Today, we derived Bell's inequality in the form in Nielsen and Chuang, and talked about the no-cloning theorem, superdense coding, and teleportation. The no-cloning theorem says that if you have one copy of an unknown quantum state | ψ 〉, you cannot produce two copies of it. To prove the no-cloning theorem, we need to show that the map
| ψ 〉 ⊗ | 0 〉 → | ψ 〉 ⊗ | ψ 〉is not unitary. We can easily do this by checking that it does not preserve inner products. This is a straightforward calculation. To be rigorous, we might also want to make sure that there isn't a way of cloning that also produces some extra garbage. This may seem like splitting hairs, but it might actually be a real concern in some situations. Surprisingly, the map that comes closest to cloning quantum states, mapping | ψ 〉 to two close but imperfect copies of | ψ 〉, really does produce extra garbage. To prove this stronger version, we need to show that the map
| ψ 〉 ⊗ | 0 〉 ⊗ | 0 〉 → | ψ 〉 ⊗ | ψ 〉 ⊗ | φ 〉is not unitary, where | φ 〉 may depend on | ψ 〉. This is also easy, as a straightforward calculation shows that this map also does not preserve inner products.
I next talked about superdense coding and teleportation. For these, we need some facts about Bell states. It is fairly easy to show that
1/√2 σx ⊗ id ( | 00 〉 + | 11 〉 ) = 1/√2 ( | 01 〉 + | 10 〉 )Thus, you can obtain each of the four Bell states by starting with 1/√2 ( | 00 〉 + | 11 〉 ), and applying one of the four Pauli matrices to the first qubit (we will take the identity matrix to be the fourth Pauli matrix, and call it σ0). If instead, we applied the Pauli matrix to the second qubit, that is, we multiplied by our Bell state by id ⊗ σd, everything would remain the same except that we would get an i rather than a −i in the case of σy.
1/√2 σz ⊗ id ( | 00 〉 + | 11 〉 ) = 1/√2 ( | 00 〉 − | 11 〉 )
1/√2 σy ⊗ id ( | 00 〉 + | 11 〉 ) = −i/√2 ( | 01 〉 − | 10 〉 )
Superdense coding is a process where Alice and Bob start out sharing a Bell state
1/√2 ( | 00 〉AB + | 11 〉AB )(any of the other three Bell states would work just as well). Here, the A and B subscripts mean that Alice holds the first qubit and Bob the second. In superdense coding, Alice sends Bob one qubit, and conveys two bits of classical information to him. The process destroys the shared Bell state. This violates Holevo's bound, one of the corollaries of which is that one qubit can carry only one bit. The catch, of course, is that for Holevo's bound to apply, Alice and Bob cannot start off sharing any qubits.
How do Alice and Bob do superdense coding? Alice takes her qubit from the Bell state, and applies the unitary transformation given by one of the four Pauli matrices to it, producing one of the four Bell states. She then sends her half of the Bell state to Bob. Now Bob holds both halves of the Bell state. As all four Bell states are orthogonal, he can identify which one he has (e.g., by measuring the commuting observables Jx2, Jy2, Jz2 as we saw in a previous lecture). Now that he knows Bell state he has, Bob can deduce which of the Pauli matrices Alice applied.
Quantum teleportation is a kind of dual procedure to superdense coding. In this case, Alice and Bob share a Bell state, and Alice holds an unknown qubit | ψ 〉. Alice can send two bits to Bob and manage thereby to let Bob prepare a qubit in the same state | ψ 〉 as her unknown qubit (the process destroys Alice's qubit, and thus does not violate the no-cloning theorem).
How does teleportation work? In teleportation, Alice performs the same measurements that Bob made in superdense coding, and Bob performs the same Pauli transforms as Alice did in superdense coding. I gave two different versions of the calculation showing that teleportation works. A third (using quantum gates) is in the textbook Nielsen and Chuang. I am going to only give the second proof in these notes. The first proof is quite similar to Nielsen and Chuang's, and the proof I am giving shows explicitly how teleportation is related to the properties of Bell states and Pauli matrices.
Suppose that Alice has an unknown qubit
| ψ 〉A1 = α | 0 〉A1 + β | 1 〉A1and Alice and Bob share two qubits in the Bell state
1/√2 ( | 00 〉A2B + | 11 〉A2B ).Here, I am being even more precise than I was in class, in that I am distinguishing between Alice's first and second qubit. The state of all three qubits is
1/√2 | ψ 〉A1 ( | 00 〉A2B + | 11 〉A2B ).Now, suppose Alice makes the Bell state measurement on the first two qubits. Her state will be projected onto one of the four Bell states, that is, she effectively applies one of the projection operators
1/√2 ( A1A2〈 00 | + A1A2〈 11 | ) (id ⊗ σd)where in σd, d is one of x, y, z, or 0. After the measurement, she obtains the residual state
1/2 ( A1A2〈 00 | + A1A2〈 11 | ) (id ⊗ σd) | ψ 〉A1 ( | 00 〉A2B + | 11 〉A2B )where the id applies to qubit A1 and the σd applies to qubit A2. The square of the amplitude on this state is the probability this projection will be the one Alice observes. In the above expression, we can multiply the | ψ 〉 by the identity to get
1/2 ( A1A2〈 00 | ψ 〉A1 + A1A2〈 11 | ψ 〉A1) (σd ⊗ idB) ( | 00 〉A2B + | 11 〉A2B )and taking the inner product gives us
1/2 ( α A2〈 0 | + β A2〈 1 | ) (σd ⊗ idB) ( | 00 〉A2B + | 11 〉A2B )Now, since the σd ⊗ id is acting on the Bell state, the identities we showed at the start of this lecture let us switch the σd and id, at the price of possibly multiplying by a global phase factor of (−1) if d=y. We thus get
(−1)δ(d,y)/2 ( α A2〈 0 | + β A2〈 1 | ) (idA2 ⊗ σdB) ( | 00 〉A2B + | 11 〉A2B )where the δ is a Kronecker delta function. We can now multiply the projector by the identity and move it past the σd, since this Pauli matrix is acting on the B qubit. This gives
(−1)δ(d,y)/2 σdB ( α A2〈 0 | + β A2〈 1 | ) ( | 00 〉A2B + | 11 〉A2B )which (computing the inner product) is the same as
(−1)δ(d,y)/2 σd(B) ( α | 0 〉B + β | 1 〉B )But the 1/2 simply means that the probability of this measurement outcome is (1/2)2 = 1/4, and we can ignore global phases, so this is the same state as
σd | ψ 〉Band the state | ψ 〉 has somehow moved from Alice to Bob. They originally shared a Bell state, but have only communicated classical bits. Now, Alice's measurement tells her what d is in σd, and with two classical bits, she can communicate this information to Bob, who can then undo the σd to obtain | ψ 〉.
Is this proof actually easier than the others? Not really. However, in some sense I think it shows what is going on more clearly; for some purposes it is easier to generalize; and the techniques used in this proof come in handy for other cases where the straightforward calculation used in the other proof are way too cumbersome.