0
$\begingroup$

I was given the following function: $$\Psi(t=0)=\frac{1}{\sqrt{2}}\left[(Y_{1,1})+i(Y_{1,-1})\right],$$ where $Y_{lm}$ refers to the standard spherical harmonic function. I am trying to come up with the expectation values for $L^2$, $L_z$ and $L_x$, but I am running into trouble.

$$\langle L^2\rangle=\left(\frac{1}{\sqrt{2}}\right)^2\hbar l(l+1)+\left(i\frac{1}{\sqrt{2}}\right)^2\hbar l(l+1)=0$$

$$\langle L_z\rangle=\left(\frac{1}{\sqrt{2}}\right)^2\hbar m+\left(i\frac{1}{\sqrt{2}}\right)^2\hbar (-m)=\hbar m$$

I am not sure how to compute $L_x$, but the above result already looks impossible, in light of: $$L^2=L_x^2+L_y^2+L_z^2.$$

I would like to know where the mistake in my reasoning is. Thank you!

$\endgroup$
7
  • 1
    $\begingroup$ Haven't you come across the operators $L_+$ and $L_-$ in your studies? You should be able to use that information to find $L_x$ so you can then compute it's expectation value. $\endgroup$
    – Triatticus
    Dec 7, 2021 at 8:36
  • $\begingroup$ I hadn't yet: we just started, and it's my first Quantum course. I am reading up on it now; but either way it is not inside (this chapter) of the course material. Thanks for the help! $\endgroup$
    – dalta
    Dec 7, 2021 at 9:32
  • 2
    $\begingroup$ note that when taking the exepectation value, you have to conjugate the bra-vector. As a general rule-of-thumb, the expectation value of a Hermitian operator is always real (though it might have matrix elements that are complex) $\endgroup$
    – user275556
    Dec 7, 2021 at 10:10
  • $\begingroup$ Thank you, yyy! So what you are you saying - in practice in this case - is that I should replace i with 1 in the above calculations? If so, that is really helpful! $\endgroup$
    – dalta
    Dec 7, 2021 at 10:20
  • 1
    $\begingroup$ replacing $i$ by $1$ will get you in trouble. what you want is not $(1/\sqrt{2})^2$ or $(i/\sqrt{2})^2)$ but $\vert i/\sqrt{2}\vert^2$ etc. $\endgroup$ Dec 7, 2021 at 14:07

2 Answers 2

3
$\begingroup$

Representing the states as $|l,m\rangle $, We know that $$L^2|l,m\rangle =l(l+1)\hbar^2|l,m\rangle $$ $$L_z|l,m\rangle=m\hbar |l,m\rangle $$ The state given by $$|\psi\rangle =\frac{1}{\sqrt{2}}(|1,1\rangle +i|1,-1\rangle) $$ Now, it's easy to find $$\langle L^2\rangle =\langle \psi|L^2|\psi\rangle, \ \ \ \ \ \langle L_z\rangle =\langle \psi|L_z|\psi\rangle $$ For $L_x$, We can use $$L_\pm =L_x\pm iL_y\rightarrow L_x=\frac{L_++L_-}{2}$$ Further, We know $$L_\pm |l,m\rangle =\hbar\sqrt{l(l+1)-m(m\pm 1)}|l,m\pm 1\rangle $$ Further, Note the orthonormality relation:- $$\langle l',m'|l,m\rangle =\delta_{l,l'}\delta_{m,m'}$$ This should suffice!!

$\endgroup$
2
$\begingroup$

When calculating expectation values of diagonal operators (like $L^2$ and $L_z$), you need to take the absolute square $|c|^2$ or $(c^*c)$ of the coefficients of $|\Psi\rangle$, not their square $(c)^2$. Then you get the correct values $$\begin{align} \langle L^2\rangle &=\langle\Psi|L^2|\Psi\rangle \\ &=\left|\frac{1}{\sqrt{2}}\right|^2\hbar^2 l(l+1)+\left|i\frac{1}{\sqrt{2}}\right|^2\hbar^2 l(l+1) \\ &=\frac{1}{2}\hbar^2 l(l+1) + \frac{1}{2}\hbar^2 l(l+1) \\ &=\hbar^2 l(l+1) \end{align}$$ and $$\begin{align} \langle L_z\rangle &=\langle\Psi|L_z|\Psi\rangle \\ &=\left|\frac{1}{\sqrt{2}}\right|^2\hbar m+\left|i\frac{1}{\sqrt{2}}\right|^2\hbar (-m) \\ &=\frac{1}{2}\hbar m - \frac{1}{2}\hbar m \\ &= 0 \end{align}$$

And likewise for calculating $\langle L_x\rangle$ and $\langle L_y\rangle$ be aware, that $\langle\Psi|$ will involve the complex conjugates.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.