Mathematics of tensor products in the Bell states

[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?

 

[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.

 

[PerilousGourd]

Thank you for your reply!

There's no reason to expand the tensor products.

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.

 

[andrewkirk]

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.

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:

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

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

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.

$$\left(\matrix{1\\0} \right) \otimes \left(\matrix{1\\0}\right) = \left(\matrix{1&1\\1&0\\0&1\\0&0}\right)$$

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.

 

[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.