# Expectation value of observable in Bell State

1. Nov 7, 2015

### ma18

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

Consider the bipartite observable

O_AB = (sigma_A · n) ⊗ (sigma_B · m)

Where n and m are three vectors and

sigma_i = (sigma_1_i, sigma_2_i, sigma_3_i)

with i = [A,B] are the Pauli vectors.

Compute using abstract and matrix representation the expectation value of O_AB in the Bell State

|ψ> = 1/sqrt(2) (|0_A 1_B> +|1_A 0_B))

for n = (0,1,0)† , b = (0,0,1)†

3. The attempt at a solution

Alright so first I computed the representation of the observable

Applying the values of n and m it comes to

O_AB = sigma_2_A ⊗ sigma_3_B

= (0 -i ⊗ (1 0
i 0) 0 -1)

= (0 0 -i 0
0 0 0 1
i 0 0 0
0 -i 0 0)

After this I am not sure how to proceed with calculating the expectation value as I don't know how to represent

|ψ> = 1/sqrt(2) (|0_A 1_B> +|1_A 0_B))

as a matrix

Any help would be much appreciated!

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2. Nov 8, 2015

### fzero

It is common to choose $|0\rangle$ and $|1\rangle$ to be orthonormal vectors given by the eigenstates of $\sigma_z$, so that
$$|0\rangle=\begin{pmatrix} 1 \\ 0 \end{pmatrix},~~|1\rangle = \begin{pmatrix} 0 \\ 1 \end{pmatrix}.$$
The components of the Bell state would be tensor products of these. You might want to check your notes to see that you are using the same convention.

3. Nov 9, 2015

### ma18

I see, so then taking the tensor product the bell state would be represented as

|ψ> = 1/sqrt(2) (|0_A 1_B> +|1_A 0_B))

= 1/sqrt(2) (0
1
1
0)

4. Nov 9, 2015

### fzero

What you've written is sometimes called the dyadic product of vectors. While it is true that the dyadic product is directly related to the tensor product of vectors, I don't believe that you want to use that here.

The properties of the tensor product of Hilbert spaces that you will want to use are that, given a pair of states belonging to individual Hilbert spaces $|\alpha\rangle_A \in \mathcal{H}_A$ and $|\beta\rangle_B \in \mathcal{H}_B$, we say that the tensor product of the states, $|\alpha\rangle_A \otimes |\beta\rangle_B$ lies in the direct product of the Hilbert spaces $\mathcal{H}_{AB} = \mathcal{H}_A\times \mathcal{H}_B$. Furthermore, given an operator $\mathcal{O}_A$ that acts on $\mathcal{H}_A$ and an operator $\mathcal{O}_B$ that acts on $\mathcal{H}_B$, then we can construct the tensor product operator $\mathcal{O}_{AB} = \mathcal{O}_A\otimes \mathcal{O}_B$, that acts on $\mathcal{H}_{AB}$ via
$$\mathcal{O}_{AB} \left( |\alpha\rangle_A \otimes |\beta\rangle_B \right) = \left(\mathcal{O}_A|\alpha\rangle_A \right)\otimes \left( \mathcal{O}_B|\beta\rangle_B\right).$$

5. Nov 10, 2015

### ma18

I see, then is my formulation of the operator also incorrect (in this context)?

6. Nov 10, 2015

### fzero

Well there is probably a way to make sense of what you've done, but you have a 4x4 matrix that won't directly act in a natural way on either the 2d vectors or that 2x2 matrix you wrote above. I would suggest working things out using the formalism in post #4. Afterwards, you can try to figure out if there's another way to reproduce those results.

7. Nov 10, 2015

### ma18

Alright, but I am unsure how to proceed with that formalism. I'm not the best at quantum but this is the last class I need just to pass in order to graduate...

8. Nov 10, 2015

### fzero

If you just had the operator $\mathcal{O}_A = (\sigma^2)_A$ and a state $|0\rangle_A$, I'm sure that you know how to compute $\mathcal{O}_A |0\rangle_A$. You could do the computation separately for each side of the tensor product in the formula for $\mathcal{O}_{AB}$ and then take the tensor product of the results. Give it a try and post back with any other questions you have.

9. Nov 11, 2015

### ma18

Huh I looked in my notes and we actually do just use the dyadic version and the 4x4 matrix to find the expectation value

In fact it specifies to use the kronecker product to represent the tensor product.

using this matrix represetnation I get 0 for the expectation value

Last edited: Nov 11, 2015
10. Nov 11, 2015

### fzero

I'm sorry, I misread your 4-vector in post #3 as a badly formatted matrix. Taking a pair of 2-vectors to build a 4-vector is precisely the tensor product rather than the dyadic product (which would have given the 2x2 matrix instead). So your method seems correct (I haven't checked the actual answer though.)

11. Nov 11, 2015

### ma18

No worries, I actually do have to do it with a matrix representation and an abstract representation anyway

12. Nov 15, 2015

### ma18

I'm still a little confused about the abstract representation :(