4
$\begingroup$

I have read about how Alice can send Bob a qubit $\alpha |0\rangle + \beta|1\rangle$ if they share an EPR pair. This gives an initial state:

$(\alpha |0\rangle + \beta|1\rangle) \otimes (\frac{1}{\sqrt{2}}|00\rangle + \frac{1}{\sqrt{2}}|11\rangle)$

The first two qubits belong to Alice, the third belongs to Bob.

The first step is for Alice to apply a controlled not from first qubit onto her half of the EPR pair. This gives the result:

$\frac{1}{\sqrt{2}} \big(\alpha (|000\rangle + |011\rangle) + \beta (|110\rangle + |101\rangle)\big)$

Next, let us say that Alice measures her second qubit. This has a 50/50 chance of resulting in a zero or a one. That leaves the system in one of two states:

$\alpha |000\rangle + \beta |101\rangle \quad\text{OR}\quad \alpha |011\rangle + \beta |110\rangle$

If Alice measures the second qubit as zero, she is in the first state. She can tell Bob: "Your half of the EPR is now the qubit I wanted to send you."

If Alice measures the second qubit as one, she is in the second state. She can tell Bob: please apply the matrix

$ \begin{bmatrix} 0 & 1\\ 1 & 0 \end{bmatrix} $

to your qubit to flip the roles of zero and one.

Hasn't Alice teleported her qubit at this point?

The only problem I see is this: Alice must continue not to measure her original qubit. If her unmeasured qubit were to be measured, that would force Bob's qubit to collapse as well.

Is this therefore why Alice needs to apply a Hadamard matrix to her first qubit? Let us apply the Hadamard to the state

$\alpha |000\rangle + \beta |101\rangle$

(This is one of the two possibilities from above). We get:

$ \frac{1}{\sqrt{2}} \big( (\alpha |000\rangle + \beta |001\rangle) (\alpha |100\rangle - \beta |101\rangle) \big) $

Alice measures her first qubit now. If it is a zero, she can tell Bob: your qubit is fine. If it is one, she can tell Bob: you need to fix it from $\alpha |100\rangle - \beta |101\rangle$ (by an appropriate rotation).

Finally, my questions are:

  1. If Alice is okay with sharing an entangled copy of the transfered qubit with Bob, can she send just the first classical bit?
  2. Is the application of the Hadamard simply to separate Alice's first qubit from Bob's qubit?
  3. It is the application of the Hadamard to Alice's first qubit, followed by the measurement, which may disturb Bob's qubit, possibly necessitating a "fixup." The second classical bit is transferred to communicate whether the fixup is needed. Am I correct?
  4. The reason Alice wants Bob to have a qubit unentangled from her own is probably because it is burdensome for Alice to keep an entangled copy from being measured. Correct?

Sorry for the very long and rambly question. I think I understand, but maybe this writeup will help someone else.

$\endgroup$
6
$\begingroup$

Your initial calculations are correct. When Alice performs her first measurement and gets a 0 outcome then, as you say, Alice and Bob are left sharing a two-qubit state $$ |\Psi\rangle=\alpha|00\rangle+\beta|11\rangle $$ (you can safely ignore the measured qubit). The problem is the statement

She can tell Bob: "Your half of the EPR is now the qubit I wanted to send you."

because it doesn't make sense. It kind of looks like the amplitudes are doing the right sort of thing, but it's not quite enough. The two qubits are entangled, and you need the second measurement to remove that entanglement.

To see this more clearly, what it should mean if Bob has received the state is that if he performs a measurement that projects onto one of two states $|\psi\rangle=\alpha|0\rangle+\beta|1\rangle$ or $|\psi^\perp\rangle=\beta|0\rangle-\alpha|1\rangle$, then Bob must always get the answer corresponding to $|\psi\rangle$. So, let's try it. We get the required answer with probability $$ \langle\Psi|(\mathbb{I}\otimes|\psi\rangle\langle\psi|)|\Psi\rangle=|\alpha|^4+|\beta|^4 $$ Since we know that $|\alpha|^2+|\beta|^2=1$, this is not generally equal to 1. (It could be if $\alpha\beta=0$). So, you can see that the outcomes of experiments that Bob performs on his half of the state $|\Psi\rangle$ do not match up perfectly with what he should get if he were experimenting with the state $|\psi\rangle$ which he should have.

Another way of thinking about it is that if Bob holds the state $|\psi\rangle$ that he's supposed to after teleportation, nothing that Alice can do should be able to change it. But when they share $|\Psi\rangle$, she certainly can change it. That's why the measurement basis of her second measurement is so important - there must be no Z component to the measurement because otherwise she would learn something about the state being sent (in other words, she'd collapse the state $|\psi\rangle$).

In summary:

  1. If Alice is okay with sharing an entangled copy of the transfered qubit with Bob, can she send just the first classical bit?

No. Bob has not received the state at this point, as I hope I've conveyed.

  1. The reason Alice wants Bob to have a qubit unentangled from her own is probably because it is burdensome for Alice to keep an entangled copy from being measured. Correct?

No, it's so that Bob actually receives the state he's supposed to.

  1. Is the application of the Hadamard simply to separate Alice's first qubit from Bob's qubit?

No. The separation doesn't happen at this point. It only happens in the act of measurement that follows. The Hadamard is important for changing the state so that Alice only removes the entanglement and does not collapse the state of $|\psi\rangle$

  1. It is the application of the Hadamard to Alice's first qubit, followed by the measurement, which may disturb Bob's qubit, possibly necessitating a "fixup." The second classical bit is transferred to communicate whether the fixup is needed. Am I correct?

Pretty much. The measurement is guaranteed to disturb the qubit, because it does change after the measurement. But, as you say, the bit of communication is required to convey how it changed, and therefore what fix is required.

$\endgroup$
  • $\begingroup$ Thank you @DaftWullie for your quick and thorough reply! (And for entertaining such an open-ended question!) I am still unpacking this equation:equation: $$\langle\Psi|(\mathbb{I}\otimes|\psi\rangle\langle\psi|)|\Psi\rangle=|\alpha|^4+|\beta|^4$$ I'm trying to learn informally through a combo of Scott Aaronson's book plus Wikipedia. My knowledge of formal notation is thus limited. I'll be studying this today! I think the middle is the density operator, and you calculate a probability by multiplying both sides by $\Psi$. But I still have work to do before I understand! Thank you! $\endgroup$ – Ned Ruggeri Oct 18 '18 at 17:13
  • $\begingroup$ Thank you! I have studied more and see you are correct! Thank you very much! This is also useful: An Introduction to Quantum Computing, Without the Physics $\endgroup$ – Ned Ruggeri Oct 25 '18 at 2:05
1
$\begingroup$

I wanted to expand on DaftWullie's accepted answer in case it is helpful to others. DaftWullie's is correct; mine is only a continuation of the same idea.

Consider the one-qubit state $\psi = \frac{1}{\sqrt{2}}(|0\rangle + |1\rangle)$. We know that applying the Hadamard matrix $H = \frac{1}{\sqrt{2}}\begin{bmatrix}1& 1 \\ 1 & -1\end{bmatrix}$ will give us $H\psi = (\frac{1}{2} + \frac{1}{2}) |0\rangle + (\frac{1}{2} - \frac{1}{2}) |1\rangle = |0\rangle$.

Of course the reason is constructive and destructive interference.

Now let's consider a two qubit system with state $\psi = \frac{1}{\sqrt{2}} (|00\rangle + |11\rangle)$. Let us say that we want to apply $H$ to the second qubit, but do nothing to the first (apply the identity $I$).

What matrix represents this transformation? The answer is $I \otimes H$. What is that? It is this matrix:

$$ \begin{bmatrix} H & 0 \\ 0 & H \end{bmatrix} = \frac{1}{\sqrt{2}}\begin{bmatrix} 1 & 1 & 0 & 0 \\ 1 & -1 & 0 & 0 \\ 0 & 0 & 1 & 1 \\ 0 & 0 & 1 & -1 \end{bmatrix} $$

Can you see why this is? It is because the the overall operation should move

$$ \begin{align} |00\rangle &\mapsto \frac{1}{\sqrt{2}}(|00\rangle + |01\rangle) \\ |01\rangle &\mapsto \frac{1}{\sqrt{2}}(|00\rangle - |01\rangle) \end{align} $$

and likewise for $|10\rangle, |11\rangle$. This is because the Hadamard application to the second qubit is not changing the first qubit.

Now, you may verify that $I \otimes H$ applied to $(\alpha |0\rangle + \beta |1\rangle) \otimes \frac{1}{\sqrt{2}}(|0\rangle + 1\rangle)$ gives you $\alpha |00\rangle + \beta|10\rangle$. When the qubits are not entangled, then you can do the intuitive thing and apply $I$ and $H$ independently to each qubit.

However, the story is totally different when the two qubits are entangled! Consider the application to $\frac{1}{\sqrt{2}} ( |00\rangle + |11\rangle)$. In this case, you get the result:

$$ \begin{bmatrix} \frac{1}{2} \\ \frac{1}{2} \\ \frac{1}{2} \\ -\frac{1}{2} \\ \end{bmatrix} $$

Can you see what happened here? Applying $I \otimes H$ no longer mean that the $|00\rangle \mapsto \frac{1}{\sqrt{2}} (|00\rangle + |01\rangle)$ part destructively interferes with the $|11\rangle \mapsto \frac{1}{\sqrt{2}} (|10\rangle - |11\rangle)$ part to cancel out the possibility that the second qubit takes on the value $|1\rangle$. That's because $|01\rangle$ and $|11\rangle$ are orthogonal basis states.

And now you can see my mistake. As DaftWullie explained. A qubit cannot be considered transmitted when it is still entangled. That's because an operation on one of the entangled qubits will not give the result that would have been expected if the qubit were not entangled.

Hope this helps! Thanks to all!

$\endgroup$

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.