# I Mathematics of tensor products in the Bell states

Tags:
1. Mar 30, 2016

### PerilousGourd

I'm having trouble with the mathematics of tensor products as applied to Bell states.

Say I have the state
\begin{align*} \left|\psi\right> &= \frac{1}{\sqrt{2}} \left(\left|0\right>_A \otimes \left|0\right>_B + \left|1\right>_A \otimes \left|1\right>_B\right) \end{align*}

How would the tensor products be expanded here? Are the states $\left|0\right>_A$ etc one dimensional?

I'd usually expand tensor products like

$$\left|\psi\right> \otimes\left|\phi\right> = \left(\matrix{\psi_1\\\psi_2}\right) \otimes \left(\matrix{\phi_1\\\phi_2}\right) = \left(\matrix{\psi_1\phi_1\\\psi_1\phi_2\\\psi_2\phi_1\\\psi_2\phi_2}\right)$$

In this case, is it just

$$\left|0\right>_A \otimes\left|0\right>_B = \left(\matrix{0_A0_B}\right)$$ ?

And $\left(0_A0_B\right)$ is equivalent to $\left|00\right>$ and so on by convention of notation (this makes me slightly uncomfortable; can any scalar be denoted a ket?), so that the original state can be written

\begin{align*} \left|\psi\right> &= \frac{1}{\sqrt{2}} \left(\left|00\right> + \left|11\right>\right) \end{align*}

Was my working here correct? Or is there some funky $\left|0\right> = \left(\matrix{1\\0}\right)$ business going on, like I've seen in places?

2. Mar 30, 2016

### andrewkirk

There's no reason to expand the tensor products. They are basis elements of the tensor product space obtained by treating A and B as a single system. The notation $|00\rangle$ is just a different way of writing the same thing as $|0\rangle{}_A\otimes |0\rangle{}_B$.
I have not seen the notation $(0_A0_B)$ before. Either it's just another form of notation for the same thing (in which case it's a ket, not a scalar), or else it's not equal to either of the other two.

3. Mar 30, 2016

### PerilousGourd

No reason to, or it actually can't be done? If it can be done, I'd love to see the process, as it would help with my intuition a lot.

How can you tell when $\otimes$ should be interpreted as a tensor product to be expanded and when it should be interpreted another way?

If $\left|0\right> = \left(\matrix{1\\0}\right)$ (where physically this represents light polarized in the horizontal direction), how do you know when to continue as
$$\left|0\right> \otimes \left|0\right> = \left(\matrix{1\\0} \right) \otimes \left(\matrix{1\\0}\right) = \left(\matrix{1&1\\1&0\\0&1\\0&0}\right)$$
which is surely not the same as $\left|00\right>$, and when to continue as
$$\left|0\right> \otimes \left|0\right> = \left|00\right>$$
?

The $0_A0_B$ notation before meant nothing unusual and was bad labelling on my part, sorry.

4. Mar 30, 2016

### andrewkirk

A vector can always be expanded in terms of a basis. A vector whose expansion is a single term in one basis will typically have a multi-term expansion in another basis. Say $|\psi\rangle$ is a basis ket for basis $B$, and $B'$ is a different basis. Then we can expand that in the $B'$ basis as:

$$|\psi\rangle=\sum_{\phi\in B'}|\phi\rangle\langle\phi|\psi\rangle$$

It's always a tensor product. Whether we want to expand it in a particular basis depends on whether we think that might help us towards our computational goal.

I'm afraid I don't know what this equality is meant to signify, as I don't know what is meant by the 4 x 2 matrix on the RHS. The space that is the tensor product of two 2D vector spaces is a 4D vector space, so its elements can be represented as 4-vectors or 2 x 2 matrices - eg $\left(\matrix{1\\0\\1\\0}\right)$ or $\left(\matrix{1&1\\0&0}\right)$. But I can't see any natural way to represent them as 4 x 2 matrices without redundancy.

5. Mar 31, 2016

### Zafa Pi

The tensor product of two 2-D unit vectors can be represented as a 4-D unit vector as you indicated where ψ1ϕ1 is regular multiplication.
The tensor product space is all 4-D unit vectors. However not every member of the tensor product space is the tensor product of two vectors, e.g. the superposition 1/√2(|00⟩+|11⟩) = 1/√2[1,0,0,1]. Such vectors (states) are said to be entangled.