Download 1. You are given one of two quantum states of a single qubit: either

 1. You are given one of two quantum states of a single qubit: either φ = 0 or ψ = cos θ 0 +
sin θ 1 . You want to make a single measurement that best distinguishes between these two states,
i.e., you want to find the best basis for
a measurement to distinguish the two states. So let’s
making
measure the qubit in basis |vi = α 0 + β 1 , |v⊥ i, where α, β are to be determined for optimal
success. For outcome |vi we guess that the qubit was in state |0i; for outcome |v⊥ i we guess that the
qubit was in state |ψi. Determine the optimal measurement basis given this procedure. You can take
α and β to be real numbers, in which case the normalization |α|2 + |β |2 = 1 implies that you can
write α and β as e.g. α = sin γ and β = cos γ.
Hint: you will need to first construct the probability of a correct measurement in this situation. You
should convince yourselves that this is given by
Pr[qubit was|0i] Pr[measure|vi|qubit was|0i] + Pr[qubit was|ψi] Pr[measure|v⊥ i|qubit was|ψi].
where, e.g.,
Pr[measure|vi|qubit was|0i] = | v 0 |2 .
If the state you are presented is either φ or ψ with 50% probability each, what is the probability
that your measurement correctly identifies the state?
Solution:
Answer: We measure the qubit in basis |vi = α 0 + β 1 , |v⊥ i. For outcome |vi we guess that the
qubit was in state |0i; for outcome |v⊥ i we guess that the qubit was in state |ψi.
The optimal measurement basis is v =cos γ 0 + sin γ 1 , where γ = 21 (θ − π2 ). Graphically,
this
means that the angle bisector between v and v⊥ is the same as that between 0 and ψ .
Let’s prove that this choice of measurement basis v , v⊥ indeed maximizes the probability that our
guess is correct:
Pr[qubit was|0i] Pr[measure|vi|qubit was|0i] + Pr[qubit was|ψi] Pr[measure|v⊥ i|qubit was|ψi]
= 12 | v 0 |2 + | v⊥ ψ |2
= 12 | v 0 |2 + 1 − | v ψ |2 .
Let v = α 0 + β 1 . We want to choose α, β to maximize
| v 0 |2 − | v ψ |2 = |α|2 − |α cos θ + β sin θ |2
= |α|2 (1 − cos2 θ ) − |β |2 sin2 θ − cos θ sin θ (ᾱβ + α β̄ ) ,
Using 1 − cos2 θ = sin2 θ and assuming without loss of generality sin θ ≥ 0, this is equivalent to
maximizing sin θ (|α|2 − |β |2 ) − cos θ (ᾱβ + α β̄ ). We may assume α, β ∈ ℜ, since for fixed magnitudes |α|, |β |, we want to choose arguments such as to either maximize (if cos θ < 0) or minimize
(if cos θ > 0) the real part of ᾱβ . Therefore let α = cos γ, β = sin γ for some angle γ. We wish to
maximize
sin θ cos 2γ − cos θ sin 2γ = sin(θ − 2γ) .
Therefore let γ = 21 (θ − π2 ). Geometrically, the vector v + v⊥ is a multiple of 0 + ψ .
The success probability with this choice of γ is cos2 ( π4 − θ2 ) = 12 (1 + sin θ ). Without the assumption
sin θ > 0, the success probability is 21 (1 + |sin θ |).
C/CS/Phys
191,
Fall
2008,
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Any two quantum states, in an arbitrary dimension, lie in a plane. By rotational symmetry
the above
q
1
argument applies: the optimal measurement gives a success probability of 2 (1 + 1 − | φ ψ |2 ).
φ and ψ are orthogonal, hence perfectly distinguishable. This
Note that this probability
is
one
if
probability is 1/2 if | φ ψ | = 1. In this case there is no optimal basis, you can do no better than just
guessing randomly.
Note that allowing multiple measurements does not increase the optimum success probability – only
the first measurement gives information about the original state. Also, a unitary transformation, followed by a measurement, is the same a measurement in a rotated basis – so allowing unitary transformations does not increase the optimum success probability either.
2. Prove that the Bell state |ψ − i is rotationally invariant: i.e. |ψ − i =
√1 (|vv⊥ i − |v⊥ vi).
2
2
|α| + |β |2 = 1), |v⊥ i = −β̄ |0i + ᾱ|1i,
Answer: For |vi = α|0i + β |1i (with normalization condition
up to choice of phase. (Indeed, hv⊥ |vi = β α − αβ = 0.) Then
|vv⊥ i − |v⊥ vi = (α|0i + β |1i) ⊗ (−β̄ |0i + ᾱ|1i) − (−β̄ |0i + ᾱ|1i) ⊗ (α|0i + β |1i)
= (−α β̄ + β̄ α)|00i + (|α|2 + |β |2 )|01i − (|β |2 + |α|2 )|10i + (β ᾱ − ᾱβ )|11i
= |01i − |10i ,
where in the last equality we used the normalization condition. Hence, |Ψ− i is rotationally invariant;
for any unitary operator U on C 2 , (U ⊗U)|Ψ− i = |Ψ− i.
3. Show that if U and V are unitary, then so is U ⊗V .
If U and V are both unitary, then each can be written in some basis as a diagonal matrix with only
complex phases eiφn along the diagonal. Since operators in a tensor product act on different Hilbert
spaces, we can diagonalize both of them independently. Let eiφn be the diagonal elements of U, and
let eiφm be the diagonal elements of V . Then we use the rules for tensor products of matrices to see
that U ⊗V is a diagonal matrix with diagonal elements eiφn eiφm . Therefore it is also a diagonal matrix
with only complex phases along the diagonal. So it is unitary.
4. Write the 4 × 4 matrix of the unitary operation on two qubits resulting from performing a Hadamard
transform on the first qubit and a phase flip on the second qubit
1
√
2
1 1
1 −1
⊗
1 0
0 −1
1 0
1 0
1
1
1 
0 −1 0 −1 =√ 

1
0
1 0
2
1
−1
0 −1
0 −1

=






1 0
1
0
1  0 −1 0 −1 

=√ 
2  1 0 −1 0 
0 −1 0
1
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Fall
2008,
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