13.9.1: Teleportation - Alice and Bob’s story

Alice and Bob, entangled a pair of qubits \(| \phi_{AB}\rangle\) when they first met, as is the tradition in the 21st century, people no longer shake hands.

\[\left|\;\phi_{A B}\right\rangle=\frac{1}{\sqrt{2}}\;\left(\left|\;0_{A}\right\rangle \otimes\left|\;0_{B}\right\rangle+\left|\;1_{A}\right\rangle \otimes\left|\;1_{B}\right\rangle\right) \tag{13.64} \]

They shared a good time, but after school, life took each of them through separate paths. However, they each kept their piece of the entangled pair (it is unlawful not to do so in the 21st century). Now, the pair looks like

\[\left|\;\phi_{A B}\right\rangle=\frac{1}{\sqrt{2}}\;\left(\left|\;0_{A}\right\rangle \stackrel{\text{far}}{\otimes}\left|\;0_{B}\right\rangle+\left|\;1_{A}\right\rangle \stackrel{\text{far}}{\otimes}\left|\;1_{B}\right\rangle\right) \tag{13.65} \]

Alice has now decided to come forward and confess to Bob her love for him. However, she is afraid of rejection and she has heard that qubits can be “teleported instantly”, so she decides that the best course of action is to send him a love letter in a qubit \(|\; \psi_L\rangle = \alpha \; |\; 0_L\rangle + \beta \;|\; 1_L\rangle\) (it could not have been otherwise, love is in the ket).

To do so, Alice carefully puts the qubit of the pair she once entangled with Bob in a composite system with her love-ket-letter \(|\; \psi_L\rangle\). The complete three-qubit system can be represented using tensor products

\[\left|\phi_{A} \psi_{L} \phi_{B}\right\rangle=\frac{1}{\sqrt{2}}\;\left(\left|\;0_{A}\right\rangle \otimes\left(\;\alpha\left|\;0_{L}\right\rangle+\beta\left|\;1_{L}\right\rangle\right) \stackrel{\text{far}}{\otimes}\left|\;0_{B}\right\rangle+\left|\;1_{A}\right\rangle \otimes\left(\;\alpha\left|\;0_{L}\right\rangle+\beta\left|\;1_{L}\right\rangle\right) \stackrel{\text{far}}{\otimes}\left|\;1_{B}\right\rangle\right) \tag{13.66} \]

Note that the order in the cross product is not relevant. It only matters when multiplying two composite systems, there we must ensure that each subsystem appears in the same position in each of the kets we multiply.

Alice has now a two-qubit system, and Bob, far away as he is, has one qubit which is entangled with one of the two qubits Alice has. Next, she takes the Bell-Analyzer-1001 (see Figure 13.8), a gift from Bob that she has cherished all this time, and uses it on her two qubits. The Bell analyzer does the following , starting

Figure 13.8: Bell analyzer.

from the initial state

\[\left|\;\phi_{A} \psi_{L} \phi_{B}\right\rangle=\frac{1}{\sqrt{2}}\;\left[\left|\;0_{A}\right\rangle \otimes\left(\;\alpha\left|\;0_{L}\right\rangle+\beta\left|\;1_{L}\right\rangle\right) \stackrel{\text{far}}{\otimes}\left|\;0_{B}\right\rangle+\left|\;1_{A}\right\rangle \otimes\left(\;\alpha\left|\;0_{L}\right\rangle+\beta\left|\;1_{L}\right\rangle\right) \stackrel{\text{far}}{\otimes}\left|\;1_{B}\right\rangle\right] \nonumber \]

The CNOT gate has the effect of coupling the love-ket-letter with Alice’s qubit, and indirectly, since Alice’s qubit is entangled with Bob’s, with Bob’s as well. If \(\psi_L\) where equal to \(|\; 0\rangle\), Alice’s other qubit would go unmodified (this is the first line in the equation below). If on the contrary it were \(|\; 1\rangle\), it would be added modulo 2 to Alice’s other qubit, or, in other words, Alice’s other qubit would be flipped (this is what happens in the second line below). Since \(\psi_L\) is itself a superposition, what it ends up happening is that a fraction of the qubit remains unmodified with amplitude \(\alpha\) and the other fraction is flipped, with amplitude \(\beta\). So in practice what the CNOT does is transfer the superposition to Alice’s Qubit

\[\begin{align*}
=\frac{1}{\sqrt{2}} \alpha\left(\left|\;0_{A}\right\rangle \stackrel {\text{far}}{\otimes}\left |\;0_{B}\right\rangle+\left|\;1_{A}\right\rangle \stackrel{\text{far}}{\otimes}\left|\;1_{B}\right\rangle\right) \otimes\left|\;0_{L}\right\rangle \\
+\frac{1}{\sqrt{2}} \beta\left(\left|\;1_{A}\right\rangle \stackrel {\text{far}}{\otimes}\left|\;0_{B}\right\rangle+\left|\;0_{A}\right\rangle \stackrel {\text{far}}{\otimes}\left|\;1_{B}\right\rangle\right) \otimes\left|\;1_{L}\right\rangle
\end{align*} \nonumber \]

At this point Alice’s and Bob’s qubit have both the information of the superposition that was originally in the love-ket-letter. The Hadamard gate produces a new superposition out of the love-ket-letter as follows

\[\begin{align*}
=\frac{1}{\sqrt{2}} \alpha\left(\left|\;0_{A}\right\rangle\stackrel{\text{far}}{\otimes}\left|\;0_{B}\right\rangle+\left|\;1_{A}\right\rangle \stackrel{\text{far}}{\otimes}\left|\;1_{B}\right\rangle\right) \otimes \dfrac{1}{\sqrt{2}}\left(\left|\;0_{L}\right\rangle+\left|\;1_{L}\right\rangle\right) \\
+\frac{1}{\sqrt{2}} \beta\left(\left|\;1_{A}\right\rangle \stackrel{\text{far}}{\otimes}\left|\;0_{B}\right\rangle+\left|\;0_{A}\right\rangle \stackrel{\text{far}}{\otimes}\left|\;1_{B}\right\rangle\right) \otimes \dfrac{1}{\sqrt{2}}\left(\left|\;0_{L}\right\rangle-\left|\;1_{L}\right\rangle\right)
\end{align*} \nonumber \]

At this point the information about the superposition in the original ket-love-letter is no longer in Alice’s hands. However, to appreciate that it is so, we need to make some manipulations and reordering of the cross products. The Expression above can be separated in two sides, what Alice has and what Bob has, and then it breaks down into four terms that have a clearer interpretation

\[\begin{align*}
&= \frac{1}{2}\left|\;0_{A}\right\rangle \otimes\left|\;0_{L}\right\rangle \stackrel{\text{far}}{\otimes}\left(\alpha\left|\;0_{B}\right\rangle+\beta\left|\;1_{B}\right\rangle\right) \\
&\quad +\frac{1}{2}\left|\;0_{A}\right\rangle \otimes\left|\;1_{L}\right\rangle \stackrel{\text{far}}{\otimes}\left(\;\alpha\;\left |\;0_{B}\right\rangle-\beta\left|\;1_{B}\right\rangle\right) \\
&\quad +\frac{1}{2}\left|\;1_{A}\right\rangle \otimes\left|\;0_{L}\right\rangle \stackrel{\text{far}}{\otimes}\left(\;\alpha\;\left |\;1_{B}\right\rangle+\beta\left|\;0_{B}\right\rangle\right) \\
&\quad +\frac{1}{2}\left|\;1_{A}\right\rangle \otimes\left|\;1_{L}\right\rangle \stackrel{\text{far}}{\otimes}\left(\;\alpha\;\left|\;1_{B}\right\rangle-\beta\left|\;0_{B}\right\rangle\right) \tag{13.67}
\end{align*} \nonumber \]

The manipulation we did leaves no doubt, all the information about the original superposition is in Bob’s side. However the information reached bob as a superposition of the right love-ket-letter and all the possible errors produced by phase change (sign) and bit flips!.

Alice realizes now that there is no way for her to avoid talking with Bob about her love-ket-leter. Bob has the information, but to recover the exact superposition from the love-ket-letter he needs to know how to unscramble it, and only Alice can tell him. The next steps are the key to faultless teleportation

• Alice measures her two qubits, she will obtain either of or \(|\; 0_A0_L\rangle, |\; 0_A1_L\rangle, |\; 1_A0_L\rangle\), or \(1_A1_L\rangle\) with equal probability.
• Upon Alice’s measurement, Bob’s qubit takes the value of one of the four possible superpositions. And so, the result of her measurement can help Bob unscramble his bit
• If she measured \(|\; 0_A0_L\rangle\), she will tell Bob not to do anything to his qubit. If she measured \(|\; 0_A1_L\rangle\), Bob will have to correct for the phase (that can be done with a \(Z\) gate). If she measured \(|\; 1_A0_L\rangle\), the states in the message have been flipped, and to unflip them Bob will have to use a bit-flip (a.k.a not, a.k.a X) gate. Finally if she measured \(|\; 1_A1_L\rangle\), Bob will have to correct both for the phase and the bit flip.

In total, Alice tells Bob to follow one of 4 possibilities, so she needs to communicate to him 2 classical bits, and then Bob will be able to read the love-ket-letter.

Now Bob better be ready to understand the love letter, because Alice no longer has it! Notice how quantum teleportation transfers the information about the superposition instantly, but Alice needs to measure her system and tell to Bob the result for him to unscramble his bit. In a sense, after teleportation, Bob is doing error correction based on the syndrome measured by Alice; the first bit of Alice informs about bit-flip errors and the second about phase errors. When we get to talk about quantum error correction, we will see that these are the two types of errors that appear in quantum communication.

It is important to appreciate that Alice can only communicate the syndrome to Bob classically, hence, instantaneous teleportation is impossible, Bob will not be able to read the love letter in less time than “far”/”speed of light”. Remember the example of the yawn-connection?, there too observers had to exchange data to ensure that communication had occurred.

Quantum teleportation has served us as a verification of the no-cloning theorem, (Alice no longer has a copy of the love letter, much like it happens with “snail” mail), and has helped us introduce the notion of error and error correction. We have also seen that instantaneous communication is not possible, because classical communication is needed to correct for the errors.