# How to find expectation value for combined state?

1. Nov 19, 2016

### anlon

1. The problem statement, all variables and given/known data
Given $\psi = AR_{21}[BY_1^1 + BY_1^{-1} + CY_1^0]$, find $\left<L_z\right>$ and $\left<L^2\right>$. (This is not the beginning of the homework problem, but I know my work is correct up to here. I am not looking for a solution, only an answer as to whether or not my method of finding expectation values is correct, thus why I have generalized the problem with these coefficients instead of using the actual coefficients from the problem.)
Here, $A$, $B$, and $C$ are constants, $Y_l^m$ is the angular wave function for an electron in a hydrogen atom in the state $\left|nlm\right>$, and similarly $R_{nl}$ is the electron's radial wave function.

2. Relevant equations
$L_z f_l^m = \hbar m f_l^m$
$L^2 f_l^m = \hbar^2 l (l+1) f_l^m$
$\left<Q\right> = \left<g\right|Q\left|g\right>$ for the expectation value of an observable $Q$ for a function $g$.

3. The attempt at a solution
Since the wave function is a linear combination of the angular wave functions, and in calculating the angular momentum the radial wave function can be ignored, let $f_l^m$ be the wave function without the radial component: $$f_l^m = A[BY_1^1 + BY_1^{-1} + CY_1^0].$$ The expectation value of $L_z$ would then be $$\left<L_z\right> = \left<f_l^m\right|\hbar m\left|f_l^m\right>$$
The problem I have is that there isn't one particular value of $m$. We instead have three values. However, if we calculate $\left<f_l^m | f_l^m\right>$, then even though the initial math would be horrible (creating nine terms where there previously were only three), because these angular wave functions are mutually orthogonal, any product of two wave functions that are not the same will be 0, meaning we can just square each wave function in place and remove the imaginary components.
Essentially, $$\left<L_z\right> = \left<f_l^m\right|\hbar m\left|f_l^m\right> = \left<ABY_1^1\right|\hbar m\left| ABY_1^1\right> + \left<ABY_1^{-1}\right|\hbar m\left|ABY_1^{-1}\right> + \left<ACY_1^0\right|\hbar m\left|ACY_1^0\right>=A^2 \left[B^2\left<Y_1^1\right|\hbar \left| Y_1^1\right> + B^2\left<Y_1^{-1}\right|-\hbar\left|Y_1^{-1}\right> + C^2\left<Y_1^0\right|0\left|Y_1^0\right>\right] = A^2(0) = 0$$
and $\left<L^2\right>$ would be found in a similar way. Is this an acceptable way to find the expectation value?

2. Nov 19, 2016

### PeroK

There's a much simpler way to get the answer.

What is the significance of these particular wave functions?

3. Nov 19, 2016

### anlon

If you're asking whether or not I see a trend, they all have the same value of $l$, with varying values of $m$. They are also all spherical harmonics.
Does it have to do with the coefficients of $Y_1^1$ and $Y_1^{-1}$ being the same?

4. Nov 19, 2016

### PeroK

Yes, those functions tell you the value of $L$ and $L_z$ that you will get. So, you can do a simple statistical process to get the expected value.

5. Nov 19, 2016

### PeroK

Suppose you have a measurable $Q$ and two normalised eigenstates $\psi_1, \psi_2$. Suppose you know the expected value of $Q$ in each of these states - let's say $q_1, q_2$ respectively.

If you have the state $\psi = A\psi_1 + B\psi_2$, then what is the expected value of $Q$ in that state?

And, what do you get in the case that $q_1 = q_2$?

Last edited: Nov 19, 2016
6. Nov 19, 2016

### anlon

I would think the expected value of $Q$ would be $$\sum_{n} \left|C_n \right|^2 q_n$$ where $C_n$ is the "amount" of the wave function $\psi_n$ contained in the total wave function. In this case, $$\left<Q\right> = \sum_{1}^{2} \left|C_n\right|^2 q_n = \left(\frac{A}{A+B}\right)^2 q_1 + \left(\frac{B}{A+B}\right)^2 q_2$$ and for the case $q_1 = q_2 = q$ you would get $\left<Q\right> = q$.

Edit: Mathematically, I don't see how this works. I'm using Griffiths as a reference, and his textbook confirms my guess that the total expectation value is found with the summation of squared coefficients with expectation values, but using this method you would get $$q\left( \frac{A^2}{(A+B)^2} + \frac{B^2}{(A+B)^2} \right) = q\left( \frac{A^2 + B^2}{(A+B)^2} \right) \neq q$$

Edit 2: My first guess would be not to square the coefficients. This works out much more nicely: $$\left<Q\right> = \sum_{n} C_n q_n$$ which leads to $$\left<Q\right> = \frac{A}{A+B}q_1 + \frac{B}{A+B}q_2$$ which does give $q$ as the expectation value when $q_1 = q_2$. However, I'm not sure how to reconcile that method with Griffiths.

Last edited: Nov 19, 2016
7. Nov 19, 2016

### PeroK

It should be $\langle Q \rangle = |A|^2q_1 + |B|^2q_2$ but the point is that if you know the expected values in each eigenstate, you can compute the expected value for a linear combination of those eigenstates.

Note that we are assuming here that $\psi_1, \psi_2$ are orthogonal eigenstates of $Q$. I forgot to say that, but you assumed it anyway!

8. Nov 19, 2016

### anlon

That makes much more sense, I was overcomplicating the problem by trying to divide the different coefficients. So for $q_1 = q_2 = q$ the expectation value would be $\left<Q\right> = q\left(\left|A\right|^2 + \left|B\right|^2 \right)$?

9. Nov 19, 2016

### PeroK

I was assuming that $\psi$ is normalised, so $|A|^2 + |B|^2 = 1$.

Anyway, you should go back to your original problem now. You know the expected values in each eigenstate and you have already spotted some things about the coefficients, so ...

10. Nov 19, 2016

### anlon

Back to the original problem, $f_l^m = A \left[ B Y_1^1 + B Y_1^{-1} + C Y_1^0 \right]$ so $$\left<L_z\right> = \sum_{m} D_m \hbar m Y_1^m = |AB|^2 \hbar Y_1^1 + |AB|^2(-\hbar) Y_1^{-1} + |AC|^2 (0) Y_1^0 = 0$$
(where $D_m$ is just the group of coefficients associated with each function $Y_l^m$)

11. Nov 19, 2016

### PeroK

Your notation is a bit confusing and you certainly shouldn't have those harmonic functions in the second equation.

$f = A \left[ B Y_1^1 + B Y_1^{-1} + C Y_1^0 \right]$ so $$\left<L_z\right> = \left<f|L_zf\right> = |AB|^2 \hbar + |AB|^2(-\hbar) + |AC|^2 (0) = 0$$

Would make more sense to me.

(It's getting late for me, so I'm going off line now.)

12. Nov 19, 2016

### anlon

I seem to keep making simple mistakes, but thank you for your patience. You've been very helpful, and I believe I can solve the rest of the problem now. Thank you for your help!