Collapse of wave function question

[VortexLattice]

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

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

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 [itex]\psi_{211}[/itex] 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.

 

[Simon Bridge]

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.

 

[VortexLattice]

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?

 

[Simon Bridge]

But that would mean that on subsequent measurement of L_x, 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 L_z you oriented it to the z axis (basically the measuring process defines what you mean by "the z axis".)

When you measure L_z 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 L_z will give you the same value... if the system originally was in a superposition of L_z 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

 

[Ilmrak]

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 [itex]\psi_{211}[/itex] you are meaning the wave function with [itex]E=E_2 \quad L^2 = 1(1+1)= 2 \quad L_z =1[/itex], 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 [itex]\{|n,L^2=2,L_z=1\rangle, n\in \mathbb{N} \}[/itex]. The only state of the starting superposition with non zero component on this space is the one VortexLattice said.

edit

[...]if the system originally was in a superposition of L_z 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 [itex][L_i,L_j]=0[/itex], but [itex][L_i,L_j]=i \epsilon_{ijk}L_k[/itex] so they do not commute and the x and y components of angular momentum actually are affected by the measurement of [itex]L_z[/itex].

Ilm

 

[VortexLattice]

But that would mean that on subsequent measurement of L_x, 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 L_z you oriented it to the z axis (basically the measuring process defines what you mean by "the z axis".)

When you measure L_z 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 L_z will give you the same value... if the system originally was in a superposition of L_z 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 L_z 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 L_z will give you the same value...".

But for subsequent measurements of L_z 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 L_z = -1,0,1...right?

I still assume the book is right and I'm missing something, but I'd like to know what.

 

[Ilmrak]

Is there maybe some misunderstanding in notation or on the exercise text?

Ilm

 

[VortexLattice]

Is there maybe some misunderstanding in notation or on the exercise text?

Ilm

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

And here is the solution:

Unless I'm missing something, they just seem to ignore the part where L_z is measured. If they only measured L = 1, this seems like it would be the right answer.

 

[Ilmrak]

As I supposed, there was a misunderstanding in the notation.

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

I hope this solves the problem^^

Ilm

 

[VortexLattice]

As I supposed, there was a misunderstanding in the notation.

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

I hope this solves the problem^^

Ilm

Wait, what the hell? First of all, I really thought [itex]L_{\pm} = L_x \pm iL_y[/itex]. Second of all, I have never ever seen this before, I thought m was always the L_z quantum number...

 

[Ilmrak]

Well, since all [itex]L_i[/itex] commute with [itex]L^2[/itex], you can choose the angular momentum component you prefer to describe your states.

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

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

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

Ilm

 

[VortexLattice]

Well, since all [itex]L_i[/itex] commute with [itex]L^2[/itex], you can choose the angular momentum component you prefer to describe your states.

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

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

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

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?

 

[Ilmrak]

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