# Quantum Mechanics II Examples

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```					              Quantum Mechanics II:
Examples

Michael A. Nielsen

University of Queensland
Goals:
1. To apply the principles introduced in the last lecture to
some illustrative examples: superdense coding, and
quantum teleportation.
2. Revised form of postulates 2 (dynamics) and 3
(measurement).
3. Introduce more elements of the Dirac notation.
4. Discuss the philosophy underlying quantum information
science.
Superdense coding
Alice                                 Bob

ab

ab

Theorist’s impression
of a measuring device
Superdense coding
Alice                        Bob

ab

ab
Superdense coding
Alice                                 Bob

X 0 = 1;      X 1 = 0
Z 0 = 0 ; Z 1 =−1

ab
00 + 11       00 + 11       00 + 11
00 : Apply I              →
2             2             2

01 : Apply Z    00 + 11       00 − 11
→
2             2          ab
10 : Apply X    00 + 11       10 + 01
→
2              2
11 : Apply XZ   00 + 11       10 − 01
→
2             2
Superdense coding can be viewed as a statement about
the interchangeability of physical resources.

1 ebit + 1 qubit of communication ≥ 2 bits of classical communication
Worked exercise: Could Alice and Bob still communicate
two bits using the superdense coding protocol if the initial
01 − 10
state shared by Alice and Bob was     2
?
Revised measurement postulate
Recall postulate 3: If we measure ψ in an orthonormal
basis e1 ,..., ed , then we obtain the result j with probability
2
P ( j ) = ej ψ       .

The measurement disturbs the system, leaving it in a state e j
determined by the outcome.

Problem: Imagine we measure a quantum system, A, in the
orthonormal basis e1 ,..., edA .

Suppose system A is part of a larger system, consisting of
two components, A and B .

How should we describe the effect of the measurement on
the larger system?
Revised measurement postulate
Recall postulate 3: If we measure ψ in an orthonormal
basis e1 ,..., ed , then we obtain the result j with probability
2
P ( j ) = ej ψ       .

The measurement disturbs the system, leaving it in a state e j
determined by the outcome.

The revised postulate replaces the orthogonal states e1 ,..., ed
with a complete set of orthogonal subspaces V1 ,...,Vm .
V = V1 ⊕V2 ⊕ ... ⊕Vm .

Example:    ψ = (α e1 + β e2 ) + γ e3
sp ( e1 , e2 , e3    ) = sp ( e
1   , e2   ) ⊕ sp ( e )
3

A general measurement can be thought of as asking the
question "which of the subspaces V1 ,...,Vm are we in?"
Revised measurement postulate
Recall postulate 3: If we measure ψ in an orthonormal
basis e1 ,..., ed , then we obtain the result j with probability
2
P ( j ) = ej ψ       .

The measurement disturbs the system, leaving it in a state e j
determined by the outcome.

Mathematically, it is convenient to describe the subspaces
V1 ,...,Vm in terms of their corresponding projectors, P1 ,...,Pm .

Example: The projector P onto sp ( e1 , e2 ) acts as
P (α e1 + β e2 + γ e3 ) = α e1 + β e2

In general, the projector P onto a subspace V acts as
the identity on that subspace, and annihilates everything
orthogonal to V .
Revised measurement postulate
Let P1 ,..., Pm be a set of projectors onto a complete set of
orthogonal subspaces of state space.
∑ j Pj   =I;   Pj Pk = δ jk Pj
This set of projectors defines a measurement.

If we measure ψ then we get outcome j with probability
Pr( j ) = ψ Pj ψ .
The measurement unavoidably disturbs the system, leaving
Pj ψ
it in the post-measurement state
ψ Pj ψ
Example: A two-outcome measurement on a qutrit
A general state of a qutrit may be written ψ = α 0 + β 1 + γ 2 .
P1 projects onto sp ( 0 , 1 ) ; and
P2 projects onto sp ( 2 ) .
α  α 
Pr(1) = ψ P1 ψ =  β  •  β  = α + β
2    2
   
γ  0 
   
2
Pr(2) = ψ P2 ψ = γ

P1 ψ     α 0 +β 1                   P2 ψ     γ 2
ψ =
'
1        =                        ψ =
'
2        =     ~ 2
ψ P1 ψ      2
α + β
2
ψ P2 ψ    γ
Example: Measuring the first of two qubits
Suppose we want to perform a measurement in the basis
e1 , e2 for the first of two qubits.
The rule is to first form the corresponding projectors
P1 , P2 onto the state space of that qubit, and then to
tensor them with the identity on the second qubit,
obtaining P1 ⊗ I and P2 ⊗ I .
Example: If the state of two qubits is
ψ = α 00 00 + α 01 01 + α10 10 + α11 11
then measuring the first qubit in the computational basis
gives the result 0 with probability
Pr ( 0 ) = ψ ( P0 ⊗ I ) ψ
= (α 00 00 + α 01 01 + α10 10 + α11 11 ) • (α 00 00 + α 01 01   )
2         2
= α 00 + α 01
Example: Measuring the first two of three qubits
Suppose we have three qubits in the state
α e1 a + β e2 b + γ e3 c + δ e4 d .
e1 , e2 , e3 , e4 is an orthonormal basis for the state
space of the first two qubits.
a , b , c , d are normalized states of the third qubit.

Measuring the first two qubits in the basis e1 , e2 , e3 , e4
gives the result 1 with probability
Pr(1) = ψ ( P1 ⊗ I ) ψ
= ψ (α e1 a   )
2
=α
Post-measurement state is e1 a .
Teleportation
Alice                   Bob
Teleportation
Alice                   Bob

01                      01
Teleportation
Alice                              Bob

00 + 11
2
α 0 +β 1
 00 + 11 
(α 0 + β 1 )          
    2    
α 000 + α 011 + β 100 + β 111
=
2

1  00 + 11  1  00 − 11             1  01 + 10  1  01 − 10 
00 =                                 10 =              −           
         +                        2    2      2    2    
2    2      2    2    
1  01 + 10  1  01 − 10           1  00 + 11  1  00 − 11 
01 =              +              11 =              −           
2    2      2    2               2    2      2    2    
Teleportation
Alice                         Bob

1  00 + 11               I
=            (α 0 + β 1 ) → (α 0 + β 1       )
2     2    

01         1  00 − 11 
+ 
2
 (α 0 − β 1 ) → (α 0 + β 1
Z
)
2    
1  01 + 10                   X
+            (α 1 + β 0   ) → (α   0 +β 1    )
2     2    
1  01 − 10                   ZX
+            (α 1 − β 0   )   → (α 0 + β 1   )
2     2    
Teleportation can be viewed as a statement about the
interchangeability of physical resources.
1 ebit + 2 classical bits of communication ≥ 1 qubit of communication

Compare with superdense coding:
1 ebit + 1 qubit of communication ≥ 2 bits of classical communication

1 qubit of communication = 2 bits of communication (Mod 1 ebit)
The fundamental question of information science
1.   Given a physical resource – energy, time, bits, space,
entanglement; and
2. Given an information processing task – data compression,
information transmission, teleportation; and
3. Given a criterion for success;
How much of 1 do I need to achieve 2, while satisfying 3?

Pursuing this question in the quantum case has led to, and
presumably will continue to lead to, interesting new information
processing capabilities.
“How to write a quant-ph”

Are there any fundamental scientific questions that can be
by quantum information science?

Knowing the rules ≠ Understanding the game
Knowing the rules of quantum mechanics
≠
Understanding quantum mechanics

What high-level principles are
implied by quantum mechanics?
Robert B. Laughlin, 1998 Nobel Lecture
“I give my class of extremely bright graduate students,
who have mastered quantum mechanics but are
otherwise unsuspecting and innocent, a take-home exam
in which they are asked to deduce superfluidity from
first principles.
There is no doubt a special place in hell being reserved
for me at this very moment for this mean trick, for the
Superfluidity, like the fractional quantum Hall effect, is
an emergent phenomenon – a low-energy collective effect
of huge numbers of particles that cannot be deduced
from the microscopic equations of motion in a rigorous
way and that disappears completely when the system is
taken apart (Anderson, 1972)”
Quantum information science as an approach
to the study of complex quantum systems

Quantum processes

Shor’s algorithm

teleportation
communication

theory of entanglement
quantum phase transitions

cryptography
quantum
error-correction

Complexity
A few quanta of miscellanea

The “outer product” notation
The spectral theorem – diagonalizing Hermitian matrices
Historical digression on measurement
The trace operation
Quantum dynamics in continuous time: an alternative
form of the second postulate
Outer product notation
Let ψ and φ be vectors.

Define a linear operation (matrix) ψ φ by
ψ φ (γ )≡ ψ φ γ

Example: 1 0 (α 0 + β 1 ) ≡ 1 α = α 1

Connection to matrices:
If a = ∑ j a j j , and b = ∑ j bj j then a b k = bk* a .

 a1                
But a2  b1* b2*
 
 1 = bk* a .
 
 
                   
 
 a1 
Thus a b = a2  b1* b2* b3*  .
                
 
 
Outer product notation

1         1    0
Example: 0 0 =   1 0  = 
 0    
0    0

0          0    0
Example: 1 1 =   0 1 = 
 1    
0    1

1 0  = 0 0 − 1 1
Example: Z = 
 0 −1

1        0    1
Example: 0 1 =   0 1 = 
 0    0
     0

0        0    0
Example: 1 0 =   1 0  = 
 1   
1    0

0 1 
Example: X =        = 0 1 + 1   0
 1 0
Exercise: Find an outer product representation for Y .
Outer product notation
One of the advantages of the outer product notation
is that it provides a convenient tool with which to
describe projectors, and thus quantum measurements.
Recall: The projector P onto sp ( e1 , e2 ) acts as
P (α e1 + β e2 + γ e3 ) = α e1 + β e2
This gives us a simple explicit formula for P , since
( e1 e1 + e2 e2 ) (α e1 + β e2 + γ e3 ) = α e1 + β e2
More generally, the projector onto a subspace spanned by
orthonormal vectors e1 ,..., em is given by
P = ∑ j ej ej .
Exercise: Suppose e1 ,..., ed    is an orthonormal basis for state
space. Prove that I = ∑ j e j    ej .
†
Exercise: Prove that a b = b a .
The spectral theorem

Theorem: Suppose A is a Hermitian matrix, A † = A. Then
A is diagonalizable,
A = Udiag ( λ1 ,…, λd )U † ,
where U is unitary, and λ1 ,…, λd are the eigenvalues of A.

But diag ( λ1 ,…, λd ) =∑ j λj j   j.

Thus A = ∑ j λj e j     e j , where ej ≡ U j is the λj
eigenvector of A, A e j = λj e j .

A = ∑ k λk Pk , where Pk is the projector onto the
λk eigenspace of A.
Examples of the spectral theorem

1 0 
Example: Z =       = 0 0 − 1 1
0 −1
0 1                       0 ± 1
Example: X =       has eigenvectors ± ≡       , with
 1 0                        2
corresponding eigenvlaues ± 1.

1 1      1 1
+ + - − − =   1 1-   1 -1
2 1    2 -1 
 


1 1 1 1  1 −1
=      −  −1 1 
2 1 1 2       
0 1 
=
 1 0

Historical digression: the measurement
postulate formulated in terms of “observables”
Our form: A complete set of projectors Pj onto orthogonal
subspaces. Outcome j occurs with probability
Pr(j ) = ψ Pj ψ .
The corresponding post-measurement state is
Pj ψ
.
ψ Pj ψ

Old form: A measurement is described by an observable,
a Hermitian operator M , with spectral decomposition
M = ∑ j λj Pj .
The possible measurement outcomes correspond to the
eigenvalues λj , and the outcome λj occurs with probability
Pr(j ) = ψ Pj ψ .
The corresponding post-measurement state is
Pj ψ
.
ψ Pj ψ
An example of observables in action
Example: Suppose we "measure Z".

Z has spectral decomposition Z = 0 0 - 1 1 , so
this is just like measuring in the computational basis,
and calling the outcomes "1" and "-1", respectively, for
0 and 1.

Exercise: Find the spectral decomposition of Z ⊗ Z .
Show that measuring Z ⊗ Z corresponds to measuring
the parity of two qubits, with the result +1 corresponding
to even parity, and the result -1 corresponding to odd
parity.

Exercise: Suppose we measure the observable M for a
state ψ which is an eigenstate of that observable. Show
that, with certainty, the outcome of the measurement is
the corresponding eigenvalue of the observable.
The trace operation
tr (A ) ≡ ∑ j Ajj
0 1                            1 0
Examples: X =      tr ( X ) = 0;         I =
0 1
tr ( I ) = 2.
 1 0                              
Cyclicity property: tr (AB ) =tr (BA ) .

tr (AB ) = ∑ j (AB ) jj = ∑ jk Ajk Bkj = ∑ jk Bkj Ajk = ∑ k (BA )kk = tr (BA )

Exercise: Prove that tr ( a b ) = a b .
An alternative form of postulate 2
Postulate 2: The evolution of a closed quantum system
is described by a unitary transformation.

ψ' =U ψ

But quantum dynamics occurs in continuous time!
An alternate form of postulate 2
The evolution of a closed quantum system is described by
Schroedinger's equation:
d ψ
i      =H ψ
dt
where H is a constant Hermitian matrix known as the
Hamiltonian of the system.

The eigenvectors of H are known as the energy eigenstates
of the system, and the corresponding eigenvalues are known
as the energies.

Example: H = ω X has energy eigenstates ( 0 + 1 ) / 2 and
( 0 − 1 ) / 2, with corresponding energies ± ω
Connection to old form
of postulate 2
The solution of Schroedinger's equation is ψ (t ) = exp( − iHt ) ψ (0)

U ≡ exp( − iHt )     ψ' =U ψ

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