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While dealing with a circling particle in an spherical symetric potential our professor said that we can replace an operator of $z$ component of angular momentum $\hat{L}_z$ with the expectation value - he denoted it just $L_z$ - of the angular momentum if $L_z$ is constant. Why is that?

So we first had this equation:

\begin{align} \underbrace{\psi (r,\varphi,\vartheta)}_{\llap{ \text{wave function in spherical coordinates}}} &= \exp\left[\hat{L}_z \frac{i}{\hbar}\, \varphi\right] \underbrace{\psi (r,0,\vartheta)}_{\rlap{\text{wave function in spherical coordinates at $\varphi=0$}}} \end{align}

and we got this one (notice that there is no operator over an $L_z$):

\begin{align} \psi (r,\varphi,\vartheta) &= \exp\left[L_z \frac{i}{\hbar}\, \varphi\right] \psi (r,0,\vartheta) \end{align}

Anyway here is the spherical coordinate system we ve been using all the time (the blue spherical aure is supposed to be a spherical potential...):

enter image description here

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4  
You can do it in e.g. $\langle \psi|\hat L_z|\psi\rangle$ if $|\psi\rangle$ is an eigenstate of $\hat L_z$, but you can not do it in e.g. $\langle \psi|[[\hat L_y,\hat L_z],\hat L_y]|\psi\rangle$, because it will make this e.v. zero, while it is actually proportional to $L_z$. You shoould provide some context. –  Peter Kravchuk Jun 5 '13 at 16:51
    
I provided more content. –  71GA Jun 5 '13 at 17:16
    
In this formula you are allowed to do that provided $\psi$ is an eigenfunction of $\hat L_z$. I guess it is what your prof. meant when he said 'angular momentum is constant'. –  Peter Kravchuk Jun 5 '13 at 17:27
1  
Yes, the only way in what you are authorized to replace an operator by an expectation value, is when your state is an eigenfunction of the operator. In this case, the "expectation value" is the eigenvalue. –  Trimok Jun 5 '13 at 17:29
    
How can i see that this is an eigenfunction? Do i have to rearange the equation? –  71GA Jun 5 '13 at 19:08

1 Answer 1

up vote 2 down vote accepted

In this context, "$L_z$ is constant" means that the operator $\hat{L}_z$ has only one eigenvalue in the space of states under consideration, let's call it $\mathcal{H}$ - in other words, for any quantum state or wavefunction $\psi$ that could occur given the constraints of the problem ($\forall\psi\in\mathcal{H}$), $\psi$ is an eigenstate of $\hat{L}_z$ with the particular eigenvalue $L_z$.

You can find a basis of eigenvectors of $\hat{L}_z$ which span $\mathcal{H}$, and all associated eigenvalues are $L_z$. The spectral theorem says that, under certain conditions (which do apply here), an operator is completely determined by its eigenvalues and eigenvectors. So any other operator $\hat{L}_z'$ which satisfies

$$\hat{L}_z'\psi = L_z\psi\ \forall\ \psi\in\mathcal{H}$$

is equivalent to $\hat{L}_z$ as long as your state space is limited to $\mathcal{H}$. Basically, for this problem, all operators satisfying $\hat{L}_z'\psi = L_z\psi$ are interchangeable.

One such operator is multiplication by the constant $L_z$,

$$\hat{L}_z' = L_z$$

So you can replace the original operator $\hat{L}_z$ by the constant $L_z$.

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I am sorry i don't understand this ... I only understand that if i have 1 possible state i need 1-dimensional Hilbert space $\mathcal{H}$. BUT WHY do i have only 1 possible state??? –  71GA Jun 6 '13 at 14:59
1  
Because the problem says you do, presumably. Your question said "if $L_z$ is constant" which means it's given that you are only dealing with one state, or with a set of multiple states that all have the same eigenvalue of $\hat{L}_z$. I can't tell you (and you didn't ask) why you are restricted to that particular set of states. But once you know that you do, for whatever reason, my answer explains why you can replace $\hat{L}_z$ with $L_z$. –  David Z Jun 6 '13 at 18:07

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