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Expectation value for angular momentum 
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#1
Jun2210, 11:43 AM

P: 150

1. The problem statement, all variables and given/known data
A wavefunction of angular momentum states is given: [tex]\psi = \frac{1}{\sqrt{7}}1,1\rangle + \frac{\sqrt{35}}{7}1,0\rangle+\sqrt{\frac{1}{7}}1,1\rangle[/tex] Calculate [tex]\langle \psi L_{\pm} \psi \rangle[/tex] and [tex]\langle 1,1L_+^2\psi\rangle[/tex] 3. Attempt at a solution. If the wavefunction and angular momentum operators were given in matrix form, I would be able to solve this, since I know how it all works in the matrix representation. But I am confused about what to do with dirac notation? I'm not really sure, should I convert the operators and kets into a matrix form... (I'm not sure how to do this either??). Griffiths, the text I'm using, didn't really go much into expectation values for angular momentum, or the matrix representation, so I would really appreciate some tips on where to go to from here. Thanks! 


#2
Jun2210, 12:47 PM

P: 286

I think that you are familiar with what result you should obtain when you have L+ (the raising operator) acting in a state? I dont remember what is it exactly but im sure there is ( taking in account that i.e 1,1> the 1 refers to l and 1 refers to m) .. I remember that when you have L+ or L the l part wont change ..



#3
Jun2210, 12:48 PM

P: 38

The basis that you have written the wavefunction in is [itex]\leftl,m\right\rangle[/itex], so consider the raising and lower operators [itex]\hat{L}_{\pm}[/itex]; do you know what these operators do when they act on those basis vectors? In other words, for
[tex] \hat{L}_{\pm}\leftl,m\right\rangle =\lambda\left l, m\pm 1\right\rangle [/tex] do you know what [itex]\lambda[/itex] would be? 


#4
Jun2210, 02:12 PM

P: 42

Expectation value for angular momentum
Using the same book.
Follow 1618##########'s suggestion. Then try computing the overlap and recall that the wavefunctions are orthonormal, so for example <1,11,1> =0. When you do the overlap, most of the terms will disappear. 


#5
Jun2210, 02:39 PM

P: 150

Thanks for the hint... yep i think i know
[tex]\lambda = \hbar \sqrt{l(l+1)m(m\pm 1)}[/tex] So... err [tex]\langle \psi L_+\psi\rangle = \langle \psi \hbar\sqrt{l(l+1)m(m+1)}\psi\rangle[/tex] Now do I apply [tex]\hbar\sqrt{l(l+1)m(m+1)}[/tex] to every basis state inside [tex]\psi[/tex]? i.e. [tex]L_+\psi \rangle = \hbar\sqrt{2}\left(\frac{1}{\sqrt{7}}1,1\rangle\right) + \hbar\sqrt{2}\left(\frac{\sqrt{35}}{7}1,0\rangle\right)+\hbar 0\left(\frac{1}{\sqrt{7}}1,1\rangle\right)[/tex] Then do i somehow "multiply" this linear combination by [tex]\langle \psi[/tex]? Sorry I don't really know if I'm going in the right direction. 


#6
Jun2210, 02:43 PM

P: 42

Going in the right direction.
So you have the operator and the ket. Now you have to multiply with the bra. So how do you find the bra form of your wavefunction? 


#7
Jun2210, 02:48 PM

P: 3,014

The problem statement has nothing to do with the title of the thread.



#8
Jun2210, 03:02 PM

P: 150

The bra is simply this right?
[tex]\langle \psi  = \frac{1}{\sqrt{7}}\langle 1,1+\frac{\sqrt{35}}{7}\langle 1,0 + \frac{1}{\sqrt{7}}\langle 1,1[/tex] So I need to find [tex]\langle \left(\frac{1}{\sqrt{7}}\langle 1,1+\frac{\sqrt{35}}{7}\langle 1,0 + \frac{1}{\sqrt{7}}\langle 1,1\right)\left(\hbar\sqrt{2}\frac{1}{\sqrt{7}}1,1\rangle + \hbar\sqrt{2}\frac{\sqrt{35}}{7}1,0\rangle\right)\rangle[/tex] And then i use something like the distributive law to multiply out the states, and if I multiply two states that are not the same, then they are orthonormal so they go to zero, and if I multiply together two states that are the same, they equal 1, right? (Dickfore I thought the title would be ok enough to describe the problem, since it does ask for the expectation values of angular momentum  related quantities. I'll try to be more specific next time) 


#9
Jun2210, 03:13 PM

P: 42

That looks exactly right to me.
Yes, each of the states are orthonormal and normalized already, so that should give you the right answer. And then to get <L>, you would do it the exact same way, just with a slightly different lambda. 


#10
Jun2210, 03:13 PM

P: 42

That looks exactly right to me.
Yes, each of the states are orthonormal and normalized already, so that should give you the right answer. And then to get <L>, you would do it the exact same way, just with a slightly different lambda. 


#11
Jun2210, 03:19 PM

P: 150

Thanks so much, I finally understand it. In retrospect it feels like I had the knowledge but just needed some prodding to put it into practice.
I guess all I need to do now is try manipulating some formulas to get [tex]\langle 1,1L_+^2\psi\rangle[/tex] into the desirable form. Just one last question, I think I know the interpretation of [tex]\langle \psiL_+\psi\rangle[/tex], it's just the expected value. But what is the physical interpretation of [tex]\langle 1,1L_+^2\psi\rangle[/tex]? 


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