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In: Computer Science

Trace how quantum cryptography will work if n = 4, and random bits are: b1 =...

Trace how quantum cryptography will work if n = 4, and random bits are:

  • b1 = 1, b2 = 0, b3 = 1, b4 = 0;
  • r1 = 1, r2 = 1, r3 = 0, r4 = 1;
  • s1 = 1, s2 = 1, s3 = 0, s4 = 0.

As a message, Alice wants to send as many first bits as possible from the string 1010, i.e.:

  • if possible, the whole string,
  • if not possible to send the whole string, the first three bits 101;
  • if this is also not possible, the first two bits 10, etc.

Solutions

Expert Solution

Alice share some pure state |φ〉. If this state is entangled, it cannot be written
as a tensor product |φA〉 ⊗ |φB〉 of separate pure states for Alice and Bob. Still, there is a way
to describe Alice’s local state as a mixed state, by tracing out Bob’s part. Formally, if A ⊗ B is
a product matrix then TrB(A ⊗ B) = A · Tr(B). By extending this linearly to matrices that are
not of product form, the operation TrB is well-defined on all mixed states (note that TrB removes
Bob’s part of the state, leaving just Alice’s part of the state). If ρAB is some bipartite state (mixed
or pure, entangled or not), then ρA = TrB(ρAB) is Alice’s local density matrix. This describes all
the information she has. For example, for an EPR-pair |φ〉 = 1 (|00〉 + |11〉), the corresponding
√2 ρAB = 1 (|00〉〈00| + |00〉〈11| + |11〉〈00| + |11〉〈11|)
density matrix is
2
= 1 (|0〉〈0| ⊗ |0〉〈0| + |0〉〈1| ⊗ |0〉〈1| + |1〉〈0| ⊗ |1〉〈0| + |1〉〈1| ⊗ |1〉〈1|), 2
and since Tr(|a〉〈b|) = 1 if a = b and Tr(|a〉〈b|) = 0 if a and b are orthogonal, we have ρA = TrB (ρAB ) = 1 (|0〉〈0| + |1〉〈1|).
2
In other words, Alice’s local state is the same as a random coin flip! Similarly we can compute Bob’s local state by tracing out Alice’s part of the space: ρB = TrA(ρAB).
The Schmidt decomposition is a very useful way to write bipartite pure states, and allows us to easily calculate the local density matrices of Alice and Bob. It says the following: for every bipartite state |φ〉 there is an orthonormal basis |a1〉, . . . , |ad〉 for Alice’s space, an orthonormal basis |b1〉, . . . , |bd〉 for Bob’s, and nonnegative reals λ1, . . . , λd whose squares sum to 1, such that
Schmidt coefficients λ1 = λ2 = 1/√2 and hence has Schmidt rank 2. 79
|φ〉 =
?d i=1
λi|ai〉|bi〉. (13.1) The number of nonzero λi’s is called the Schmidt rank of the state. For example, an EPR-pair has

The existence of the Schmidt decomposition is shown as follows. Let ρA = TrB?(|φ〉〈φ|) be Alice’s local density matrix. This is Hermitian, so it has a spectral decomposition ρ?= μ |a 〉〈a | with
Aiiii orthonormal eigenvectors |ai〉 and nonnegative real eigenvalues μi. Note i μi = Tr(ρA) = 1. Since the {|ai〉} form an orthonormal set, we can extend it to an orthonormal basis for Alice’s d-dimensional space (adding additional orthonormal |ai〉 and μi = 0). Hence there are cij such that
μicij|ai〉|j〉, | φ 〉 = ?d √
i,j =1
?where the |j〉 are the computational basis states for Bob’s space. Define λi = √μi and |bi〉 = j cij|j〉. This gives the decomposition of |φ〉 of Eq. (13.1). It only remains to show that {|bi〉} is
an orthonormal set, which we do as follows. The density matrix version of Eq. (13.1) is
?d i,j =1
We know that if we trace out the B-part from |φ〉〈φ|, then we should get ρA = ?i λ2i |ai〉〈ai|, but that can only h?appen if 〈bj|bi〉 = Tr(|bi〉〈bj|) = 1 for i = j and 〈bj|bi〉 = 0 for i ̸= j. Hence the |bi〉 form an orthonormal set.


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