# Homework Help: The Hilbert-Schmidt inner product and entanglement

1. Jan 5, 2016

### Emil_M

1. The problem statement, all variables and given/known data

Suppose $R$ and $Q$ are two quantum systems with the same Hilbert space. Let $|i_R \rangle$ and $|i_Q\rangle$ be orthonormal basis sets for $R$ and $Q$. Let $A$ be an operator on $R$ and $B$ an operator on $Q$. Define $|m\rangle := \sum_i |i_R\rangle |i_Q\rangle$. Show that \begin{align} \mathrm{tr}(A^\dagger B)=\langle m| (A \otimes B)|m\rangle, \end{align}
where the multiplication on the left hand side is of matrices, and it is understood that the matrix elements of $A$ are taken with respect to the basis $|i_R\rangle$ and those for $B$ with respect to the basis $|i_Q\rangle$.
2. Relevant equations
See above
3. The attempt at a solution
I've tried the following

$A=\sum_{i,j} a_{ij} |i_R\rangle \langle j_R|$ and $B=\sum_{k,l} b_{kl} |k_R\rangle \langle l_R|$

Then

\begin{align} A^\dagger B &= \left(\sum_{i,j} a^*_{ji} |i_R\rangle \langle j_R|\right)\left(\sum_{k,l} b_{kl} |k_R\rangle \langle l_R| \right)\\ &=\sum_{ijkl} a^*_{ji}b_{kl} |i_R\rangle |k_Q\rangle \langle j_R| \langle l_Q| \end{align}

But that doesn't seem to be the right track, is it?

Thanks for your help!

2. Jan 5, 2016

### TSny

I believe you meant to have Q's instead of R's in the expression for B.

These expressions represent the operators A and B. However, in the expression $\mathrm{tr}(A^\dagger B)$ , A and B are just matrices where the matrix elements of A are $a_{ij} = \langle i_R| A| j_R\rangle$ and the matrix elements of B are $b_{ij} = \langle i_Q| B |j_Q\rangle$. So, if you were handed these two matrices, how would you find $\mathrm{tr}(A^\dagger B)$?

[Aside: Note that you should have written

\begin{align} A^\dagger B &= \left(\sum_{i,j} a^*_{ji} |i_R\rangle \langle j_R|\right)\left(\sum_{k,l} b_{kl} |k_Q\rangle \langle l_Q| \right)\\ &=\sum_{ijkl} a^*_{ji}b_{kl} |i_R\rangle \langle j_R|k_Q\rangle \langle l_Q| \end{align}

where $\langle j_R|k_Q\rangle$ is just a number.

But you won't need these expressions. ]

3. Jan 5, 2016

### Emil_M

Thanks for your help!

Well the matrix element of $A^\dagger B$ is $a^*_{ji} b_{jk}$, so the trace is given by $tr(A^\dagger B)=a^*_{ji} b_{ji}$ isn't it?

4. Jan 5, 2016

### TSny

Yes (with repeated indices summed).

It appears that equation (1) of the first post is not quite correct. Maybe I'm overlooking something, but it seems to me that the dagger on A should just be a transpose.

Last edited: Jan 7, 2016
5. Jan 5, 2016

### Emil_M

Last edited by a moderator: May 7, 2017
6. Jan 5, 2016

### TSny

In the equation $\mathrm{tr}(A^\dagger B)=\langle m| (A \otimes B)|m\rangle$, it appears to me that the left hand side involves complex conjugates of the matrix elements of $A$, whereas the right hand side does not. So, it doesn't look correct to me. But, I feel like I might be overlooking something. I will try to bring this thread to the attention of other helpers.

7. Jan 5, 2016

### Emil_M

Thank you so much for taking the time to help me!

8. Jan 6, 2016

### TSny

I put out a request to other helpers. While we wait to see if someone else comes in, let's see if we agree on the right hand side:

$\langle m| (A \otimes B)|m\rangle$.

What did you get when you evaluated this?

9. Jan 6, 2016

### Emil_M

I'm not really sure how to solve this.

What I got for the matrix element is $a_{ij}b_{kl}$

That would make the trace $a_{ii} b_{ii}$...

10. Jan 6, 2016

### TSny

I don't believe that's what you should get.
Can you show more detail for what you did after you substituted $|m \rangle = \sum_i |i_R \rangle |i_Q \rangle$ into $\langle m| (A \otimes B)|m\rangle$?

Here, the "tensor product" $A \otimes B$ means that $A$ operates only on $|i_R \rangle$ and $B$ operates only on $|i_Q \rangle$ when $A \otimes B$ operates on $|i_R \rangle |i_Q \rangle$.

Last edited: Jan 6, 2016
11. Jan 6, 2016

### andrewkirk

I think the problem will be easier with a change of notation. The problem presents itself as one about operators on Hilbert spaces, rather than about physical systems. The two Hilbert Spaces are the same, so let's not confuse things by giving them two different names $R$ and $Q$. Instead let R and Q be names for the two different bases for the single Hilbert space.
For the problem to be meaningful at all, the two bases must have the same cardinality, which may be finite or infinite. Let the index set for that cardinality be $N$. Then we can label our index vectors as $|r_i\rangle$ and $|q_i\rangle$ for basis R and Q respectively, where $i\in N$.
Then, if we use $A$ and $B$ to denote matrices, rather than linear operators, we can write
the $i,j$ element of $A^\dagger B$ as $\sum_{k\in N}a^*_{ki}b_{kj}$.

So the trace on the LHS is:
$$\sum_{k,j\in N}a^*_{kj}b_{kj}$$
In the new notation we have
$$|m\rangle = \sum_{k\in N} |r_k\rangle|q_k\rangle$$.

Use that to write out the RHS of the sought equality and let's see what happens.

Last edited: Jan 6, 2016
12. Jan 6, 2016

### andrewkirk

Hmmm. When I evaluate the RHS I get
$$\sum_{k,j\in N}a_{kj}b_{kj}$$

So it looks to me like the equation (1) in the OP may be wrong. I think it should be
$$tr(A'B)=\langle m| (A \otimes B)|m\rangle$$
where the prime indicates transposition (but not conjugation).

I also see that Tsny has already pointed this out. I agree with Tsny.

13. Jan 6, 2016

### andrewkirk

I feel I should also point out that the problem statement contains horrendous abuse of notation. It says that $A$ and $B$ are linear operators, which are basis-independent. But then, in the LHS of the equation it uses the same symbols to denote the matrices of those operators, with respect to two different bases!
Since linear operators have traces too, the expression $tr(A^\dagger B)$ has a well-defined meaning where $A,B$ are linear operators, and that will give a different value from what they want in the LHS of (1), because of the change of basis of only one of the matrices.
Yet it looks like on the RHS of (1) they want to treat $A$ and $B$ as linear operators rather than matrices.

14. Jan 6, 2016

### TSny

Thanks for joining the discussion!

I also found the notation confusing at first, until I clearly distinguished A and B as matrices specified in the problem statement on the LHS of (1) and operators on the RHS.

I see nothing wrong with using different bases on the LHS of (1). The LHS is well-defined. However, its value does depend on the particular choice of bases.

The RHS also depends on the choice of bases. Note that the RHS does not represent the trace of the operator $A \otimes B$. The operator $A \otimes B$ is an operator that acts on the tensor product of the Hilbert spaces of systems R and Q. A possible set of basis vectors for the tensor product space would be all the vectors of the form $|r_i>|q_j>$ (in your notation). But the RHS of (1) only sums over the particular vectors $|r_i>|q_i>$ which is just a subset of the basis vectors. So, the RHS is not a “full trace” and therefore the RHS can depend on the choice of bases.

Anyway, I agree with your evaluation of the RHS that you gave in post #12.

15. Jan 7, 2016

### Emil_M

Thank you both for investing your time in helping me!

Could you give me a pointer how to evaluate the RHS?

16. Jan 7, 2016

### andrewkirk

The general approach is that $(A\otimes B)( |c\rangle\otimes|d\rangle)$ is defined to mean $(A|c\rangle)\otimes(B|d\rangle)$. In QM, $|c\rangle\otimes|d\rangle$ is usually written $|c\rangle|d\rangle$, or sometimes $|c,d\rangle$.

With a bit of manipulation involving adjoints and bras vs kets, this leads to

$$\langle c,d|(A\otimes B)|c,d\rangle= \langle c|A|c\rangle \langle d|B|d\rangle$$
where the two items on the RHS are just complex numbers that are multiplied by one another.

Now to apply this in your problem, you first need to write out the ket $|m\rangle$ as a sum of kets of the form $|c,d\rangle$, and similarly for the bra. Then you insert that in the above formula and use linearity to expand and then simplify.

17. Jan 12, 2016

### Emil_M

I still can't get it quite right I'm afraid.

What I do is the following:

\begin{align} \langle m|(A \otimes B) | m \rangle &:= \sum_n \langle n_R | \left( \sum_{i,j} a_{i,j} |i\rangle \langle j | \right) | n_R \rangle \cdot \sum_{n} \langle n_Q | \left( \sum_{k,l} b_{kl} |b\rangle \langle k | \right) |n_Q \rangle\\ &= \sum_{n,i,j} a_{ij} \underbrace{\langle n_R | i \rangle}_{\delta_{n_R,i}} \underbrace{\langle j|n_R\rangle}_{\delta_{j,n_R}} \cdot \sum_{n,k,l} b_{kl} \underbrace{\langle n_Q | k \rangle}_{\delta_{n_Q,k}} \underbrace{\langle l|n_Q\rangle}_{\delta_{l,n_Q}}\\ &=\sum_j a_{jj} \cdot \sum_l b_{ll} \end{align}

Thanks for your help!

18. Jan 12, 2016

### TSny

In the expression $|m\rangle = \sum_i |i_R\rangle |i_Q\rangle$, $i$ is just a dummy summation index. You could just as well write is as $|m\rangle = \sum_j |j_R\rangle |j_Q\rangle$.

If $\langle m|$ and $|m\rangle$ occur in the same expression, you should use different summation indices.

There is no need to write $A$ as $\sum_{i,j} a_{i,j} |i\rangle \langle j |$. Similarly for $B$.

19. Jan 12, 2016

### Fightfish

Your expectation value expression is not done correctly. It should be:
$$\langle m|(A \otimes B) | m \rangle = \left[\sum_{\alpha} \langle \alpha_{R}| \langle \alpha_{Q}|\right] \left(A \otimes B\right) \left[\sum_{\beta} | \beta_{R}| \rangle |\beta_{Q}\rangle\right] = \sum_{\alpha}\sum_{\beta} \langle \alpha_{R}| A | \beta_{R} \rangle \langle \alpha_{Q}| B | \beta_{Q} \rangle$$

20. Jan 12, 2016

### Emil_M

Thanks!

I got hung up on the fact that I wanted to prove a trace identity on the LHS, so I didn't realize $\langle m|$ and $|m\rangle$ didn't have to be the same indices.

21. Jan 26, 2016

### Emil_M

Does anybody have any idea how to get the conjugation correct?

22. Jan 26, 2016

### andrewkirk

Are you asking how to derive the formula in the matheplanet link, as quoted in your OP? If so, I think the answer is that there is no way to do that because [I am almost positive that] the formula is wrong. The LHS should use a transpose sign rather than an adjoint sign (ie it should be $tr(A'B)$ not $tr(A^\dagger B)$). So there is no conjugation to be done.

23. Jan 27, 2016

### Emil_M

I just confirmed, that it was indeed an error in the Nielsen, Chuang.

Thanks again for your help!