In: Computer Science
Trace how quantum cryptography will work if n = 4, and random bits are:
As a message, Alice wants to send as many first bits as possible from the string 1010, i.e.:
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.
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|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.