Angular momentum multiple choice question

[yamata1]

Homework Statement

Hello,
Here is a multiple choice question I would like to be clarified.
Suppose that the angular momentum of a system can take the values 0, 1, 2. One carries out a measurement of $J_z$,the state of the system will:
$a$-Be perfectly known if the result is 0
$b$-Be perfectly known if we measure $J^2$
$c$-Be perfectly known if no matter what the result is
$d$-will never be known
$e$-Be perfectly known if ,and only if the result is $\mp2\hbar$

The Attempt at a Solution

After the answers I got in my previous thread https://www.physicsforums.com/threads/angular-momentum-state-of-a-system.933967/#post-5899253 I would say only $b$ is correct but I dislike the use of the term "perfectly" as we can't know $J_x$ and $J_y$.
Thank you for your remarks.

 

[Orodruin]

"Perfectly" well known means that there is only one possible state that you have restricted the system to be in. It does not mean that you will have a definite value for all possible measurements.

To be honest, I think the problem gives you too little information. That the system can only have spin 0, 1, or 2 does not necessarily mean that it is split into a single irrep of each spin. For example, take the case of four spin 1/2 particles. The representation of the total system is
$$
2\otimes 2 \otimes 2 \otimes 2 = 2\otimes 2 \otimes (1 \oplus 3) = 2\otimes(2\times 2 \oplus 4) = 2\times 1 \oplus 3\times 3 \oplus 5,
$$
i.e., the total spin state is composed into two singlet states, three triplet states, and one quintuplet state. The total spin of the system can thus be 0, 1, or 2, but generally measuring the state to have overall angular momentum zero will only tell you that it is in one of the singlets.

 

[yamata1]

"Perfectly" well known means that there is only one possible state that you have restricted the system to be in. It does not mean that you will have a definite value for all possible measurements.

To be honest, I think the problem gives you too little information. That the system can only have spin 0, 1, or 2 does not necessarily mean that it is split into a single irrep of each spin. For example, take the case of four spin 1/2 particles. The representation of the total system is
$$
2\otimes 2 \otimes 2 \otimes 2 = 2\otimes 2 \otimes (1 \oplus 3) = 2\otimes(2\times 2 \oplus 4) = 2\times 1 \oplus 3\times 3 \oplus 5,
$$
i.e., the total spin state is composed into two singlet states, three triplet states, and one quintuplet state. The total spin of the system can thus be 0, 1, or 2, but generally measuring the state to have overall angular momentum zero will only tell you that it is in one of the singlets.

Could you explain how the tensor product and direct sum are used here ? I have never used them together in quantum mechanics.Is my answer correct ?

 

[Orodruin]

The state space of four spin-1/2 particles is the tensor product of the state spaces for each individual particle. Using the SU(2) rules for decomposing the tensor product into irreducible representations, you can find which irreps (i.e., spin multiplets) the state space consists of. In the case of SU(2), a doublet tensor product with any multiplet is particularly simple, i.e., $2 \otimes N = (N-1) \oplus (N+1)$. The multiplications I used (i.e., $2\times$) denote the number of times that particular multiplet appears.

Is my answer correct ?

It depends on how you interpret the question. As I said in #2, the question is not particularly well defined. That the total spin can be 0, 1, or 2 generally does not mean that there is only one singlet, one triplet, and one quintuplet. Although I suspect that this is what the person who constructed the problem is fishing for, the problem itself is not formulated in that way.

 

[yamata1]

The state space of four spin-1/2 particles is the tensor product of the state spaces for each individual particle. Using the SU(2) rules for decomposing the tensor product into irreducible representations, you can find which irreps (i.e., spin multiplets) the state space consists of. In the case of SU(2), a doublet tensor product with any multiplet is particularly simple, i.e., $2 \otimes N = (N-1) \oplus (N+1)$. The multiplications I used (i.e., $2\times$) denote the number of times that particular multiplet appears.

Is my answer correct ?

It depends on how you interpret the question. As I said in #2, the question is not particularly well defined. That the total spin can be 0, 1, or 2 generally does not mean that there is only one singlet, one triplet, and one quintuplet. Although I suspect that this is what the person who constructed the problem is fishing for, the problem itself is not formulated in that way.

Thank you for this clarification. Is there a way to interpret this ambiguous question such that any answer other than $b$ is correct ?

 

[Orodruin]

Thank you for this clarification. Is there a way to interpret this ambiguous question such that any answer other than $b$ is correct ?

Yes.

 

[yamata1]

Yes.

How so ?Could $d$ be correct ?

 

[Orodruin]

How so ?Could $d$ be correct ?

d can never be correct as you could always imagine a set of measurements that would uniquely identify the state so "never" is a kind of harsh statement. If you restrict the never to "without doing further measurements" it could be made correct by making sure that regardless of what the $J_z$ measurement gives, there would be several possible states.

 

[yamata1]

d can never be correct as you could always imagine a set of measurements that would uniquely identify the state so "never" is a kind of harsh statement. If you restrict the never to "without doing further measurements" it could be made correct by making sure that regardless of what the $J_z$ measurement gives, there would be several possible states.

Could $a$,$c$ or $e$ ever be correct ?

 

[Orodruin]

What do you think?

 

[yamata1]

What do you think?

Measuring $J_z$ does not give us the eigenvalues for the other angular momentum operators,so I think not.

 

[Orodruin]

Measuring $J_z$ does not give us the eigenvalues for the other angular momentum operators,so I think not.

This is not the criterion you should be looking at as I said in #2. $J_z$ does not commute with $J_x$ or $J_y$ and so you cannot measure them simultaneously. The question is whether or not your measurement restricts your state to a one-dimensional subspace of the state space.

 

[yamata1]

This is not the criterion you should be looking at as I said in #2. $J_z$ does not commute with $J_x$ or $J_y$ and so you cannot measure them simultaneously. The question is whether or not your measurement restricts your state to a one-dimensional subspace of the state space.

since $J_z|j,m>=m|j,m>$ and $J^2|j,m>=j(j+1)|j,m>$ regardless of $m$ we still don't have $j$ ,it cannot be enough to determine a one-dimensional subspace of the state space ?

 

[Orodruin]

it cannot be enough to determine a one-dimensional subspace of the state space ?

This is not necessarily true.

 

[yamata1]

If j can take values 0,1or 2 and $-j<m<j$ ,how can knowing m be enough to determine a one-dimensional subspace of the state space ?

 

[Orodruin]

Consider if any measurement of $J_z$ would lead to a unique identification of a one-dimensional subspace. If so, what values?

 

[yamata1]

So $e$ is also correct because $\mp2\hbar$ implies j can only be $2\hbar$ but "if,and only if" would exclude $b$.

 

[Orodruin]

e could be correct, it is if your state space only contains one copy of j=2. The if and only if is also ambiguous since b implies taking more measurements. The bottom line is that the question is not well written.

 

[yamata1]

e could be correct, it is if your state space only contains one copy of j=2. The if and only if is also ambiguous since b implies taking more measurements. The bottom line is that the question is not well written.

Thank you for your explanations.The same test asked what the wavelength of the photon emmited for the transition between first and second excited states of the hydrogen atom was.The choices were :
$a$- < 652 nm
$b$- >652 nm
$c$- equal to 652 nm
$d$- >241 nm
$e$- equal to 241nm
The Balmer red alpha emission is 656 nm !