Addition of Angular Momentum - Measurement Probability

[Tangent87]

Say we have two particles with angular momentum basis states $$|j_1~m_1\rangle$$ and $$|j_2~m_2\rangle$$ and the combined system is in the state:

$$|0~0\rangle=\sqrt{\frac{1}{6}}\left(|1~1\rangle |1~-1\rangle+|1~-1\rangle |1~1\rangle\right)$$

What is the probability that measurements of the z-component of angular momentum for either constituent particle will give the value of 1?

I think it's 1/6 because we just square the coefficient of |1 1> for one of the particles. Is that correct?

 

[ideasrule]

Are you sure that's the right state? If you have two particles with angular momentum 1, you should get
$$|0~0\rangle=\sqrt{\frac{1}{3}}\left(|1~1\rangle |1~-1\rangle+|1~-1\rangle |1~1\rangle\right + |1~0\rangle |1~0\rangle)$$

Note that the probabilities add to 1, as they should.

 

[Tangent87]

Are you sure that's the right state? If you have two particles with angular momentum 1, you should get
$$|0~0\rangle=\sqrt{\frac{1}{3}}\left(|1~1\rangle |1~-1\rangle+|1~-1\rangle |1~1\rangle\right + |1~0\rangle |1~0\rangle)$$

Note that the probabilities add to 1, as they should.

No I'm not sure, but here's my reasoning. The combined state for |2 0> is:

$$|2~0\rangle=\sqrt{\frac{1}{6}}\left(|1~1\rangle |1~-1\rangle+|1~-1\rangle |1~1\rangle\right)+\sqrt{\frac{2}{3}}|1~0\rangle |1~0\rangle$$

Therefore knowing that the combined state |1 0> must be antisymmetric under interchange of the two particles, orthogonal combination gives:

$$|1~0\rangle=\sqrt{\frac{1}{6}}\left(|1~1\rangle |1~-1\rangle-|1~-1\rangle |1~1\rangle\right)$$

Then finally knowing that the combined state |0 0> must be symmetric under interchange of the two particles, orthogonal combination gives:

$$|0~0\rangle=\sqrt{\frac{1}{6}}\left(|1~1\rangle |1~-1\rangle+|1~-1\rangle |1~1\rangle\right)$$

I don't see how you can keep the |1 0>|1 0> term if the combined state |1 0> is to be antisymmetric?

 

[AndyUrquijo]

I concur with ideasrule. The system state $$|0~0\rangle$$ of two particles with momentum 1 simply "is" the linear conbination

$$\sqrt{\frac{1}{3}}\left(|1~1\rangle |1~-1\rangle+|1~-1\rangle |1~1\rangle\right + |1~0\rangle |1~0\rangle)$$

In this case you have a 1/3 chance of finding a particle with zth component 1. If you measure both particles at the same time then you get a 2/3 chance of finding one of them with zth component 1 (the other with -1). Obviously they can't both have momentum 1 at the same time.

 

[Tangent87]

I concur with ideasrule. The system state $$|0~0\rangle$$ of two particles with momentum 1 simply "is" the linear conbination

$$\sqrt{\frac{1}{3}}\left(|1~1\rangle |1~-1\rangle+|1~-1\rangle |1~1\rangle\right + |1~0\rangle |1~0\rangle)$$

In this case you have a 1/3 chance of finding a particle with zth component 1. If you measure both particles at the same time then you get a 2/3 chance of finding one of them with zth component 1 (the other with -1). Obviously they can't both have momentum 1 at the same time.

But why is it that? What is wrong with my logic in my previous post?

 

[Tangent87]

Hello?

 

[AndyUrquijo]

Sorry, I didn't want to argue on something I'm not sure about. I simply don't understand your use of pair exchange symmetry for external angular momentum... it may be simple lack of knowledge of mine tough... Orthogonal combination?? Note that the ket you obtain is not properly normalized (1/6 + 1/6 = 1/3, not 1)

What I think may be wrong with your logic is taking a well-defined $$|2~0\rangle$$ ket (with clebsh-gordan coefficients) and somehow do your "ortogonal combinations" to get to the $$|2~0\rangle$$ ket, which has already well defined coeficients. And without any apprent reason to do so. This cofficients come from some recursion relations (Sadly I can't remember them now). But I don't think symmetry had any role there. In other words I don't think I agree with your claims of symmetry/antisimmetry.

Pair exchange symmetry IIRC is important for bosons and fermions, spin interactions, etc.

In any case I suggest you wait for someone with more clarity on this subject, I guess I already forgot half of what I learned in my undergrad...

 

[Tangent87]

Sorry, I didn't want to argue on something I'm not sure about. I simply don't understand your use of pair exchange symmetry for external angular momentum... it may be simple lack of knowledge of mine tough... Orthogonal combination?? Note that the ket you obtain is not properly normalized (1/6 + 1/6 = 1/3, not 1)

What I think may be wrong with your logic is taking a well-defined $$|2~0\rangle$$ ket (with clebsh-gordan coefficients) and somehow do your "ortogonal combinations" to get to the $$|2~0\rangle$$ ket, which has already well defined coeficients. And without any apprent reason to do so. This cofficients come from some recursion relations (Sadly I can't remember them now). But I don't think symmetry had any role there. In other words I don't think I agree with your claims of symmetry/antisimmetry.

Pair exchange symmetry IIRC is important for bosons and fermions, spin interactions, etc.

In any case I suggest you wait for someone with more clarity on this subject, I guess I already forgot half of what I learned in my undergrad...

Hi Andy, thanks for replying. I've made a bit more progress on this now as I've realized that to get to |1 0> I have to take the orthogonal combination of |2 1> so that we'll obtain the top state |1 1> and then apply J_ to get |1 0> ! But from this I'm still not sure as to how to get to |0 0> as if we take the orthogonal combination I still won't get the much needed
|1 0>|1 0> term.

Everything you've said makes sense and I can see why what you wrote HAS to be the |0 0> state but I'm still a little fuzzy on how you arrived at that, it's okay to just "magic" it from somewhere in a simple case of two j=1 particles but what I want to know is the general method you use to get to |J-1 M> from |J M> as I will then be able to apply it in more complicated cases. Thanks very much.

 

[ideasrule]

Therefore knowing that the combined state |1 0> must be antisymmetric under interchange of the two particles, orthogonal combination gives:

$$|1~0\rangle=\sqrt{\frac{1}{6}}\left(|1~1\rangle |1~-1\rangle-|1~-1\rangle |1~1\rangle\right)$$

This is correct, except it isn't normalized.

Then finally knowing that the combined state |0 0> must be symmetric under interchange of the two particles, orthogonal combination gives:

$$|0~0\rangle=\sqrt{\frac{1}{6}}\left(|1~1\rangle |1~-1\rangle+|1~-1\rangle |1~1\rangle\right)$$

You forgot the |10>|10> term. Since exchanging the two particles in that state simply gives you back the original state, the equation is still symmetric if you include |10>|10>

I don't see how you can keep the |1 0>|1 0> term if the combined state |1 0> is to be antisymmetric?

You're right: there is no |10>|10> term for the combined state |10>.

 

[ideasrule]

Hi Andy, thanks for replying. I've made a bit more progress on this now as I've realized that to get to |1 0> I have to take the orthogonal combination of |2 1> so that we'll obtain the top state |1 1> and then apply J_ to get |1 0> ! But from this I'm still not sure as to how to get to |0 0> as if we take the orthogonal combination I still won't get the much needed
|1 0>|1 0> term.

Everything you've said makes sense and I can see why what you wrote HAS to be the |0 0> state but I'm still a little fuzzy on how you arrived at that, it's okay to just "magic" it from somewhere in a simple case of two j=1 particles but what I want to know is the general method you use to get to |J-1 M> from |J M> as I will then be able to apply it in more complicated cases. Thanks very much.

I'm sorry for taking so long to reply. The first thing I have to say is that I was wrong in my first post: the second term in the combined state should be negative, not positive. This doesn't change the final answer, however.

The coefficients in front of the spin states are called Clebsch-Gordan coefficients, and are given by a rather complicated recursion relationship. Usually if you need the coefficients, you just consult a table to look them up instead of deriving them from scratch. Here is one such table: http://en.wikipedia.org/wiki/Table_of_Clebsch-Gordan_coefficients

 

[Tangent87]

You forgot the |10>|10> term. Since exchanging the two particles in that state simply gives you back the original state, the equation is still symmetric if you include |10>|10>

Ok now we're getting somewhere, I can see why there must be a |10>|10> term because 0 and 0 is the other combination of the m_i that add to give 0, but according to that wikipedia table it has coefficient $$-\sqrt{\frac{1}{3}}$$ how is this determined by inspection? Because in an exam I have neither that table nor the massive recurrence relation for them. I can see that normalisation will give us the $$\sqrt{\frac{1}{3}}$$ but where does the minus come from? As it will be symmetric whether it is plus OR minus!

 

[ideasrule]

First, are you very sure that you're expected to know how to derive the Clebsch-Gordan coefficients on an exam without the table or the recurrence relation? I rather doubt that you do.

I don't think there is a way to derive the coefficients without recursion, but for this question it happens to be very simple. First, let's assume the wavefunction looks like:

|00>=a|11>|1 -1> + b|10>|10> + c|1 -1>|1 1>

By symmetry arguments, we know a=c. By normalization, we know a^2+b^2+c^2=1. To derive the third equation, we use J₊ and J_-, the raising and lowering operators, where J₊=J₊¹ + J₊² and similarly for J-. J₊¹*f(j,m)=sqrt(j(j+1)-m(m+1))*f(j,m+1) and similarly for J_-¹, except the m(m+1) is replaced by m(m-1).

For this particular problem, we apply J+ until everything on the right is converted to |11>|11>. We would then have an equation relating a, b, and c. We can do the same thing with J-, but it would give the same equation. We now have 3 equations relating a, b, and c, and can solve the system of linear equations.

Ok now we're getting somewhere, I can see why there must be a |10>|10> term because 0 and 0 is the other combination of the m_i that add to give 0, but according to that wikipedia table it has coefficient $$-\sqrt{\frac{1}{3}}$$ how is this determined by inspection? Because in an exam I have neither that table nor the massive recurrence relation for them. I can see that normalisation will give us the $$\sqrt{\frac{1}{3}}$$ but where does the minus come from? As it will be symmetric whether it is plus OR minus!

 

[Tangent87]

First, are you very sure that you're expected to know how to derive the Clebsch-Gordan coefficients on an exam without the table or the recurrence relation? I rather doubt that you do.

I don't think there is a way to derive the coefficients without recursion, but for this question it happens to be very simple. First, let's assume the wavefunction looks like:

|00>=a|11>|1 -1> + b|10>|10> + c|1 -1>|1 1>

By symmetry arguments, we know a=c. By normalization, we know a^2+b^2+c^2=1. To derive the third equation, we use J₊ and J_-, the raising and lowering operators, where J₊=J₊¹ + J₊² and similarly for J-. J₊¹*f(j,m)=sqrt(j(j+1)-m(m+1))*f(j,m+1) and similarly for J_-¹, except the m(m+1) is replaced by m(m-1).

For this particular problem, we apply J+ until everything on the right is converted to |11>|11>. We would then have an equation relating a, b, and c. We can do the same thing with J-, but it would give the same equation. We now have 3 equations relating a, b, and c, and can solve the system of linear equations.

Ah yes I see, that's a nice trick with applying the J_+ until everything on the RHS is |11>|11>. Thanks again for your help!