5
$\begingroup$

Reading about superdense coding I came upon a calculation I can not understand.

We have an EPR entangled pair of qubits $\frac{1}{\sqrt2}(|00\rangle + |11\rangle)$ and we want to apply a Pauli X gate to the first of the entangled qubits. On wikipedia, there is an example showing that such an operation would lead to the state $\frac{1}{\sqrt2}(|10\rangle + |01\rangle)$. I can't see an intuition behind it. I know that the Pauli X gate operates as a quantum NOT, but at the same time I do not know how to write it algebraically.

How can a single qubit from an entangled pair be represented? Is it $\frac{1}{\sqrt2}(|0\rangle + |1\rangle)$ for the pair above? And if so, why does the Pauli X gate change it at all (if we wrote the operation using matrices, the qubit would be represented as $\left[\frac{1}{\sqrt2} \frac{1}{\sqrt2}\right]$, which multiplied by the Pauli X gate matrix would not change at all...)

$\endgroup$
5
$\begingroup$

I came to a conclusion that we do not actually take away a single qubit from an entangled pair and operate it, but operate on both qubits with Identity operator on the one we do not want to change.

So basically to apply Pauli X gate to the first qubit from an entangled pair, we apply X tensored with $I$ to the entangled pair to get the result. That is, to operate a not on the first qubit, we evolve the system by $$ \sigma_x \otimes I = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} \otimes \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1\\ 1 & 0 & 0 & 0\\ 0 & 1 & 0& 0 \end{pmatrix} $$

Intuitively it is like applying a single operation to the first qubit and identity to the second one (effectively not changing its state).

Thus if we start with the entangled state $|\psi\rangle \frac{1}{\sqrt2}(|00\rangle + |11\rangle)$ we end up with $$ (\sigma_x \otimes I)|\psi\rangle = \begin{pmatrix} 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1\\ 1 & 0 & 0 & 0\\ 0 & 1 & 0& 0 \end{pmatrix} \cdot \frac1{\sqrt{2}} \begin{pmatrix}1 \\0 \\0 \\1 \end{pmatrix} = \frac1{\sqrt{2}} \begin{pmatrix}0 \\1 \\1 \\0 \end{pmatrix} = \frac1{\sqrt{2}}(|01\rangle+|10\rangle) $$

$\endgroup$
  • $\begingroup$ In the first displayed equation, you got the $4\times4$ matrix wrong (it's right in the second displayed equation). The matrix you wrote would be a controlled NOT. $\endgroup$ – celtschk Dec 10 '14 at 21:38
  • $\begingroup$ @celtschk a general state in this case is $a|00\rangle + b|01\rangle + c|10\rangle + d|11\rangle$, isn't? So, what is the first qubit here? Is it $a|00\rangle + b|01\rangle$ ? Thanks $\endgroup$ – Hilder Vítor Lima Pereira Dec 4 '16 at 13:30
  • $\begingroup$ @Vitor: The first qubit isn't in a pure state, that is, its state cannot be described by a state vector. You have to describe it with a density matrix. See my answer for details. Note that what you wrote down is not really a general state, but a general pure state for two qubits. And $a|00\rangle+b|01\rangle$ is again a two-qubit state. $\endgroup$ – celtschk Dec 4 '16 at 14:19
  • $\begingroup$ @celtschk ok. Nielsen and Chuang's book sometimes uses concepts that will be rigorously defined later, this is why I'm still confused about that. But I will keep reading it and I hope everything becomes clear. Thank you very much! $\endgroup$ – Hilder Vítor Lima Pereira Dec 4 '16 at 14:44
3
$\begingroup$

While you have found the answer to your main question yourself, let me answer the sub-question you asked as part of the main question:

How can a single qubit from an entangled pair be represented?

A single qubit from an entangled pair cannot be represented by a state vector, but has to be represented by a density matrix. More exactly, it is represented by the partial trace over the other qubit of the entangled state's density matrix.

For example, for the state $$\left|\phi^+\right\rangle = \frac{1}{\sqrt{2}}\left(\left|00\right\rangle + \left|11\right\rangle\right)$$ the corresponding two-qubit density matrix is the projector onto that state: $$\rho_{\phi^+}=\begin{pmatrix} \color{green}{\tfrac{1}{2}} & \color{green}{0} & \color{red}{0} & \color{red}{\tfrac{1}{2}}\\ \color{green}{0} & \color{green}{0} & \color{red}{0} & \color{red}{0}\\ \color{blue}{0} & \color{blue}{0} & 0 & 0\\ \color{blue}{\tfrac{1}{2}} & \color{blue}{0} & 0 & \tfrac{1}{2} \end{pmatrix}$$ Here I've marked with the same colour the elements which belong to the same indices for the first qubit; for example, the $\color{green}{\text{green}}$ numbers are those matrix elements where both indices of the first qubit are $\color{green}{0}$.

Now the partial trace over the second subsystem just means the trace over those submatrices: $$\operatorname{tr}_B \rho_{\phi^+} = \begin{pmatrix} \color{green}{\tfrac{1}{2}} & \color{red}{0}\\ \color{blue}{0} & \tfrac{1}{2} \end{pmatrix}$$ Note that this is the completely mixed one-qubit state. That is, the single qubit has a completely undetermined state.

Note that also the state of the second qubit is completely mixed. Which shows that the information about the entangled state is not in the individual qubits. This can also be seen in the fact that the information in half of the density matrix elements is completely ignored by the partial trace (this includes in particular the upper-right and lower-left $\frac{1}{2}$ elements, which are exactly what distinguish that entangled state from a separable equal mixture of the states $\left|00\right\rangle$ and $\left|11\right\rangle$).

$\endgroup$
2
$\begingroup$

@3yakuya

I think you made a mistake writing the multiplication between Pauli Matrix X tensor product with Identity matrix, you wrote

$$ \sigma_x \otimes I = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} \otimes \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 0 & 1 & 0 & 0\\ 1 & 0 & 0 & 0\\ 0 & 0 & 0 & 1\\ 0 & 0 & 1& 0 \end{pmatrix} $$

the correct one is

$$ \sigma_x \otimes I = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} \otimes \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1\\ 1 & 0 & 0 & 0\\ 0 & 1 & 0& 0 \end{pmatrix} $$

Claim: Why I say this is true; because according to the definition of Tensor Product, they apply as following, and Tensor product doesn't follow the commutative rule; so for example $ A \otimes B =! B \otimes A $; if that were true, then both the above products are correct..

https://en.wikipedia.org/wiki/Tensor_product#Tensor_product_of_linear_maps

so the answer would be the same (this is not always true)

$|\psi\rangle = \frac{1}{\sqrt2}(|00\rangle + |11\rangle)$ $$ (\sigma_x \otimes I)|\psi\rangle = \begin{pmatrix} 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1\\ 1 & 0 & 0 & 0\\ 0 & 1 & 0& 0 \end{pmatrix} \cdot \frac1{\sqrt{2}} \begin{pmatrix}1 \\0 \\0 \\1 \end{pmatrix} = \frac1{\sqrt{2}} \begin{pmatrix}0 \\1 \\1 \\0 \end{pmatrix} = \frac1{\sqrt{2}}(|10\rangle+|01\rangle) $$


Note: 1 - I would like to comment in the "comment box" but I couldn't so this is why I write this in "Answer box" ..

2 - I would see how @3yakuya will respond because may be he has an argument.

$\endgroup$
  • 1
    $\begingroup$ You are exactly right! I edited my answer, the matrices were incorrect (I must have written them looking at some bad notes, as those equations outcomes were correct although they surely couldn't have been.) Thank you very much! $\endgroup$ – 3yakuya Jun 21 '15 at 20:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.