here it is along with the sentence directly below it:


\hat{L}_{z}\psi(x,y,z)=\frac{1}{\sqrt{2}}\hat{L}_{z}Y_{2,0}+\sqrt{\frac{2}{5}}\hat{L}_{z}(Y_{2,-1}-Y_{2,1})=-\hbar\sqrt{\frac{2}{5}}\hat{L}_{z}(Y_{2,-1}+Y_{2,1})

"This shows that psi(x,y,z) is not an eigenstate of L_z."

This doesn't seem correct to me since:

\hat{L}_{z}Y_{l,m}(\theta,\phi)=m\hbar{Y_{l,m}(\theta,\phi)}

Where Y_l,m are the spherical harmonics.

Should the last term on the right hand side not include the operator L_z? Otherwise, it seems we would get an hbar^2.

If it is correct, how does this show that psi(x,y,z) is not an eigenstate of L_z?

 

[vela]

You're right. That's a typo. The L_z shouldn't be there anymore.

 

[CalcYouLater]

Ok, and the reason it is not an eigenstate is because the first term (m=0) vanishes and the act of the operator does not result in a multiple of the original state.

 

[vela]

Right.

 

[CalcYouLater]

Thanks a lot for all the help.