Quantum Black Hole and Information

Soo-Jong Rey @ copyright

Lecture (12): Entanglement

1 Quantum Entanglement

This week, we will study quantum entanglement, multiple qubit systems and quantum

gates (the quantum counterpart of digital gates0. These are essential ingredients for

quantum computation.

One-Qubits

One qubit is a quantum state in two-dimensional vector space, so one-qubit Hilbert space

is H1 = V2 ' Z2. In the standard basis, it is spanned by orthonormal basis vectors

V2 = {|0〉, |1〉}, 〈0|0〉 = 〈1|1〉 = 1, 〈0|1〉 = 〈1|0〉 = 0.

A general qubit state is described by a vector

|ψ〉 = c0|0〉+ c1|1〉 ∈ V2, |c0|2 + |c1|2 = 1.

Since V2 is two-dimensional, a physical measurement would result in two mutually

exclusive result, |0〉 state with 100 % or |1〉 state with 100 %.

Exercise [Hadamard gate]

The qubit basis |0〉, |1〉 is the more familiar basis. From these, construct a new basis which

has equal mixture of qubit states. Such change of basis is called the Hadamard gate.

(Solution) The Hadamard gate H rotates the qubit basis

H : |0〉, |1〉 −→ |+〉, |−〉, H =

1/√

2 1/√

2

−1/√

2 1/√

2

(1.1)

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where

|+〉 =|0〉+ |1〉√

2|−〉 =

|0〉 − |1〉√2

. (1.2)

For both Hadamard states, if we make a measurement in qubit basis, we have equal

probability of | ± 1/√

2|2 = 1/2. This confirms that Hadamard basis is an equal mixture

of qubit basis.

Exercise 1 [measurement and rotation and measurement]

Suppose I start with the qubit state (|0〉+ |1〉)/√

2, make a measurement to an orthogonal

state, then make a Hadamard rotation by π/4, and finally make another measurement to

an orthogonal state. What are possible outcomes?

The initial qubit state is |ψ〉 = (|0〉 + |1〉)/√

2. Orthogonal state measurement would

thus yield either |0〉 or |1〉 with probability |1/√

2|2 = 1/2. In case the qubit is in the

state |0〉 or |1〉, the Hadamard rotation yields

|0〉 → 1√2

(|1〉+ |0〉)

|1〉 → 1√2

(|1〉 − |0〉).

For either states, second measurement will yield |1〉 or ±|0〉 with probability |±1/√

2|2 =

1/2.

two qubits

Consider putting together two qubits. In linear algebra, such operation is called tensor

product. We can temporarily label the two qubits as 1 and 2. The two-qubit Hilbert

space is H2 = V2 ⊗ V2 = (Z2)2. The total number of possible orthogonal states is 2 times

2 = 4. They are

|00〉 ≡ |0〉1 ⊗ |0〉2, |01〉 ≡ |0〉1 ⊗ |1〉2, |10〉 ≡ |1〉1 ⊗ |0〉2, |11〉 ≡ |1〉1 ⊗ |1〉2. (1.3)

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A general qubit state is then given by

|ψ〉 = α|00〉+ β|01〉+ γ|10〉+ δ|11〉

|α|2 + |β|2 + |γ|2 + |δ|2 = 1. (1.4)

We now measure two qubit system. First, if we decide to measure both qubits, we

will get 4 possibilities: (00) result with probability |α|2, or (01) with probability |β|2, or

(10) with probability |γ|2, or (11) with probability |δ|2. After the measurement, the qubit

state is in |00〉, or |01〉, or |10〉, or |11〉.

Second, if we decide to measure only the first qubit but not the second, I will get

(0) result with probability |α|2 + |β|2 or (1) result with probability |γ|2 + |δ|2. After the

measurement, the qubit state is in

1√|α|2 + |β|2

(α|0〉+ β|1〉), 1√|γ|2 + |δ|2

(γ|0〉+ δ|1〉).

The quantum rule is, ”anytime you ask the Universe a question, it makes up its mind;

anytime you do NOT ask a question, it postpones making up its mind for as long as it

can.”

We now transform the two-qubit system. First, we apply a NOT gate to the first or

the second qubit.

NOT1 : |00〉 → |10〉, |01〉 → |11〉, |10〉 → |00〉, |11〉 → |01〉.

NOT2 : |00〉 → |01〉, |01〉 → |00〉, |10〉 → |11〉, |11〉 → |10〉.

In the basis (1.3) of H2, they are realizable by (4× 4) matrices:

NOT1 =

0 0 1 0

0 0 0 1

1 0 0 0

0 1 0 0

, NOT2 =

0 1 0 0

1 0 0 0

0 0 0 1

0 0 1 0

acting on |ψ〉 =

α

β

γ

δ

.

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So, we find that

NOT1|ψ〉 =

γ

δ

α

β

, NOT2|ψ〉 =

β

α

δ

γ

.

(dis)entangled quantum state

In the two-qubit system, depending on values of α, β, γ, δ, the quantum state |ψ〉 may or

may not be factored into the first qubit state and the second qubit state.

• If the quantum state vector |ψ〉 of multi-qubit is factorizable, the state is disen-

tangled or separated.

• If the quantum state vector |ψ〉 of multi-qubit is non-factorizable, the state is en-

tangled.

Examples of disentangled state are

|01〉 = |0〉 ⊗ |1〉

1

2

(|00〉+ i|01〉+ i|10〉 − |11〉

)=

1

2

(|0〉+ i|1〉

)⊗(|0〉+ i|1〉

).

We see that the unentangled states as above are expressed as product of two independent

quantum systems, so the two systems are not related in any manner and each does not

know the presence of the other.

Examples of entangled state are

|Bell pair〉 =1√2

(|10〉 − |01〉)

|Symm1〉 =1√2

(|00〉+ |11〉)

|Symm2〉 =1√3

(|00〉+ |01〉+ |10〉) . (1.5)

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We see that the entangled states as above have the property that once we know the

quantum state of one system, quantum state of the other system follows immediately.

For instance, in the first case, if we know first qubit is in |0〉, then it follows immediately

that the second qubit is in |1〉 state.

It is said that the entanglement is the most distinguishing feature of quantum physics

from the classical physics. In classical physics, two bits that are separated by a wide

distance can be treated separately. In quantum physics, two qubits that are separated by

a wide distance cannot be treated separately if the two qubits were entangled.

Like quantum states can be entangled, linear operators acting on quantum states can

also be entangled. An example is so-called controlled-NOT(CNOT), which maps

controlled−NOT :

|a〉 ⊗ |b〉 → |a〉 ⊗ |a⊕ b〉

|00〉 → |00〉

|01〉 → |01〉

|10〉 → |11〉

|11〉 → |10〉

, so

1 0 0 0

0 1 0 0

0 0 0 1

0 0 1 0

(1.6)

is the matrix realization acting on the basis (1.3).

Problem [controlled NOR]

Find explicit map of so-called controlled NOR, which maps |a〉|b〉 to |a〉|a⊗ b〉.

(Solution) Working out the map explicitly,

controlled−NOR :

|00〉 → |00〉

|01〉 → |00〉

|10〉 → |10〉

|11〉 → |11〉

so

1 1 0 0

0 0 0 0

0 0 1 0

0 0 0 1

(1.7)

I see that this matrix has zero determinant, and has rank-3. So, this matrix does not have

an inverse matrix. Therefore, operation using controlled-NOR represents ”irreversible”

quantum operation.

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In multi-qubit system, entanglement is the single most important aspect of

quantum state vector for the realization of quantum computing.

how to create entanglement?

The idea of quantum computer is to put together qubits and entangle them to perform

computations. So, operations need to be devised for entangling qubits so that they are

not just a bunch of disjoint quantum bits laying around. What are such operations? The

simplest such operation is the CNOT gate followed by Hadamard gate.

First, let’s study NOT1 or NOT2 gates. Acting on a disentangled 2-qubit state

|Ψ〉 = (α|0〉+ β|1〉)⊗ (γ|0〉+ δ|1〉)

We get

NOT1|Ψ〉 = (α|1〉+ β|0〉)⊗ (γ|0〉+ δ|1〉)

NOT2|Ψ〉 = (α|0〉+ β|1〉)⊗ (γ|1〉+ δ|0〉)

So, we only obtain disentangled states. Second, we now apply CNOT or CNOR gate and

obtain

CNOT|Ψ〉 = α|0〉 ⊗ (γ|0〉+ δ|1〉) + β|1〉 ⊗ (δ|0〉+ γ|1〉)

CNOR|Ψ〉 = α(γ + δ)|0〉 ⊗ |0〉+ β|1〉 ⊗ (γ|0〉+ δ|1〉) (1.8)

and cannot be further brought to factorized states. Therefore, these outcomes are entan-

gled quantum states.

The reason why NOT-gates do not produce entanglement while CNOT and CNOR

gates do is clear: NOT-gates operates to each single qubit without reference to other

qubits, whereas CNOT and CNOR gates operate to each single qubit with influence of

other qubits. Since the operation on one qubit depends on the state of the other qubits,

it is evident that entanglement is created.

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In the above results, one may ask if (1.8) can be a disentangled state; for example,

we may set α = 0 or β = 0. Though correct, such result, however, is not generic and

depends on particular orientation in the Hilbert space. In other words, if we change the

initial state vector slightly from this particular orientation, the resulting states in (1.8)

will always be entangled states.

is entanglement measurable?

We learned that all physical observables, i.e. quantities that can be measured by some

experiment, are described by Hermitian matrices in the Hilbert space. An example

is the Hamiltonian H, which measures the energy. Is entanglement a physical observ-

able? Suppose it is, and divide the Hilbert space H into two Hilbert subspaces HA,HB,

H = HA ⊕ HB. Then, there must exist a projection operator PE that acts on both

subspaces simultaneously such that 〈ψ|PE|ψ〉 is nonzero if ψ is entangled and zero if ψ

is un-entangled. But such an observable or linear operator cannot exist. The reason is

simply because any entangled state can be decomposed to a linear combination of disen-

tangled state. For instance, take the entangled states (1.8). It is clear they are a linear

superposition of individual, disentangled states such as |00〉, |11〉, |01〉, |10〉. Consider now

the projection operator PE acting on these states. If we assign that PE acting on any

disentangled state is zero, then

PE|ψ1〉 = PE

(1√2

(|00〉+ i|11〉))

=1√2

(PE|00〉+ i PE|11〉)

= 0. (1.9)

The trouble is that a linear operator such as PE ought to act on states linearly.

We note that the above observation does not exclude construction of a ”specific”

projection operator which tests whether a given state is a particular entangled state |ψo〉

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or not. It is the standard projection operator Po = |ψo〉〈ψo|, but this operator does not

offer me anything useful for testing whether some other states |ψ1〉, |ψ2〉, · · · are entangled

or not.

Summarizing, the fact that linear operator acts linearly in Hilbert space and that

entangled states are always decomposable into a sum of disentangled states are mutually

incompatible. It is said an outstanding question in quantum computing is to devise a

method for measuring and quantifying entanglement of a given state.

Classical and Quantum Correlations

Let’s recall correlation in classical statistics. If we have events A,B, the correlation (

AKA cross-correlation) between them is the ”conditional probability”

C(A,B) = P (B|A)P (A)∣∣∣P (A)=1

= P (A|B)P (B)∣∣∣P (B)=1

.

So, how do we generate correlations? There are two ways: one is quantum correlations

and another is classical correlations. To illustrate this, we start with the Bell pair

|Bell pair〉 =1√2

(|1〉A ⊗ |0〉B − |0〉A ⊗ |1〉B).

This is a vector in 4-dimensional Hilbert space. We know from linear algebra class that

we can construct 4× 4 matrix

ρBell(A,B) = |Bell pair〉〈Bell pair|

=1

2[(|1〉〈1|)A ⊗ (|0〉〈0|)B + (|0〉〈0|)A ⊗ (|1〉〈1|)B]

− 1

2[(|1〉〈0|)A(|0〉〈1|)B + (|0〉〈1|)A(|1〉〈0|)B] (1.10)

This matrix is called density matrix. A density matrix is a matrix of size (n×n) where

n is the dimension of the Hilbert space.

We learned from previous week’s study that the last line corresponds to interference

of the quantum state. Since interference is the hallmark of quantum particle not present

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in classical particle, the classical limit of quantum state corresponds to the limit this last

line is dropped out. This process of eliminating the interference or the coherence is called

decoherence.

Entangled versus Disentangled

• Multiparticle pure states are vectors in the Hilbert space

H = H1 ⊗H2 ⊗ · · · × Hn.

• If a state is the form of product state |ψ1〉⊗· · · |ψn〉, the state is said distentangled.

• If not the form of product state, the state is said to be entangled.

• The total Hilbert space H is very large. Most of its vectors are not product states.

Pure versus Mixed States

• The dual Hilbert space H∗ is the space of linear maps from H to C. Its vectors are

Hermitian conjugates 〈ψ|.

• It is possible to form the space of density matrix operators on H:

D(H) = H⊗H∗

• A vector of the form |ψ〉〈ψ| is called a pure state

• If not of that form, the state is called a mixed state.

• A mixed state is always obtainable from a pure state by ”decoherence” process of

washing out coherence phase.

The notion of pure versus mixed states and the notion of entangled and disentangled

states are not related at all. So, qubit states can be classified into four different classes,

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(i) pure and entangled, (ii) pure and disentangled, (iii) mixed and entangled, and (iv)

mixed and disentangled.

Example [pure and mixed states from Bell pair]

For the above Bell pair, the pure state corresponds to (1.10)

ρpure(A,B) =1

2

[(|1〉〈1|)A ⊗ (|0〉〈0|)B + (|0〉〈0|)A ⊗ (|1〉〈1|)B

−(|1〉〈0|)A(|0〉〈1|)B − (|0〉〈1|)A(|1〉〈0|)B]

whereas a mixed state corresponds to the density matrix where the second line is dropped

off:

ρmixed(A,B) =1

2

[(|1〉〈1|)A ⊗ (|0〉〈0|)B + (|0〉〈0|)A ⊗ (|1〉〈1|)B

]From the structure of the linear sum, I see that this mixed state density matrix represents

a configuration in which A = 1, B = 0 ”or” A = 0, B = 1 with equal ”probability”. This

should be contrasted to the pure state density matrix, which represents a configuration

in which A = 1, B = 0 ”or” A = 0, B = 1 with equal ”amplitude”.

As a final issue, let us ask if an entangled state can be mapped to a disentangled state

or vice versa by a change of qubit basis.

Problem [Hadamard gate on pure / mixed states]

Suppose I apply the Hadamard gate to the qubit basis

UH =

1/√

2 1/√

2

−1/√

2 1/√

2

i.e. UH |0〉 =1√2

(|0〉+ |1〉), UH |1〉 =1√2

(|0〉 − |1〉).(1.8)

How does the density matrices of pure and mixed states change and what does that

mean?

(Answer) The density matrix is transformed under Hadamard gate as

ρ(A,B) −→ (UH,A ⊗ UH,B)ρ(A,B)(UH,A ⊗ UH,A)†.

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This has the effect of replacing 0, 1 qubit orientation to a linear combination of 0, 1 as

(1.8). So,

|Bell pair〉 → 1√2

[1

2(|0〉+ |1〉)A(|0〉 − |1〉)B −

1

2(|0〉 − |1〉)A(|0〉+ |1〉)B

]=

1√2

[|0〉A|1〉B − |1〉A|0〉B

]= |Bell pair〉.

So, Bell pair state is invariant under the Hadamard gate operation. Since density matrix

of pure state is |Bell〉〈Bell|, it is obvious that it does not change under the Hadamard

gate operation.

Problem [Invariance of Bell Pair State]

Consider a gate rotating from the computational basis 0, 1 to a rotated basis simultane-

ously on every qubit

|0〉 → (cos θ|0〉+ sin θ|1〉); |1〉 → (− sin θ|0〉+ cos θ|1〉).

Show that the Bell-pair state remains the Bell-pair state under this rotation.

(Solution) The important point is that every qubits are rotated by the same angle. So,

|01〉 → − sin θ cos θ|00〉+ cos2 θ|01〉+ sin2 θ|10〉+ sin θ cos θ|11〉

|10〉 → − sin θ cos θ||00〉 − sin2 θ|01〉+ cos2 θ|10〉+ sin θ cos θ|11〉.

Subtracting, I see that Bell-pair state goes back to Bell-pair state.

Einstein-Podolsky-Rosen Paradox

Einstein, Podolsky, Rosen (EPR) gave a specific argument pointing to incompleteness

of the quantum-mechanical description of the microscopic world. Consider a system of

2-qubits in the Bell-pair state

|Ψ〉 =1√2

(|0〉A|1〉B − |1〉A|0〉B) .

The conditions imposed by EPR to the system are

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• perfect correlations When the states of qubits A and B are measured along a

common given orientation, the outcomes of A and B will be opposite.

• locality Since at the time of measurement the qubits A and B no longer interact,

no real change can take place in the B qubit in consequence of anything that may

be done to the A qubit.

Based on these, the EPR argument then goes as follows:

• (a) If the outcomes of a measurement can be predicted without disturbing the mea-

sured qubit, then the measurement has a definite outcome, whether this operation

is ”actually” performed or not.

• (b) For a pair of qubits in the Bell-pair state, the outcome of a measurement of A

qubit can be determined without disturbing B qubit and vice versa.

Then, by measuring the quantity corresponding to P (|0〉A), the value of the quantity

corresponding to P (|1〉A) can be found as well. By (b), the values of the same two

properties of the second qubit can be obtained ”without disturbing it”. This is because

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Figure 1: Illustration of the EPR paradox for two entangled atoms. Atom I and atom II

are pair produced from the vacuum and separated far apart so that there is no time for

an interaction between them.

the Bell-pair state describes perfect anti-correlation between A and B qubits. By (a), these

values of the B qubit are ”definite”. Now, recalling important property that the Bell-pair

state is invariant under arbitrary unitary rotation of computational basis, I conclude that

the value of the states of both A and B must be definite for all values of the rotation.

Based on the above lines of reasoning, EPR argues that the description of the sys-

tem in current formulation of quantum mechanics is incomplete. EPR concluded that

there are two possible resolutions. Either there was some interaction between the qubits,

even though they were separated, or the information about the outcome of all possible

measurements was already present in both particles. Since the first is in contradiction

with special relativity, EPR concluded that the second option is preferred. In the second

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option, however, the information about the outcome must be encoded in some ”hidden

parameters”. The necessity of such parameters is the basis that EPR called that present

formulation of quantum mechanics is incomplete.

Bell’s Theorem

The first fundamental theorem is the Bell’s theorem or Bell’s inequality. This theorem

refutes the option EPR preferred (incompleteness of quantum mechanics) in their paradox.

This inequality is a test to distinguish quantum correlations (where coherence and

entanglement is extremely important) from classical correlations (where coherence is par-

tially or completely lost) by comparing the correlation along different qubit orientation

0, 1.

Consider the physical observable represented by the Hermitian matrix

C(a,b) = (a · ~σ)A ⊗ (b · ~σ)B =

a3 a1 + ia2

a1 − ia2 −a3

A

⊗

b3 b1 + ib2

b1 − ib2 −b3

B

where a,b are two arbitrary unit vectors and ~σ is so-called Pauli matrices

σ1 =

0 1

1 0

, σ2 =

0 i

−i 0

, σ3 =

1 0

0 −1

.

Then, Bell’s inequality asserts that, for any choice of unit vectors a,b, a′,b′, classically

correlated system obeys

B :=∣∣∣C(a,b) + C(a,b′) + C(a′,b)− C(a′,b′)

∣∣∣ ≤ 2. (1.0)

If B > 2 for some choice of a,b, a′,b′, then the system is quantum correlated.

It should be noted that the converse is not true: There are mixed states that are

entangled yet satisfy B ≤ 2 for all set of unit vectors.

entanglement measures for pure states

Starting from disentangled product state, entangled state can be prepared by exchanging

a certain amount of quantum information. This quantum information can take the form

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of Bell pairs, shared between A,B. The unit of entanglement is ”entangled bit” or ”ebit”:

1 Bell pair = 1 bit of entanglement = 1 ebit

The average number of Bell pairs per copy needed to prepare a large number of copies of

the pure state |ΨAB〉 is given by so-called entanglement entropy or entanglement of

formation

E = TrA

(ρA log2

1

ρA

)where ρA = TrB|ΨAB〉〈ΨAB|. (1.1)

Here, the reduced density matrix ρA of subsystem A is obtained by tracing over the

subsystem B. Note that the entanglement entropy is partition-specific: even if entangled

for the partition into A,B, the total system may well be separable or disentangled for

some other partition into A,B.

Problem [entanglement entropy of Bell pair]

Consider a system of 2-qubits. A normalized 2-qubit state has the form

|Ψ〉 =1∑

i=0

1∑j=0

γij|i〉A|j〉B, Trγγ† = 1.

Compute the entanglement entropy E . Show that it is maximum for a Bell-pair state.

(Solution) The density matrix takes the form

ρ(A,B) =1∑

i,j=0

1∑k,l=0

γijγ†kl|i〉A|j〉BB〈k|A〈l|.

Taking trace over B qubit, the reduced density matrix of A qubit is

ρA =1∑

i,`=0

(γγ†)i`|i〉AA〈`|.

To compute the trace of −ρA log2 ρA, the idea is to diagonalize ρA and then get the trace

by summing over the diagonal entries. This is what I learned now in linear algebra class.

The characteristic equation Det(γγ† − λI) reads

λ2 − λ[(γγ†)11 + (γγ†)22] + Detγγ† = 0.

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Noting that the bracket in the second term is Trγγ† = 1, the eigenvalues are

λ± =1

2± 1

2

√1− 4Detγγ†.

So, the entanglement entropy is

E = −λ+ log2 λ+ − λ− log2 λ− = F(1

2+

1

2

√1− 4Detγγ†)

where the function F(x) is defined by

F(x) = −x log2 x− (1− x) log2(1− x).

The quantity

C = 2√

Detγγ† = 2√

Detρreduced (1.-5)

is called the concurrence of the 2 qubits. Let us evaluate this for a few pure states. For

a Bell pair, the matrix γ is given by

γBell =

0 +1/√

2

−1/√

2 0

→ CBell = 1.

For a product state |Ψ〉 = |00〉,

γproduct =

1 0

0 0

→ Cproduct = 0.

We conclude that a Bell pair has unit concurrence and carries one bit of entan-

glement, while both of them vanish for a product state.

For mixed states, consider a reduced density matrix given by

ρ =

(1 + ε)/2 0

0 (1− ε)/2

→ C =√

1− ε2 ≤ 1.

Consider another reduced density matrix given by

ρ =

1/2 α

α∗ 1/2

→ C =√

1− 4|α|2 ≤ 1.

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We conclude that mixed states have concurrence less than unity.

Yet another quantity is the Bell correlator B. If we define maximal Bell correlator as

Bmax = Maxa,b,a′,b′B

it is related to the concurrence by

Bmax = 2√

1 + C2.

We see that the Bell’s inequality is saturated for product state, while it is violated for

any nonzero concurrence state. I also see that the Bell’s inequality is violated for any

entangled state.

entanglement measure of mixed states

We first recall that, for a mixed state, the density matrix can be taken as a mixture of

(not necessarily orthogonal) pure states

ρ =∑n

pn|Ψn〉〈Ψn|, pn > 0,∑n

pn = 1.

A mixed state for A,B subsystems is disentangled if its density matrix can be put

to the above form with |Ψn〉 = |Φn〉A|Φ′n〉B for all n. Then, the entanglement entropy E

defined over all possible choices of pure state basis by

E = MinΨn,pn

∑n

pnE(Ψn).

vanishes if the state is disentangled.

So, what is the meaning of the entanglement entropy of a mixed state defined as

above? Firstly, it is an upper bound to the average number of Bell pairs per copy that

it costs to prepare many copies ρ ⊗ ρ ⊗ · · · of the state ρ. This so-called ”entanglement

cost” is additive, viz. E(ρ ⊗ ρ) = 2E(ρ), etc. Secondly, it is also an upper bound to

the average number of Bell pairs per copy one can extrat or ”distill” from many copies

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of an entangled state using only local operations and classical communications. While

the production process is reversible for pure states, meaning that E equals the distillable

entanglement, this is not the case for mixed states. The distillable entanglement can be

less than E , so a fraction of the Bell pairs can be lost in the conversion from Bell pairs to

entangled mixed states and back to Bell pairs.

Entanglement Monogamy Theorem

Monogamy is one of the most fundamental properties of entanglement. If two qubits A

and B are maximally quantum correlated, they cannot be correlated at all with a third

qubit C. In general, there is a trade-off between the amount of entanglement between qubit

A and qubit B and the same qubit A and qubit C. This is mathematically expressed by

the Coffman-Kundu-Wootters (CKW) monogamy inequality.

Theorem 1 [Entanglement Monogamy Theorem]

Suppose a tripartite system consisting of subsystems A,B,C. Denote concurrences be-

tween A and B, respectively, A and C as CAB, CAC , while the concurrence between A and

(BC) as CA(BC). Then, they must satisfy the following inequality:

(CAB)2 + (CAC)2 ≤ (CA(BC))2.

The inequality can be extended to the case of n-qubits.

More generally, the monogamy inequality can be expressed in terms of entanglement

measure E. For any tripartite state of systems A,B1, B2,

E(A|B1) + E(A|B2) ≤ E(A|B1B2).

If the above inequality holds in general, then it can be immediately generalized by induc-

tion to the multipartite case:

E(A|B1) + E(A|B2) + · · ·+ E(A|BN) ≤ E(A|B1B2 · · ·BN).

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No-Communication Theorem

We found the entanglement very weird. Suppose we have the 2-qubit state |00〉 + |11〉.

What happens if we measure ”only” the first qubit and ”not” the second? With prob-

ability 1/2 each, we get |00〉 and |11〉. This looks amazing since, by ”only” measuring

the state of the first qubit, we ”can” know what state the second qubit is in! It does not

matter how far the second qubit is physically separated from the first one. We can put

one at my laptop and another at the end of the Universe. Still, we ”can” know ”instantly”

what state the qubit at the end of the Universe is in by only knowing the state in my

laptop.

Isn’t this violation of Einstein’s special relativity, which says that speed of light is

the maximum speed? According to my father, NO. The reason is that if I try to send a

message faster than light to the second qubit, the result would be unpredictable. This

leads to the following theorem

Theorem 2 [No Communication Theorem]

In 2-qubit system as above, operation done ”only” to the first qubit can not affect the

probability of outcome of any measurement ”only” on the second qubit.

No-Cloning Theorem

Is it possible to xerox-copy a quantum state? If yes, that would be very nice since we

know we only have one chance to measure a quantum state, and duplicated copy can be

used for multiple measurement. Suppose we have a qubit system and have xerox-copied

once. Then, this generates 2-qubit system

α|0〉+ β|1〉 −→ (α|0〉+ β|1〉)⊗ (α|0〉+ β|1〉) = α2|00〉+ αβ|01〉+ αβ|10〉+ β2|11〉.

Theorem 3 [No-Cloning Theorem]

The xerox-copies of an arbitrary unknown quantum state is forbidden in quantum me-

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chanics since it is ”nonlinear” in the amplitudes α, β and goes outside linear algebra (the

superposition principle).

We note that cloning and entanglement are different. We saw earlier that using Con-

trolled NOT gate or Walsh-Hadamard gate we can put two qubits entangled. This is not

cloning. No well-defined quantum state can be attributed to a subsystem of an entangled

state. Cloning is a process whose result is a separable state with identical factors.

(Proof) Suppose the state of a quantum system A is |Ψ〉A, which I wish to xerox-copy.

In order to copy it, I take a system B with the same state space and initial state |0〉B. This

initial state must be blank and independent of |Ψ〉A, of which I have no prior knowledge.

The state of the composite system (AB) is then described by the following tensor product:

.

|Ψ〉AB = |Ψ〉A ⊗ |0〉B (1.-16)

There are only two ways to manipulate the composite system. We could perform an

observation, which irreversibly collapses the system into some eigenstate of the observable,

corrupting the information contained in the qubit. This is obviously not what I want.

Alternatively, I could control the Hamiltonian of the system, and thus the time evolution

operator U = exp(i∫H(t)dt) up to some fixed time interval, which yields a unitary

operator. Then U acts as a xerox-copier provided that

U |Ψ〉A|0〉B = |Ψ〉A|Ψ〉B (1.-15)

for all possible states |Ψ〉 in the state space. Let’s choose two arbitrary initial states

|ψ〉 and |φ〉. Since U is a unitary operation acting on the total Hilbert space of AB, it

preserves the inner product. So,

B〈φ|A〈φ|ψ〉A|ψ〉B = B〈0|A〈φ|U †U |ψ〉A|0〉B = B〈0|A〈φ|ψ〉A|0〉B. (1.-14)

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and since quantum mechanical states are assumed to be normalized, it follows that

〈φ|ψ〉 = (〈φ|ψ〉)2 → 〈φ|ψ〉 = 0 or 1 (1.-13)

for arbitrary |ψ〉 and |φ〉. So, either φ = ψ or φ ⊥ ψ. This is not the case for two

arbitrarily chosen states. We conclude that a unitary operation U cannot clone a general

quantum state. This proves the no-cloning theorem.

n Qubits

The explosion of quantum computing started from studying multi qubits, so we will first

study them.

Recall that we have 2 amplitudes for a qubit. If we have n qubits, then we would have

2n amplitudes. So, a quantum state is a vector in the Hilbert space of dimension 2n:

|Ψ(n)〉 =∑x∈Zn

cx|x〉. (1.-12)

What does this mean? To keep track of the state of n numbers, Nature apparently has to

write down this list of 2n complex numbers. They are the components of a vector in the

Hilbert space of n qubits. If we have n = 1, 000 qubits, that means there are 21000 ∼ 10300

complex coefficients cx’s to compute. This is more number than the total atoms in the

current Universe.

Interference

Waves can be regarded as a ’vector’ in Hilbert space: if we superpose two waves ψ1(x, t)

and ψ2(x, t), the ”amplitude” of superposed wave is obtained by adding them:

|ψtotal〉 = |ψ1〉+ |ψ2〉. (1.-11)

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The ”intensity” of superposed wave is proportional to the norm of this wave vector

Itotal(x, t) ∝∣∣∣|ψtotal〉

∣∣∣2 = 〈ψtotal|ψtotal〉

=∣∣∣|ψ1〉

∣∣∣2 +∣∣∣|ψ2〉

∣∣∣2 + (〈ψ1|ψ2〉+ 〈ψ2|ψ1〉)

= I1 + I2 + (〈ψ1|ψ2〉+ 〈ψ1|ψ2〉∗). (1.-12)

The last term is called ”interference” between the two waves, ψ1, ψ2. Its value takes

maximum I1 + I2 and minimum 0. So, total intensity can be as large as 2(I1 + I2) for

constructive interference or can be zero for destructive interference. In classical mechanics,

this is the hallmark of waves as opposed to particles. This is the hallmark of quantum

mechanics since quantum mechanics equates particles with waves (de Broglie’s material

wave hypothesis).

The destructive interference is precisely what quantum computing exploits. The goal

is to arrange computational paths so that the paths leading to wrong answers interfere

destructively while the paths leading to right answers interfere constructively. In this

arrangement, a key point is that the paths that will interference destructively must lead

to identical outcomes in ’every’ respect. I saw this kind of arrangement in physics course,

’diffraction grating’ two light paths that differ by 1/2 wavelength will interfere to zero,

three light paths that successively differ by 1/3 wavelength will interfere to zero, · · · , n

light paths that successively differ by 1/n wavelength will interfere to zero.

Quantum gates

Suppose we have a quantum computer operating with n qubits. How do we arrange the

destructive interference by a sequence of quantum evolution? Here, quantum evolution

refers to a sequence of unitary transformations called ”quantum gates” and these gates

form so-called ”quantum circuit”.

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As an example, consider the quantum computer in the initial state

|ψ(0)〉 = |0〉 ⊗ |0〉 ⊗ · · · ⊗ |0〉. (1.-11)

We then apply the Hadamard gate to the first qubit, then a CNOT gate to the second

qubit. The sequence gives the following entangled state:

|ψ(0)〉 → (Hadamard1)→ 1√2

(|000 · · · 〉+ |100 · · · 〉)

→ (CNOT2) → 1√2

(|000 · · · 〉+ |110 · · · 〉) . (1.-11)

In principle, we can infinitely many quantum gates, corresponding to infinitely many

possible unitary matrices. Among them, can we find more elementary or fundamental

gates? Indeed, yes.

Definition: universal set of quantum gates

These are special gates which act only on just 1-, 2- or at most 3-qubits and whose infinite

variety of combinations involving only real numbers can approximate any unitary matrix

to arbitrary precision.

If we take Hadamard gate and CNOT gate as the universal set, the system can be

simulated efficiently with a classical computer. This was proven by Gottesman and Knill.

On the other hand, if we take Hadamard gate and controlled-controlled NOT (CCNOT)

gate:

CCNOT : |x〉|y〉|z〉 → |x〉|y〉|z ⊕ xy〉 (1.-10)

as the universal set, the system can approximate real-valued unitary matrices to arbitrary

precision.

The choice of universal set is not unique. For instance, another set which simulates

quantum computing is CNOT gate and Rotation(π/8)-gate:

R(π/8) :

cos(π/8) sin(π/8)

− sin(π/8) cos(π/8)

. (1.-9)

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Quantum Teleportation

Based on the fundamental theorems, we now want to show how Alice can ’teleport’ an

arbitrary qubit |ψ〉 to Bob, despite sending only classical information. Alice and Bob

share beforehand an entangled Bell state:

|BellAB〉 =1√2

(|00〉+ |11〉). (1.-8)

Alice stays on Physics Building at SNU and hold qubit A, while Bob goes to Gangnam

and holds qubit B. Alice is given another qubit X and she wants to send it to Bob, whose

state will be denoted as |ψ〉 = α|0〉+ β|1〉. The goal of my protocol is to end with |ψ〉 in

Bob’s qubit B.

(1) Alice begins by applying a CNOT gate from her qubit X to her qubit A. What is the

state of the 3-qubit system XAB after this?

The starting state is

(α|0〉+ β|1〉) 1√2

(|00〉+ |11〉) =1√2

(α|000〉+ α|011〉+ β|100〉+ β|111〉). (1.-7)

After applying a CNOT from X to A, the state changes to

1√2

(α|000〉+ α|011〉+ β|110〉+ β|101〉). (1.-6)

(2) Next, Alice applies a Hadamard gate to her qubit X. What is the resulting 3-qubit

state?

1

2(α|000〉+ α|100〉+ α|011〉+ α|111〉+ β|010〉 − β|110〉+ β|001〉 − β|101〉). (1.-5)

(3) Next, Alice measures both her qubits X,A in the standard basis. What is the prob-

ability that she gets the measurement outcome |11〉? If she does, what 1-qubit state is

Bob holding in B?

The probability that Alice gets |11〉 equals the norm-squared of (α|111〉−β|110〉)/2. Ths

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equals (|α|2 + |β|2)/4 = 1/2. Then, Bob’s state becomes proportional to (α|1〉 − β|0〉.

(4) Bob’s state from (3) is almost like the state |ψ〉 = α|0〉 + β|1〉 that Alice wanted to

send, but not quite. What 2 × 2 unitary operation can Bob apply to convert his state

into |ψ〉.

Bob needs to turn −β|0〉+ α|1〉 into α|0〉+ β|1〉. This is done with

U =

0 1

−1 0

, U

−βα

=

α

β

. (1.-4)

(5) The operations Bob needs to perform to get |ψ〉 depend on the outcome of Alice’s

measurement. So, Alice must communicate to Bob which measurement outcome she got

for Bob to know how to recoever |ψ〉. Why does any protocol to teleport a qubit requires

communication from Alice to Bob?

If the protocol worked without communication, Alice could instantaneously send a mes-

sage to Bob by teleporting a qubit, say, in either |0〉 or |1〉, in violation of the speed of

light communication bound and the No-Communication Theorem.

(6) What if the state |ψ〉 could involve unlimited information? For example, Alice could

put entire text of Encyclopedia as the binary expansion of α!. Is it possible that the

teleportation protocol let Bob actually recover that much information, given that Alice

and Bob can both perform unitary transformation with infinite precision?

No, that is impossible. Bob can only measure his qubit ”once”, after which the state

vector collapses to a basis state. So, Bob cannot recover more than one bit of information

from it.

(7) It might seem that Alice has sent a copy of |ψ〉 to Bob, in violation of the No-Cloning

Theorem. Why is this not the case?

During the protocol, Alice destroyed her state |ψ〉 by measuring the qubit X. Alice can

no longer recover it except by getting it from Bob. So, |ψ〉 was not duplicated because

only one copy now exists.

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