Expectation value of angular momentum

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fluidistic
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Homework Statement


A particle is under a central potential. Initially its wave function is an eigenfunction ##\psi## such that ##\hat {\vec L ^2} \psi = 2 \hbar ^2## , ##\hat L_3 \psi =0##.
Calculate the expectation value of ##\hat {\vec L}## for all times.

Homework Equations


##\frac{d}{dt}<\hat {\vec L}>=\frac{i}{\hbar } [\hat H , \hat {\vec L}]##.
I get that the quantum numbers of the eigenfunction are ##l=1##, ##m=0##. Not sure if that helps.

The Attempt at a Solution


Using the relevant equation, I get that ##\frac{d}{dt}<\hat {\vec L}>=0## because H commutes with L because it's a central potential problem. This means that the expectation value of the operator L is constant in time.
So I can calculate it at t=0, and use the information that at ##t=0##, ##\Psi (x,t=0)## is an eigenfunction (of the Hamiltonian but also of L^2 and L_z apparently).
So here's my problem: ##<\hat {\vec L}>(0)=<\psi , \hat {\vec L} \psi>##. I don't know how to calculate this. I've checked in Nouredine Zetilli's book because its said to have many solved problems but I couldn't find a single problem where one have to calculate the expected value of a vector operator such as ##\hat {\vec L}##.
Using intuition I guess I should rewrite that expectation value in terms of the expectation values of L^2 and L_z (=L_3) but I don't know how I could do this.

Here's an attempt: ##\hat {\vec L^2}=L_x^2+L_y^2+L_z^2##.
##\Rightarrow (L_x^2+L_y^2) \psi =2\hbar ^2 \psi##. Stuck here.

Here's another attempt: I want the expectation value of ##<L_x \hat i + L_y \hat j + L_z \hat k>##. I know I can't write this as ##<L_x>+<L_y>+<L_z>##. So I'm stuck here.

Any help is appreciated.
 

Answers and Replies

  • #2
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Here's another attempt: I want the expectation value of ##<L_x \hat i + L_y \hat j + L_z \hat k>##. I know I can't write this as ##<L_x>+<L_y>+<L_z>##.

Well, obviously not, since it's a vector :smile:

But you can write it as [itex]\langle L_x \rangle \hat i+ \langle L_y \rangle \hat j + \langle L_z \rangle \hat k[/itex].
 
  • #3
fluidistic
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Well, obviously not, since it's a vector :smile:

But you can write it as [itex]\langle L_x \rangle \hat i+ \langle L_y \rangle \hat j + \langle L_z \rangle \hat k[/itex].

I see thanks, with the last term being equal to 0.
So I am left with ##<\hat {\vec L}>=<L_x>\hat i + <L_y> \hat j##. I'm not sure on how to proceed.
 
  • #4
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Try expressing ##L_x## and ##L_y## in terms of ladder operators.
 
  • #5
fluidistic
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Try expressing ##L_x## and ##L_y## in terms of ladder operators.
Oh right...
##L_x \psi =L_x |1,0>=\frac{\hbar}{\sqrt 2} (|1,1>+|1,-1>)##. Similarly ##L_y=\frac{\hbar}{\sqrt 2 i}(|1,1>+|1,-1>)##.
Then if the relation ##<L_x>=\frac{\hbar}{\sqrt 2}(<|1,1>>+<|1,-1>>)## is true, then ##<L_x>=0## and ##<L_y>=0##.

This would make that ##<\hat { \vec L}>(t)=0 \hat i + 0 \hat j + 0 \hat k##.
So I am learning that the expectation value of a "vector operator" is actually a vector... I would have thought that the expectation value of anything would be a scalar!!!
 

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