Download Mixed, pure, and entangled quantum states. Density matrix

Mixed, pure, and entangled quantum states.
Density matrix
Quantum Information and Quantum Optics course, 2013
Göran Johansson, Thilo Bauch, Jonas Bylander
Chalmers, MC2
September 24, 2013
The density operator or density matrix is a more general way of describing
the state of a quantum system than that provided by the wave function or state
vector. It allows us to describe situations where the state vector is not precisely
known, and states other than pure quantum states. It will be useful later, when
we discuss quantum decoherence.
1
A mixed vs. a pure state
What is the difference between saying that a two-state system is in a pure state
1
|Ψ(π/2, 0)i = √ (|0i + |1i) ,
2
(1)
and that it has a probability P0 = 0.5 to be in state |0i and a probability
P1 = 0.5 to be in state |1i (i.e. a mixed state)? (|0i and |1i are in the z basis.)
An example of how these states are different can be seen by rotating by
π/2 around the y axis: The pure state transforms as Ry (−π/2)|Ψ(π/2, 0)i =
|Ψ(0, 0)i = |0i, yielding unit probability for the reading out the qubit in state
|0i when measured in the z basis. Performing the same operation on the mixed
state yields the state
√
Ry (−π/2)|0i = |Ψ(−π/2, 0)i = (|0i − |1i) / 2
with probability 0.5, and
√
Ry (−π/2)|1i = |Ψ(π/2, 0)i = (|0i + |1i) / 2
with probability 0.5. Now reading out the mixed state gives either state, |0i or
|1i, with equal probability. A good way to describe this difference is by using
the density matrix, which is a quantum mechanical analogy of a probability
distribution.
1
2
The density matrix
We define the density operator as
X
ρ̂ =
pi |Ψi ihΨi |,
(2)
i
where pi is the probability
P for the qubit to be in state |Ψi i. Choosing a arbitrary
basis {|bj i} (with 1 = j |bj ihbj |), we get the density matrix with the elements
ρkl =
X
pi hbk |Ψi ihΨi |bl i.
(3)
i
Expectation values for any operator Ô with matrix elements Okl = hbk |Ô|bl i
are given by
X
X XX
hÔi =
pi hΨi |Ô|Ψi i =
pi
hΨi |bl ihbl |Ô|bk ihbk |Ψi i =
(4)
i
=
i
XX
k
ρkl Olk =
l
X
k
l
(ρ̂ · Ô)kk = Tr(ρ̂Ô),
(5)
k
where Tr is the trace of the matrix, i.e., a summation over the diagonal elements.
The density-matrix is Hermitian, ρkl = ρ∗lk , as seen from Eq. 3, and it has
trace unity since
X X
X X
Tr(ρ̂) =
pi
hbk |Ψi ihΨi |bk i =
pi
hΨi |bk ihbk |Ψi i =
i
=
X
pi hΨi |Ψi i =
i
2.1
i
k
X
k
pi = 1.
(6)
i
One density matrix can represent many different state
mixtures
Compare the density matrix (in the z-basis as usual) for a fifty-fifty mixture of
spin-up and spin-down in the z-direction,
ρ̂↑↓ =
1
1
[|0ih0| + |1ih1|] ≡ 1,
2
2
(7)
with that for an equal mixture of spin-up and spin-down in the x-direction,
1 1
1
1
1
√ (|0i + |1i) √ (h0| + h1|) + √ (|0i − |1i) √ (h0| − h1|) =
ρ̂→← =
2
2
2
2
2
1
=
[|0ih0| + |1ih1| + |0ih1| + |1ih0| + |0ih0| + |1ih1| − |0ih1| − |1ih0|] =
4
1
=
[|0ih0| + |1ih1|] = ρ̂↑↓ .
(8)
2
2
2.2
How to recognize a pure state
The density operator of a pure state can be written ρ̂ = |ΨihΨ|, so it’s clear
that ρ̂2 = |ΨihΨ|ΨihΨ = |ΨihΨ| = ρ̂, so Tr(ρ̂2 ) = 1. It can be shown that for
a mixed state 0 < Tr(ρ̂2 ) < 1. We check for the mixed state in the previous
section
1 1 0
1
1
1 1 0
2
·
= Tr(1) = .
(9)
Tr(ρ̂ ) = Tr
2 0 1
2 0 1
4
2
3
The Bloch-sphere for mixed states
The set of matrices {1, σx , σy , σz } form a complete basis for 2 x 2 Hermitian
matrices. Therefore, an arbitrary single qubit density-matrix can be written
ρ̂ =
1 + r x σ x + r y σ y + rz σ z
1
≡ (1 + r · σ),
2
2
(10)
where r = (rx , ry , rz ) is an arbitrary real vector of length |r| ≤ 1. We see that
Tr(ρ̂) = 1 since Tr(σx ) = Tr(σy ) = Tr(σz ) = 0. For the projection onto the x
axis, e.g., we find using the trace-formula for expectation values, that
1
1
hσx i = T r(ρ̂σx ) = Tr
1σx + rx σx2 + ry σy σx + rz σz σx = Tr
r x 1 = rx ,
2
2
(11)
since Tr(σi σj ) = 0 for i 6= j. Similarly hσy i = ry and hσz i = rz . For pure states
the mapping is identical to the one we discussed last week, and |r| = 1, i.e.,
the points are on the surface of the Bloch sphere. The fifty-fifty mixed state
discussed above has the density matrix ρ̂ = 1/2 (see Eq. 7). This corresponds
to the vector r = (0, 0, 0). In general we see that
h
i
1 2
2
Tr(ρ̂2 ) = Tr
1 + 2 (rx σx + ry σy + rz σz ) + (rx σx + ry σy + rz σz )
=
4
!
1 + rx2 + ry2 + rz2
1 + |r|2
1 =
,
(12)
= Tr
4
2
so that only the pure states are on the surface of the sphere, whereas mixed
states are inside the sphere.
4
Interpretation of the density matrix elements
What is the probability to find the qubit in the state |0i when it is described
by a density matrix ρ̂? This probability is given by the expectation value of
the projection operator P̂0 = |0ih0|. For a pure state |Ψi = c0 |0i + c1 |1i this
gives the usual hΨ|P̂0 |Ψi = (c∗0 h0| + c∗1 h1|) |0ih0| (c0 |0i + c1 |1i) = |c0 |2 . For an
arbitrary single qubit density matrix we find
1 0
ρ00 ρ01
P0 = T r(P̂0 ρ̂) = T r
·
= ρ00 ,
(13)
0 0
ρ10 ρ11
3
and similarly P1 = ρ11 . (P̂0/1 = (1 ± σz )/2) So the probability to find the qubit
in a certain state is given by the diagonal elements.
The off-diagonal elements give the amount of coherence between the states.
Under the influence of a noisy environment the coherence is easily killed, as we
will see in the next lecture. The environment tries to make the density matrix
diagonal in some basis. The qubit states then evolves from a pure state into
a mixed state. On the Bloch sphere this is seen as the disappearance of the
components rx and ry , if the environment makes the density matrix diagonal in
the z-direction.
5
Reduced density matrix for composite systems
Decoherence is a loss of quantum correlations, which limits the functionality of
qubits in information processing. It appears due to the interaction of a quantum
system with its environment. The simplest case of an environment is another
qubit coupled to the system of interest. When the entire system is described
by a density matrix, the properties of the individual qubit taken by itself are
described by a reduced density matrix, taken by tracing over (averaging out) the
environment. This illustrates how decoherence appears in a quantum mechanical
system.
We can write any pure two-qubit state in the basis of two single qubits, using
the tensor product, {|01 i, |11 i} ⊗ {|02 i, |12 i}, as
|Ψi = c00 |01 i|02 i + c01 |01 i|12 i + c10 |11 i|02 i + c11 |11 i|12 i,
(14)
with normalization
|c00 |2 + |c01 |2 + |c10 |2 + |c11 |2 = 1.
(15)
We want to operate on and measure qubit 1 (qubit 2 is now the environment).
We therefore consider theexpectation
value of an operator (measurement) acting only on qubit 1, Ô = Ô1 ⊗ 1̂2 :
i
c∗00 h01 |h02 | + c∗01 h01 |h12 | + c∗10 h11 |h02 | + c∗11 h11 |h12 | Ô1 ⊗ 1̂2 ×
h
i
× c00 |01 i|02 i + c01 |01 i|12 i + c10 |11 i|02 i + c11 |11 i|12 i
(16)
hΨ|Ô|Ψi =
h
Because the operator does nothing to the second qubit, only the brackets with
equal states for the second qubit will contribute, e.g.
h01 |h02 | Ô1 ⊗ 1̂2 |11 i|02 i = h01 |Ô1 |11 ih02 |02 i = h01 |Ô1 |11 i,
(17)
while
h01 |h12 | Ô1 ⊗ 1̂2 |11 i|02 i = h01 |Ô1 |11 ih12 |02 i = 0.
4
(18)
Out of the original 16 terms, we are left with eight
hΨ|Ô|Ψi = [c∗00 h01 | + c∗10 h11 |] Ô1 [c00 |01 i + c10 |11 i] +
+ [c∗11 h11 | + c∗01 h01 |] Ô1 [c11 |11 i + c01 |01 i] ,
(19)
which we should regroup into
hΨ|Ô|Ψi = h01 |Ô1 |01 i |c00 |2 + |c01 |2 + h01 |Ô1 |11 i [c∗00 c10 + c∗01 c11 ] +
+ h11 |Ô1 |01 i [c∗10 c00 + c∗11 c01 ] + h11 |Ô1 |11 i |c10 |2 + |c11 |2 . (20)
We can now construct a reduced density matrix for the first qubit only
|c00 |2 + |c01 |2 c∗10 c00 + c∗11 c01
ρ1 =
,
c∗00 c10 + c∗01 c11 |c10 |2 + |c11 |2
(21)
which reproduces hΨ|Ô|Ψi = Tr(ρ1 O1 ). We can check that Tr(ρ) = 1 and that it
is Hermitian, so it is a valid density matrix. It can also, more easily, be obtained
by tracing out the second qubit from the pure two-qubit density matrix, using
the state given in Eq. 14,
ρ̂1
= T r2 (|ΨihΨ|) =
1
X
hj2 | [|ΨihΨ|] |j2 i =
j=0
= h02 |ΨihΨ|02 i + h12 |ΨihΨ|12 i =
= |01 ih01 | |c00 |2 + |c01 |2 + |01 ih11 | [c∗10 c00 + c∗11 c01 ] +
+ |11 ih01 | [c∗00 c10 + c∗01 c11 ] + |11 ih11 | |c10 |2 + |c11 |2 .
(22)
1
For a product state |Ψi = |Ψ1 i|Ψ2 i = c0 |01 i + c11 |11 i c20 |02 i + c21 |12 i we have
c00 = c10 c20 , c01 = c10 c21 , c10 = c11 c20 and c11 = c11 c21 . Putting this into Eq. 21,
and using the normalization |c20 |2 + |c21 |2 = 1, gives the ordinary pure-state
density matrix for qubit
choosing a Bell state |Ψi =
√
√ 1, ρ1 = |Ψ1 ihΨ1 |. Instead
(|01 i|02 i + |11 i|12 i) / 2, i.e. c00 = c11 = 1/ 2 and c01 = c10 = 0 gives
1
0
ρ1 = 2 1 ,
(23)
0 2
which we recognize as the density matrix of a mixed state with equal probabilities for spin-up and spin-down in any arbitrary direction, which maps onto the
origin of the Bloch sphere r = (0, 0, 0). So even if the two qubits together are
in a pure state, qubit number 1 will behave like a mixed state.
6
Time-evolution of the density matrix
Remembering the Schrödinger equation ih̄|Ψ̇(t)i = Ĥ|Ψ(t)i and its transpose
complex conjugate version −ih̄hΨ̇(t)| = hΨ(t)|Ĥ (Ĥ † = Ĥ) we find
i
X
X h
˙
ih̄ρ̂(t)
= ih̄
pi ∂t [|Ψi (t)ihΨi (t)|] =
pi ih̄|Ψ̇i (t)ihΨi (t)| + ih̄|Ψi (t)ihΨ̇i (t)| =
i
i
X
h
i
h
i
pi Ĥ|Ψi (t)ihΨi (t)| − |Ψi (t)ihΨi (t)|Ĥ = Ĥ ρ̂ − ρ̂Ĥ = Ĥ, ρ̂ . (24)
i
5
This is called von Neumann’s equation.
If we know the time-evolution operator for some pulse Û (t) so that |Ψ(t)i =
Û (t)|Ψ(0)i, we get the effect on the density matrix as
X
X
ρ̂(t) =
pi |Ψi (t)ihΨi (t)| =
pi Û (t)|Ψi (0)ihΨi (0)|Û † (t) = Û (t)ρ̂(0)Û † (t).
i
i
(25)
Note that the von Neumann equation is analogous to Liouville’s equation for
the time evolution of a probability distribution in statistical mechanics. (This
equation also looks similar to the Heisenberg equation, but note that it is not
the same: in the Heisenberg picture, the equation describes the evolution of the
operators in time, whereas here, in the Schrdinger picture, it is the state vectors
that are time dependent.)
7
Reading suggestions
Gerry and Knight, appendix A; Nielsen and Chuang, pages 98-108 and chapter
8; Sakurai pages 174-184.
6