## Lecture 3

### The No-Cloning Theorem

Today, we talked about the no-cloning theorem and quantum teleportation. Recall that last time, we gave Bell's inequality. If you have a probability distribution on variables P, Q, R and S, which take on values ±1, then we have

(QS + RS + RT - QT) ≤ 2.
| ψ ⟩ = 1/√2 | 01 ⟩ − | 10 ⟩
and let
Q = &sigmazA
R = &sigmaxA
S = − (&sigmazB + &sigmaxB ) / √2
T = (&sigmazB − &sigmaxB ) / √2
then we have
E QS = E RS = E RT = − E QT = 1/√2 .
This seems to imply that you must either give up the idea of locality or of classical probability. Locality says that a measurement of A's qubit cannot affect B's qubit. Giving up classical probability means that the manipulation of the random variables Q, R, S and T we performed to produce Bell's inequality is not permissible because you cannot simultaneously measure Q and R because these operators do not commute.

A natural question is: if you have non-local effects — this "spooky action at a distance" that disturbed Einstein — can you use it to communicate faster than light. What we will first show is that if you can "clone" quantum bits, you can use this process to communicate faster than light. We will next prove that you cannot "clone" quantum bits. Later in the term, when we introduce density matrices, we will sketch a proof that quantum entanglement never lets you transmit information faster than light.

Suppose you could clone a qubit. That is, given one copy of a qubit | ψ ⟩, you could make many copies | ψ ⟩⊗ n. Consider a qubit in the state
1/√2 ( | 01 ⟩ − | 10 ⟩ )
If Alice measures in the basis | 0 ⟩, | 1 ⟩, then Bob holds a qubit which with probability 1/2 is in the state | 1 ⟩ and with probability 1/2 is in the state | 0 ⟩. Similarly, if Alice measures in the basis | + ⟩, | − &rang, where
| + &rang = 1/√2 ( | 0 ⟩ + | 1 ⟩ )      and      | − ⟩ = 1/√2 ( | 0 ⟩ − | 1 ⟩ ),
Bob will get a qubit which is in one of the two states | + ⟩ and | − ⟩ with equal probability.

We will show that if you can clone a state, then Bob can distinguish the two above probability distributions (equal probabilities of | + ⟩, | − ⟩ vs. equal probabilities of | 0 ⟩ and | 1 ⟩) from each other. How he does this is he first makes many copies of the qubit Alice sent him. He then measures all these copies in, say, the { | 0 ⟩,| 1 ⟩ } basis. If he originally held either a | 0 ⟩ or a | 1 ⟩, then all these measurements will result in the same outcome. If he originally held either a | + ⟩ or a | − ⟩, then each clone will have a 50-50 chance of giving a | 0 ⟩ or a | 1 ⟩, and if there are many clones, then with high probability, there will be two different measurement outcomes. Thus, if quantum mechanics allowed cloning of states, Alice and Bob could use entangled pairs of qubits to transmit information faster than light. This should be viewed as good evidence that quantum mechanics does not allow a qubit to be cloned. We will prove this theorem later in the term, once we have introduced the concept of density matrices.

The no-cloning theorem says that if you have one copy of an unknown quantum state | ψ ⟩, you cannot produce two copies of it. That is, given | ψ ⟩ | 0 ⟩, we need to 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: the inner product of | &phi ⟩A | 0 ⟩B and | ψ ⟩A | 0 ⟩B is ⟨ φ | ψ ⟩, and the inner product of | &phi ⟩A | &phi ⟩B and | &psi ⟩A | &psi ⟩B is ⟨ φ | ψ ⟩ 2. These inner products are only equal if | φ ⟨ and | ψ ⟩ are equal or are orthogonal, in which case | φ ⟩ and | ψ ⟩ are essentially classical, and cloning is allowed.

This actually isn't enough to show the no-cloning theorem. To be rigorous, we 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 is 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. (I may give this near-cloner as an exercise on the homework.) To prove this stronger version, we need to show that the map
| ψ ⟩ ⊗ | 0 ⟩ ⊗ | 0 ⟩ → | ψ ⟩ ⊗ | ψ ⟩ ⊗ | θ ψ
is not unitary, where | θ ψ ⟩ may depend on | ψ ⟩. This is also easy, as a straightforward calculation shows that this map also does not preserve inner products.

## Teleportation

I next talked about teleportation. A corollary of the no-cloning theorem is that a quantum state (i.e., quantum information) cannot be transmitted over the telephone. Suppose that Alice has an unknown quantum state | φ ⟩. If she could send information over the telephone that was sufficient for Bob to recreate it, then Bob could recreate two copies. (Since classical information is duplicatable, Bob can simply write down the information and create as many copies as he wants using the same procedure he used for one copy.) However, if Bob and Alice share an entangled bit in the EPR state 1/&radic2; ( | 01 ⟩ - | 10 ⟩ ). Alice can indeed send a qubit in an unknown state | φ ⟩ over the telephone. This procedure is called teleportation. What Alice does is make a joint measurement on her bit of the EPR state and her unknown state | φ ⟩. She sends the result to Bob. The knowledge of this result then lets him perform a transformation on his half of the qubit which puts it in the state | φ ⟩. Notice that by measuring the qubit in state | φ ⟩, Alice has destroyed its state, so the information in it is not cloned. Further, as Bob must wait to receive the classical outcome of Alice's measurement, teleportation cannot be used to transmit information faster than light.

To analyze teleportation, we need some facts about four states of qubits known as the Bell states. It is straightforward to show that

1/√2 ( I ⊗ σx ) ( | 01 ⟩ − | 10 ⟩ )   =   − 1/√2 ( σx ⊗ I ) ( | 01 ⟩ − | 10 ⟩ )   =     1/√2 ( | 00 ⟩ − | 11 ⟩ ),
1/√2 ( I ⊗ σz ) ( | 01 ⟩ − | 10 ⟩ )   =   − 1/√2 ( σz ⊗ I ) ( | 01 ⟩ − | 10 ⟩ )   =   − 1/√2 ( | 01 ⟩ + | 10 ⟩ ),
1/√2 ( I ⊗ σy ) ( | 01 ⟩ − | 10 ⟩ )   =   − 1/√2 ( σy ⊗ I ) ( | 01 ⟩ − | 10 ⟩ )   =   − i /√2 ( | 00 ⟩ + | 11 ⟩ ).
These states, along with the original EPR state 1/√2 ( | 01 ⟩ &minus | 10 ⟩ ), are known (up to phases ±1 ±i) as the four Bell states, or the Bell basis. It is easy to check that they are orthonormal, and so indeed form a basis.

As illustrated above, you can obtain all of the four Bell states by starting with any of the four Bell states and applying either the identity or one of the three Pauli matrices to the first qubit. For uniformity, we call the identity matrix the fourth Pauli matrix, and label it σ0). If instead, we had applied the Pauli matrix to the second qubit, that is, we multiplied by our Bell state by I ⊗ σd, everything would remain the same except that we would possibly get some factors of −1.

In quantum teleportation, 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. We will call it Alice's first qubit, or A1.

| ψ ⟩A1 = α | 0 ⟩A1 + β | 1 ⟩A1
and Alice and Bob share two qubits in the Bell state
1/√2 ( | 01 ⟩A2B − | 10 ⟩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 ( | 01 ⟩A2B − | 10 ⟩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⟨ 01 | − A1A2⟨ 10 | ) (IA1 ⊗ σd)
where in σd, d is one of x, y, z, or 0. After the measurement, she obtains the residual state
1/2 ( A1A2⟨ 01 | − A1A2⟨ 10 | ) (IA1 ⊗ σd) | ψ ⟩A1 ( | 01 ⟩A2B − | 10 ⟩A2B )
where the I 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.

Now, I | ψ ⟩ = | ψ ⟩, and we can move the first qubit | ψ ⟩ past the &sigmad, which operates on the second qubit to get,

1/2 ( A1A2⟨ 01 | ψ ⟩A1A1A2⟨ 10 | ψ ⟩A1) (σd ⊗ IB) ( | 01 ⟩A2B − | 10 ⟩A2B ).
Now remember that 1/√2 ( | 01 ⟩ &minus | 10 ⟩ ) was equal to e/√2 ( | v v ⟩ − | v v ⟩ ) for any | v ⟩, | v ⟩ where ⟨ v | v ⟩ = 0. Thus, we have
1/√2 ( A1A2&lang 01 | − A1A2&lang 10 | ) | ψ ⟩A1 ) = 1/√2 A2⟨ &psi |
Substituting this in the above expression gives
1/2 A2⟨ ψ | (σd ⊗ IB) ( | 01 ⟩A2B − | 10 ⟩A2B )
Now, since the σd ⊗ I is acting on the Bell state, the identities we showed at the start of this lecture let us switch the σd and I, at the price of possibly multiplying by a global phase factor. We thus get, up to this global phase
1/2 A2⟨ ψ | (IA2 ⊗ σd) ( | 01 ⟩A2B − | 10 ⟩A2B ).
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
1/2 σd A2⟨ ψ | ( | 01 ⟩A2B − | 10 ⟩A2B )
which (computing the inner product) is the same as
1/2 σd | ψ ⟩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 ⟩B
and 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.