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!

  • 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
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    $\begingroup$ 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


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!!


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.


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