Collapse of wave function question

VortexLattice
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Hi all, I'm doing a practice question in which we have a hydrogen atom in the state:

\psi = (2\psi_{100} + \psi_{210} + \sqrt{2}\psi_{211} + \sqrt{3}\psi_{21 -1})/\sqrt{10}

It says that, now a measurement is taken and we find the angular momentum variables to be L = 1 and L_z = 1. The question is: immediately after the measurement, what is the wave function?

Now, I thought that we definitely observed only \psi_{211} here, because that is the only one with these values. So, I thought that collapses the wave function to this eigenstate, and it basically stays there.

However, it appears I'm wrong. The answer I have says that there's now a new wave function comprised of all three of the L = 1 states, and they do a bit of math to figure out their coefficients.

Can anyone help?

Thanks!

PS: I never found out an answer to this question, if anyone could help me here.
 
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Measuring one component fixed that one in a particular state, but the others remain in superposition. iirc the picture you should have in your text/notes has moment precessing in the x-y plane.
 
Simon Bridge said:
Measuring one component fixed that one in a particular state, but the others remain in superposition. iirc the picture you should have in your text/notes has moment precessing in the x-y plane.

I know the drawing you're talking about, but I'm still not sure what you're saying here. I thought that if we just measured L^2 (essentially L), then after the measurement it could still be in a combination of states with that value of L^2, but it says we find L_z also. Doesn't that mean we specifically measured ψ_211? How can the others still be in superposition?
 
But that would mean that on subsequent measurement of Lx, say, you'd get nothing right? But that's not what happens is it?

Anyway ... in the vector model you've seen, which is handy for visualizing this, the angular momentum is a vector with some angle ... when you measured Lz you oriented it to the z axis (basically the measuring process defines what you mean by "the z axis".)

When you measure Lz you are measuring only one component of the overall vector - that won't make the other components go away. It just stops that component from being uncertain ... subsequent measurements of Lz will give you the same value... if the system originally was in a superposition of Lz states then you'll have just collapsed that part of the wavefunction... since you made no measurements on the x and y components, those parts are unaffected*.

-------------------
* this works for commuting observables
http://farside.ph.utexas.edu/teaching/qmech/lectures/node70.html
http://nonlocal.com/hbar/commuting.html
 
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VortexLattice said:
I know the drawing you're talking about, but I'm still not sure what you're saying here. I thought that if we just measured L^2 (essentially L), then after the measurement it could still be in a combination of states with that value of L^2, but it says we find L_z also. Doesn't that mean we specifically measured ψ_211? How can the others still be in superposition?

If in your notation with \psi_{211} you are meaning the wave function with E=E_2 \quad L^2 = 1(1+1)= 2 \quad L_z =1, then I'd say you are right, this is the wave function after the measurement of your exercise.

The measurement projects the state of the atom on the space spanned by \{|n,L^2=2,L_z=1\rangle, n\in \mathbb{N} \}. The only state of the starting superposition with non zero component on this space is the one VortexLattice said.

edit
Simon Bridge said:
[...]if the system originally was in a superposition of Lz states then you'll have just collapsed that part of the wavefunction... since you made no measurements on the x and y components, those parts are unaffected*.

-------------------
* this works for commuting observables
http://farside.ph.utexas.edu/teaching/qmech/lectures/node70.html
http://nonlocal.com/hbar/commuting.html

Exactly, it would work if it was [L_i,L_j]=0, but [L_i,L_j]=i \epsilon_{ijk}L_k so they do not commute and the x and y components of angular momentum actually are affected by the measurement of L_z.

Ilm
 
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Simon Bridge said:
But that would mean that on subsequent measurement of Lx, say, you'd get nothing right? But that's not what happens is it?

Anyway ... in the vector model you've seen, which is handy for visualizing this, the angular momentum is a vector with some angle ... when you measured Lz you oriented it to the z axis (basically the measuring process defines what you mean by "the z axis".)

When you measure Lz you are measuring only one component of the overall vector - that won't make the other components go away. It just stops that component from being uncertain ... subsequent measurements of Lz will give you the same value... if the system originally was in a superposition of Lz states then you'll have just collapsed that part of the wavefunction... since you made no measurements on the x and y components, those parts are unaffected*.

-------------------
* this works for commuting observables
http://farside.ph.utexas.edu/teaching/qmech/lectures/node70.html
http://nonlocal.com/hbar/commuting.html

Hmmm, but from what you said it still seems like I'm right... You said "When you measure Lz you are measuring only one component of the overall vector - that won't make the other components go away. It just stops that component from being uncertain ... subsequent measurements of Lz will give you the same value...".

But for subsequent measurements of Lz to give the same value (1, in this case), it seems like the wave function has to be in the ψ211 state! If it were in a superposition of the three different L = 1 states (for E = 2), subsequent measurements could give values of Lz = -1,0,1...right?

I still assume the book is right and I'm missing something, but I'd like to know what.
 
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Is there maybe some misunderstanding in notation or on the exercise text?

Ilm
 
Ilmrak said:
Is there maybe some misunderstanding in notation or on the exercise text?

Ilm

Just to make sure, here is the problem (part e):

Df1E6.png


And here is the solution:

UBm60.png


Unless I'm missing something, they just seem to ignore the part where Lz is measured. If they only measured L = 1, this seems like it would be the right answer.
 
As I supposed, there was a misunderstanding in the notation.

The book says that L_z= \frac{1}{2}(L_+ + L_- ), this means that the quantum number m is the eigenvalue of L_y (L_- lowers m by 1).

I hope this solves the problem^^

Ilm
 
  • #10
Ilmrak said:
As I supposed, there was a misunderstanding in the notation.

The book says that L_z= \frac{1}{2}(L_+ + L_- ), this means that the quantum number m is the eigenvalue of L_y (L_- lowers m by 1).

I hope this solves the problem^^

Ilm

Wait, what the hell? First of all, I really thought L_{\pm} = L_x \pm iL_y. Second of all, I have never ever seen this before, I thought m was always the L_z quantum number...
 
  • #11
Well, since all L_i commute with L^2, you can choose the angular momentum component you prefer to describe your states.

Usually your definition of L_± is right, because is usually used L_z to describe the states. The useful definition of L_± is as the operators that increase (or lowers) the value of m. So if by m you mean the eigenvalue of L_y the operators L_± have to change if they have to increase or lower the value of L_y.

To be brief, since the name of the axis is arbitrary, you can simply change x→ z ,\: y→ x, \: z→ y (to maintain the right handed orientation).

Then L_± = L_z ±i L_x and so L_z= \frac{1}{2}(L_+ + L_-).

Ilm
 
  • #12
Ilmrak said:
Well, since all L_i commute with L^2, you can choose the angular momentum component you prefer to describe your states.

Usually your definition of L_± is right, because is usually used L_z to describe the states. The useful definition of L_± is as the operators that increase (or lowers) the value of m. So if by m you mean the eigenvalue of L_y the operators L_± have to change if they have to increase or lower the value of L_y.

To be brief, since the name of the axis is arbitrary, you can simply change x→ z ,\: y→ x, \: z→ y (to maintain the right handed orientation).

Then L_± = L_z ±i L_x and so L_z= \frac{1}{2}(L_+ + L_-).

Ilm

Ok, I see that because of the symmetry, that makes sense. But it seems like nowhere in the question does it say that m is the y component of angular momentum, and it's been z everywhere else I've seen it. Is there something I'm not seeing where they make it clear that m is the y component, in the question?
 
  • #13
Absolutely not.

I can only see it from the answer given by the book.

I suppose it's the standard notation of the book?

Ilm
 
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