Verifying the "quantum teleportation identity" in $\mathbb{C}^2 \otimes \mathbb{C}^2$ (Bell basis)

Question

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Let $\{\psi_{i,j} : i, j = 0,1\}$ be the Bell basis of $\mathbb{C}^2 \otimes \mathbb{C}^2$. Let $\psi \in \mathbb{C}^2$. Let $X = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}$ and $Z = \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix}$.

In particular, this means: $\psi_{00} = \frac{1}{\sqrt 2}(\ket{00} + \ket{11})$, $\psi_{01} = \frac{1}{\sqrt 2}(\ket{01} + \ket{10})$, $\psi_{10} = \frac{1}{\sqrt 2}(\ket{00} - \ket{11})$, and $\psi_{11} = \frac{1}{\sqrt 2}(\ket{01} - \ket{10})$. Also, $\psi = \alpha \ket{0} + \beta \ket{1}$ for some $\alpha, \beta \in \mathbb{C}$.

I am given the following identity:

$\psi \otimes \psi_{00} = \frac{1}{2}(\psi_{00} \otimes \psi + \psi_{01} \otimes X \psi + \psi_{10} \otimes Z \psi + \psi_{11} \otimes XZ \psi)$

However, I am not sure where exactly this identity is coming from (I am unclear on how distributivity works with tensor products, and I seem to be making some algebraic mistakes somewhere).

All that is really clear to me is that $X \psi = \alpha \ket{1} + \beta \ket{0}$, $Z \psi = \alpha \ket{0} - \beta \ket{1}$, and $XZ \psi = \alpha \ket{1} - \beta \ket{0}$.

Would anyone be able to give a rundown on how to expand the left hand side of the equation to get the right? Or, how to work backwards from the right hand side to get back to the left?

Edit: The rough work I did is added below.

The LHS is: \begin{align} \psi \otimes \psi_{00} & = (\alpha \ket{0} + \beta \ket{1} )\otimes \frac{1}{\sqrt 2}(\ket{00} + \ket{11}) \\ & = \alpha \ket{0} \otimes \frac{1}{\sqrt 2}(\ket{00}) + \alpha \ket{0} \otimes \frac{1}{\sqrt 2}(\ket{11}) \\ & \qquad + \beta \ket{1} \otimes \frac{1}{\sqrt 2}(\ket{00}) + \beta \ket{1} \otimes \frac{1}{\sqrt 2}(\ket{11}) \end{align}

The RHS is:
\begin{align} \psi_{00} \otimes \psi & = \frac{1}{\sqrt 2}(\ket{00} + \ket{11}) \otimes (\alpha \ket{0} + \beta \ket{1}) \\ & = \frac{1}{\sqrt 2}(\ket{00}) \otimes \alpha \ket{0} + \frac{1}{\sqrt 2}(\ket{00}) \otimes \beta \ket{1} \\ & \qquad + \frac{1}{\sqrt 2}(\ket{11}) \otimes \alpha \ket{0} + \frac{1}{\sqrt 2}(\ket{11}) \otimes \beta \ket{1} % \\ & \\ % \psi_{01} \otimes X\psi & = \frac{1}{\sqrt 2}(\ket{01} + \ket{10}) \otimes (\alpha \ket{1} + \beta \ket{0}) \\ & = \frac{1}{\sqrt 2}(\ket{01}) \otimes \alpha \ket{1} + \frac{1}{\sqrt 2}(\ket{01}) \otimes \beta \ket{0} \\ & \qquad + \frac{1}{\sqrt 2}(\ket{10}) \otimes \alpha \ket{1} + \frac{1}{\sqrt 2}(\ket{10}) \otimes \beta \ket{0} % \\ & \\ % \psi_{10} \otimes Z\psi & = \frac{1}{\sqrt 2}(\ket{00} - \ket{11}) \otimes (\alpha \ket{0} - \beta \ket{0} ) \\& = \frac{1}{\sqrt 2}(\ket{00}) \otimes \alpha \ket{0} + \frac{1}{\sqrt 2}(\ket{00}) \otimes (-\beta) \ket{1} \\ & \qquad + \big(-\frac{1}{\sqrt 2}(\ket{11})\big) \otimes \alpha \ket{0} - \frac{1}{\sqrt 2}(\ket{11}) \otimes \beta \ket{1} % \\ & \\ % \psi_{11} \otimes Xz\psi & = \frac{1}{\sqrt 2}(\ket{01} - \ket{10}) \otimes (\alpha \ket{1} - \beta \ket{0}) \\ & = \frac{1}{\sqrt 2}(\ket{01}) \otimes \alpha \ket{1} + \frac{1}{\sqrt 2}(\ket{01}) \otimes (-\beta) \ket{0} \\ & \qquad + \big(-\frac{1}{\sqrt 2}(\ket{10})\big) \otimes \alpha \ket{1} - \frac{1}{\sqrt 2}(\ket{10}) \otimes \beta \ket{0} \end{align}

I don't think this is correct, because it seems all the terms with $\beta$ cancel out, but clearly those need to be part of the LHS.

Edit 2: Also, does the fact that the LHS and RHS have tensor products in "reverse" order matter at all? Or does it not matter since $X \otimes Y \simeq Y \otimes X$?

Answer 1

I am unclear on how distributivity works with tensor products

The same way that it works everywhere else: \begin{align} |u⟩ \otimes (\alpha |x⟩ + \beta |y⟩) & = \alpha |u⟩ \otimes |x⟩ + \beta |u⟩ \otimes |y⟩, \\ (\alpha|u⟩ + \beta|v⟩) \otimes |x⟩ & = \alpha|u⟩ \otimes |x⟩ + \beta|v⟩ \otimes |x⟩. \end{align} If it doesn't distribute over vector addition, then we simply do not use the name "product".

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As for the rest, it's just plain algebra - substitute everything in, crank the algebra handle, and verify that both sides reduce to the same huge expression when boiled down to their individual components in the computational basis.

If the problem is that all the terms in $\beta$ are cancelling out, then the thing to do is to design methods to diagnose your computations explicitly. Thus, as an example, what happens to your calculations if you set $\alpha=0$ and $\beta=1$? Presumably you'll get a zero somewhere, but where, and why? This kind of simplification is generally the kind of tool that helps you pinpoint which algebraic errors led to incorrect cancellations.

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Regarding your second edit,

Edit 2: Also, does the fact that the LHS and RHS have tensor products in "reverse" order matter at all? Or does it not matter since $X \otimes Y \simeq Y \otimes X$?

No, it doesn't matter, but it's not because $X \otimes Y$ and $Y \otimes X$ are isomorphic. (They are indeed isomorphic, but that isomorphism is explicitly not being called upon here.) The reason why it doesn't matter is that you don't just have two tensor factors, you have three - the two qubits in the initial entangled state, and the qubit whose state you want to teleport. Mathematically, the key property isn't commutativity, but associativity: $$ (X \otimes Y) \otimes Z \simeq X \otimes (Y \otimes Z), $$ with a unique natural isomorphism, so in physics we just drop the $\simeq$ pretense and just write $=$. That means, in particular, that there's no need for complicated notation like $|01⟩ \otimes |0⟩$ - you can just write $|010⟩$. It is vital that you keep careful track of the order in which you're using the tensor-product factors, but as long as you do that, you can just mush them together into one single blob without any problem.