# Problem about spherical angle operators

1. Aug 11, 2011

### issacnewton

Hi

Here's the problem I am trying to do.

a) Is the state $\psi (\theta ,\phi)=e^{-3\imath \;\phi} \cos \theta$

an eigenfunction of $\hat{A_{\phi}}=\partial / \partial \phi$ or of
$\hat{B_{\theta}}=\partial / \partial \theta$ ?

b) Are $\hat{A_{\phi}} \;\mbox{and} \;\hat{B_{\theta}}$ hermitian ?

c)Evaluate the expressions $\langle \psi \vert \hat{A_{\phi}} \vert \psi \rangle$
and $\langle \psi \vert \hat{B_{\theta}} \vert \psi \rangle$

Now $\hat{A_{\phi}}$ has imaginary eigenvalues , so its not hermitian.
I could show that $\psi$ is an eigenfunction of square of $\hat{B_{\theta}}$. I have been able to show that the commutator of A and B is zero.
So with this information, how do I check the hermiticity of B ?

for part c) , since there are two state variables , I am little confused about how to go
about it ? any guidance will be appreciated... thanks

2. Aug 11, 2011

### kuruman

Do you know how to represent the braket as an integral? If so, then to prove hermiticity you need to show that <ψ|Βψ> = <ψΒ|ψ>, i.e. that operator B gives the same result when it operates either to the left or to the right.

3. Aug 11, 2011

### issacnewton

but this wavefunction has two variables, so how do I evaluate these amplitudes ?

4. Aug 11, 2011

### kuruman

You do a double integral over two-dimensional spherical space.

5. Aug 11, 2011

### mathfeel

Commutator is irrelevant here. What is an eigenfunction of B? What is the corresponding eigenvalue? Is it real?
What Kuruman said.

6. Aug 11, 2011

### issacnewton

$$\langle \psi \vert \hat{B_{\theta}} \vert \psi \rangle = \int \langle \psi \vert \theta ,\phi \rangle \langle \theta , \phi \vert \hat{B_{\theta}} \vert \psi \rangle d\theta \;d\phi$$

is this the correct way ? so what next ? these kets in $\theta$ and $\phi$ , sre those eigenfunctions of B ?

7. Aug 11, 2011

### vela

Staff Emeritus
That's almost right. You need to throw a factor of sin θ in there because the infinitesimal element of solid angle is $d\Omega = \sin\theta\,d\theta\,d\phi$.

8. Aug 12, 2011

### issacnewton

ok vela

so that would be

$$\langle \psi \vert \hat{B_{\theta}} \vert \psi \rangle = \int \langle \psi \vert \theta ,\phi \rangle \langle \theta , \phi \vert \hat{B_{\theta}} \vert \psi \rangle \sin \theta \;d\theta \;d\phi$$

so now what ? how do I evaluate this ?

9. Aug 12, 2011

### kuruman

I guess you don't know how to represent a braket as an integral.

$<\psi|B|\psi>=\int \psi^*(B\psi) sin \theta \: d \theta \: d\phi$

You operate on the wavefunction with B to get a new function then integrate as indicated.

10. Aug 12, 2011

### issacnewton

Hi kuruman

to go from my equation to your equation , we need to assume the following

$$\langle \theta , \phi \vert \hat{B_{\theta}} \vert \psi \rangle = \hat{B_{\theta}} \langle \theta , \phi \vert \psi \rangle$$

am i right ? if so , is the above equality always true , for all operators ?

Last edited: Aug 12, 2011
11. Aug 12, 2011

### kuruman

You really need to beef up your understanding of Dirac bra-ket notation. When we write
B|ψ> we mean "operate on function ψ with operator B", i.e. find a new function
$f(\theta,\phi)=\frac{\partial}{\partial \theta}\psi(\theta,\phi)$

When we write <ψ|Β|ψ>,
<ψ| stands for ψ*(θ,φ)
Β|ψ> stands for f(θ,φ)
and closing the braket stands for "do the integral" $\int \psi^*(\theta,\phi)f(\theta,\phi) \: sin\theta \:d\theta\:d\phi$

12. Aug 12, 2011

### vela

Staff Emeritus
Or to put it a little more formally,$\langle \theta,\phi \vert \psi \rangle$ is the representation of $\vert \psi \rangle$ in the $\vert \theta,\phi \rangle$ basis. That is, $\langle \theta,\phi \vert \psi \rangle = \psi(\theta,\phi)$. Similarly, you have $$\langle \theta,\phi \lvert \hat{B}_\theta \rvert \psi \rangle = \frac{\partial}{\partial\theta} \psi(\theta,\phi)$$
Putting it all together, you get$$\langle \psi \vert \hat{B_{\theta}} \vert \psi \rangle = \int \langle \psi \vert \theta ,\phi \rangle \langle \theta , \phi \vert \hat{B_{\theta}} \vert \psi \rangle \sin \theta \,d\theta \,d\phi = \int \psi^*(\theta ,\phi) \left[\frac{\partial}{\partial\theta} \psi(\theta,\phi)\right] \sin \theta \,d\theta \,d\phi$$

13. Aug 12, 2011

### issacnewton

kuruman , i was looking for more formal explanation. i am just using completeness property.

which is what i was saying exactly in post # 10 .

$$\langle \theta , \phi \vert \hat{B_{\theta}} \vert \psi \rangle = \hat{B_{\theta}} \langle \theta , \phi \vert \psi \rangle$$

is it not ?

i remember studying somewhere about the derivation of such relationship. i will give an example about the momentum operator $\hat{P}$

$$\langle x \vert \hat{P} \vert \psi \rangle =\int \langle x \vert \hat{P} \vert p \rangle \langle p \vert \psi \rangle dp$$

$$= \int p \langle x \vert p \rangle \langle p \vert \psi \rangle dp$$

now we know

$$\langle x \vert p \rangle = \frac{1}{(2\pi \hbar)^{1/2}}\; e^{ipx/ \hbar}$$

putting this in the integral ,

$$= \int p \;\; \frac{1}{(2\pi \hbar)^{1/2}}\; e^{ipx/ \hbar} \langle p \vert \psi \rangle dp$$

$$= \frac{\partial}{\partial x}\int \frac{\hbar}{i} \frac{1}{(2\pi \hbar)^{1/2}}\; e^{ipx/ \hbar} \langle p \vert \psi \rangle dp$$

$$= \frac{\hbar}{i} \frac{\partial}{\partial x} \int \langle x \vert p \rangle \langle p \vert \psi \rangle dp$$

$$= \frac{\hbar}{i} \frac{\partial}{\partial x} \langle x \vert \psi \rangle$$

this is what i was talking about. is there any such process by which we can show that ,

$$\langle \theta , \phi \vert \hat{B_{\theta}} \vert \psi \rangle = \hat{B_{\theta}} \langle \theta , \phi \vert \psi \rangle$$

thanks

14. Aug 12, 2011

### vela

Staff Emeritus
To be a bit pedantic, sort of, but not exactly. I have $\partial/\partial\theta$ on the right-hand side, not $\hat{B}_\theta$. Consider $\hat{B}_\theta$ as some abstract operator while $\partial/\partial\theta$ is its representation in the $\vert\theta,\phi\rangle$ basis. So you have$$\langle \theta , \phi \vert \hat{B}_\theta \vert \psi \rangle = \frac{\partial}{\partial \theta} \langle \theta , \phi \vert \psi \rangle$$
Though it's common, I see writing $\hat{B}_\theta = \partial/\partial\theta$ as kind of sloppy if you want to stick with the formalism of inserting complete sets, etc. It makes sense to apply $\partial/\partial\theta$ to the wave function of $\theta$, but not to an abstract ket.

It's just like how $\hat{p}$ is the momentum operator in a general sense while $p$ and $(\hbar/i) \partial/\partial x$ are its representations in, respectively, the momentum and position bases. Say you have some ket $\vert \psi \rangle$. Then you have \begin{align*}
\langle x \lvert \hat{p} \rvert \psi \rangle &= \frac{\hbar}{i}\frac{\partial}{\partial x}\langle x \vert \psi \rangle = \frac{\hbar}{i}\frac{\partial}{\partial x} \psi(x) \\
\langle p \lvert \hat{p} \rvert \psi \rangle &= p\langle p \vert \psi \rangle = p\psi(p)
\end{align*}
It wouldn't make sense to use $(\hbar/i) \partial/\partial x$ when you're working in the momentum basis and vice-versa. Once you choose a basis, you know how to represent the operator and you have also chosen which wave function to use.

So getting back to your original question, I think saying $\hat{B}_\theta = \partial/\partial\theta$ pretty much tells you that you're working in a particular basis and that
$$\langle \theta , \phi \vert \hat{B}_\theta \vert \psi \rangle = \frac{\partial}{\partial \theta} \langle \theta , \phi \vert \psi \rangle$$
holds.

15. Aug 12, 2011

### vela

Staff Emeritus
You could also look at this way:
$$\langle \theta , \phi \vert \hat{B}_\theta \vert \psi \rangle = \langle \theta , \phi \vert \hat{B}_\theta \psi \rangle = \frac{\partial}{\partial \theta} \psi(\theta,\phi)$$

16. Aug 13, 2011

### issacnewton

so vela ,

is this the definition of the the way operator acts on the ket ?

so when an operator acts on the ket $\vert \psi \rangle$ one way to evaluate it
is to take the inner product of this ket with some bra $\langle x \vert$ and then
use the representation of the operator in the basis of this bra. i am using the book,
"Quantum Mechanics: Concepts and Applications" by Nouredine Zettili and he uses
the operators in a sloppy way. Can you suggest any book or online resource where the formalism of quantum mechanics is presented the way it should be.

17. Aug 13, 2011

### vela

Staff Emeritus
I wouldn't say it's a definition, but yeah, what you're describing is essentially what it boils down to. When you multiply a ket by $\langle x \vert$ on the left, you're finding the representation of the ket in that basis. This is what I was getting at in post #15. When you act on $\vert \psi \rangle$ with $\hat{B}_\theta$, you get the state $\vert \hat{B}_\theta \psi \rangle$. Then when you multiply by $\langle \theta, \phi \vert$, you're choosing a basis. In that basis, the state is represented by the derivative with respect to θ of the $\psi(\theta,\phi)$.

You could also say
\begin{align*}
\hat{A} \vert \psi \rangle &= \int dx \vert x \rangle \langle x \vert \hat{A} \vert \psi \rangle \\
&= \int dx \vert x \rangle \int dx'\langle x \vert \hat{A} \vert x'\rangle\langle x' \vert \psi \rangle
\end{align*}
which you can interpret as the representation of operator A in the x basis multiplied by the representation of the state ψ in the same basis.

In any case, don't get too caught up in the mathematical formalism. What kuruman said back in post #11 is what you need in a practical sense.

Last edited: Aug 13, 2011
18. Aug 14, 2011

### issacnewton

wikipedia article

talks about the "abuse of notation" in case of operators