Quantum Mechanics 1 Lecture 13: Quantum Theory and Classical physics

Quantum Mechanics 1 Lecture 13:

Quantum Theory

and Classical physics

Massimiliano Grazzini

University of Zurich

Outline

Quantum information

- pure vs mixed states The density matrix

Entanglement EPR experiment

Hidden variables and the Bell theorem

- quantum mechanics vs local realism

Quantum information

The theory of quantum information deals with how a quantum system can be used to accumulate, transmit and elaborate information

In the following we just want to discuss the simplest questions in the case of the simplest quantum system, the two-state system (qubit)

How can we extract the information in quantum state?

How can we code it ?

How can we determine the state of a quantum system ?

Bit Qubit

|χ⟩ = a₊| + ⟩ + a₋| − ⟩ _{a+, a− ∈ ℂ}

with

In Lecture 10 we have seen that an arbitrary normalised spin state

can be written as

|χ⟩ = cos(θ/2)| + ⟩ + e^iϕ sin(θ/2)| − ⟩

with

and

This parametrization in terms of only two real parameters is possible because the overall phase is irrelevant

This holds for a state

of a general two-level system

We now ask ourselves how can we determine the state of the system having many copies of the system in the same state

but not knowing how it was prepared

Suppose that we can measure an observable and let us choose the basis

corresponding to its eigenvectors

A| + ⟩ = λ₊| + ⟩ A| − ⟩ = λ₋| − ⟩

The expectation value of this observable on the state

is

⟨ψ |A|ψ⟩ = ⟨A⟩ = P₊λ₊ + P₋λ₋

where

P₊ = |a₊|² P₋ = |a₋|²

These probabilities are fully determined by the value of and by the normalisation condition

Therefore we can fix the coefficient or, equivalently, the angle , but we

cannot say anything on the relative phase, since a measurement of does not depend on it

a₊ θ

A

To learn more about our system we have to measure other observables which do not commute with

Indeed if commutes with we can write

A′| + ⟩ = λ′₊| + ⟩ A′| − ⟩ = λ′₋| − ⟩

and thus

which still allows us to determine the probabilities but not the relative phase

The density matrix

Where the trace is defined as

For a system in a state

the density matrix is the hermitian operator

ρ_ψ ≡ |ψ⟩⟨ψ |

This matrix allows us to easily write the expectation value of an observable :

we have

⟨A⟩ = Tr (Aρ)

To proceed further it is useful to introduce the density matrix

Tr(O) = ∑

i

⟨e_i|O|e_i⟩

and

is an orthonormal basis of the system

Indeed since the trace is independent on the basis, we can choose a basis of eigenstates of and we find

Tr(Aρ) = ∑

i

⟨e_i|Aρ|e_i⟩ = ∑

i

⟨e_i|A|ψ⟩⟨ψ |e_i⟩ = ∑

i

λ_i|⟨e_i|ψ⟩|² = ⟨A⟩

The density matrix has the property of having unit trace

Tr(ρ) = ∑

i

⟨e_i|ρ|e_i⟩ = ∑

i

⟨e_i|ψ⟩⟨ψ |e_i⟩ = ∑

i

|⟨e_i|ψ⟩|² = 1

The density matrix allows us to describe the case in which we don’t have

complete information on the state of a quantum system, but we just know the probabilities of the states

We assume that we have an ensemble of states and we distinguish two cases

The system or objects under investigation are all in the same state : we say that the system is in a pure state or pure ensemble (for example: beam of spin 1/2

particles all with the same polarisation)

|ψ⟩

Pure state

We can investigate such ensemble of identically prepared states Suppose that

|ψ⟩ = ∑

i

c_i|e_i⟩

where the orthonormal states

correspond to values of the observable

If we have identical copies of the state and we measure we can call the number of times we obtain the result

The larger is and the better the ratio

will approach

|c_i|² = lim

N→∞

N_i N

In this limit the expectation value of and the statistical average coincide

⟨ψ|A|ψ⟩ = ∑

i

|c_i|²a_i = lim

N→∞

1

N ∑

i

N_ia_i

Mixed state

Our system is a statistical superposition of states: we say that the system is in a mixed state, or a mixed ensemble (for example: beam of spin-1/2 particles of which we do not know the polarisation)

The system is in each of the states with probability : the density matrix is

defined as

ρ ≡ ∑

i

P_i|ψ_i⟩⟨ψ_i|

Statistical average

Note that a mixed state actually corresponds to a classical mixture of pure states, not to their quantum superposition, and that the

need not be orthogonal

The pure state is obtained back in the case in which for a given and

for

P_j = 0 j ≠ i

The density matrix defined in this way allows us to compute the average of an observable as

Tr(Aρ) = ∑

i

⟨e_i|Aρ|e_i⟩ = ∑

ij

⟨e_i|A|ψ_j⟩⟨ψ_j|e_i⟩P_j = ∑

ij

⟨ψ_j|e_i⟩⟨e_i|A|ψ_j⟩P_j

= ∑

j

⟨ψ_j|A|ψ_j⟩P_j

that is, the weighted average of the expectation values over the states ,

which we assume to be normalised

The density matrix is a semi-positive operator: indeed for any state

we have

⟨ψ |ρ|ψ⟩ = ∑

i

P_i⟨ψ |ψ_i⟩⟨ψ_i|ψ⟩ = ∑

i

P_i|⟨ψ |ψ_i⟩|² ≥ 0

A measurement of always produces the result

Tr(Aρ_pure) = 12 Tr

[(0 1

1 0) (1 1

1 1)] = ⟨A⟩_ψpure = 1

We can illustrate the difference between a pure and a mixed state with a simple example

Let us consider the pure state defined as

2 (| + ⟩ + | − ⟩)

ρ_pure = |ψ_pure⟩⟨ψ_pure| = 12 (1 1 1 1)

This is an eigenvector with eigenvalue 1 of the operator

1 0) A|ψ_pure⟩ = |ψ_pure⟩

Pure vs mixed states

Let us now consider a mixed state, with 50% probability of being in and

50% probability of being in

The corresponding density matrix is

Indeed we have

pure) = ⟨B⟩_ψpure = 0 Tr(Bρ_mixed) = 1

2 ⟨ + |B| + ⟩ + 1

2 ⟨ − |B| − ⟩ = 0 ρ_mixed = 12 | + ⟩⟨ + | + 12 | − ⟩⟨ − | = 12 (1 0

0 1)

The density matrix is

A measurement of gives in this case

Tr(Aρ_mixed) = 12⟨ + |A| + ⟩ + 12⟨ − |A| − ⟩ = 0

The result for the average value of is different !

Therefore a suitable measurement allows us to distinguish a statistical mixture from a quantum superposition of the same states

We note, however, that if we now consider the operator which is diagonal on the basis we have chosen a measurement of gives the same results in both states

B = (1 0 0 −1) B

What are the differences in the density matrix in the case of pure and mixed states ?

This is the special case of a general result: in the case of absence of information the density matrix is a multiple of the identity

We conclude that

pure state

Tr(ρ²) = 1 _Tr(ρ²_{) < 1}

mixed state and that for a pure state the density matrix is a projection operator It is easy to see that in a pure state we have

and thus

Indeed

On the other hand in a mixed state we have

Tr(ρ²) = Tr ∑

i

P_i|ψ_i⟩⟨ψ_i| × ∑

j

P_j|ψ_j⟩⟨ψ_j| = ∑

i,j,k

⟨e_k|ψ_i⟩⟨ψ_i|ψ_j⟩⟨ψ_j|e_k⟩P_iP_j

= ∑

i,j

|⟨ψ_i|ψ_j⟩|²P_iP_j = ∑

i

P_i² + ∑

i≠j

|⟨ψ_i|ψ_j⟩|²P_iP_j < ∑

i

P_i² + ∑

i≠j

P_iP_j

= ∑

i,j

P_iP_j =

(∑

i

P_i )

2

= 1

where we have used

if

and the normalisation of the states

Tr(ρ²) < 1

We now ask ourselves what are the most general measurement and density

operator for our two-level system

In the case of a two-level system these objects can be fully characterised: both must indeed be

hermitian matrices

A general observable is an arbitrary hermitian operator , while the most general density matrix is an arbitrary hermitian operator with unit trace

A

hermitian matrix is fully determined by real parameters

Indeed the diagonal elements must be real and the off diagonal elements must be one the complex conjugate of the other

M_ii = M*_ii i = 1,2 M₁₂ = M*₂₁

We can thus introduce four linearly independent hermitian matrices and write as a linear combination of them

M

A convenient choice is represented by the Pauli matrices and the identity matrix

σ₁ = (0 1

1 0) σ₂ = (0 −i

i 0 ) σ₃ = (1 0 0 −1) σ₀ = I = (1 0

0 1)

The most general observable is thus

A = a₀I + a ⋅ σ

The most general density matrix takes the same form with the additional constraint that

Tr(ρ) = 2a₀ = 1

ρ = 12 (I + n ⋅ σ) = 12 ( 1 + n₃ n₁ − in₂ n₁ + in₂ 1 − n₃ )

We can write it in the form

Since is a semi-positive hermitian operator its eigenvalues must be real and positive or vanishing, which means that

det(ρ) ≥ 0

On the other hand for a pure state must be a projection operator, which means that, in our case, in some basis it must take the form

ρ = (1 0 0 0)

And, therefore for a pure state

det(ρ) = 0

(1)

Now, in the parametrisation the determinant of is

det(ρ) = 1

4 (1 − |n|²)

Therefore, since

the density matrices can be described as

ρ = 12 (I + n ⋅ σ) |n|² ≤ 1

This is the Bloch sphere, i.e. the sphere of unit radius we have encountered in Lecture 10, whose surface ( and ) describes the most general pure state

ρ_mixed = 12 | + ⟩⟨ + | + 12 | − ⟩⟨ − |

We can now go back to the case of the mixed state considered before

In this case there is 50% of probability of being in and 50% probability of being in

: the density matrix is proportional to the identity

ρ_mixed = 12 I

Which corresponds to the case

in the parametrisation

Therefore to determine the it is enough to measure three observables that are linearly independent when expanded in terms of Pauli matrices

In the case of a pure state two such measurements are sufficient, since the third component can be fixed through the condition

We can now answer our original question on how can we determine the state of a system assuming we can prepare many copies of it

The measurement of an observable whose operator is a Pauli matrix gives

Tr(ρσ_i) = Tr (1

2 (I + n ⋅ σ) σ_i) = n_i

𝒫_± = 1

2 (I ± n ⋅ σ)

The above discussion also lead us to the conclusion that for

the operator

is the projector on a

state along the direction

(2)

Entanglement

Its state vector lives in the direct product space of two Hilbert spaces

|ψ⟩ = ∑

m,n

c_mn|m₁⟩ ⊗ |n₂⟩ |m₁⟩ ∈ ℋ₁ |n₂⟩ ∈ ℋ₂

We note that, while the basis vectors are a direct product of basis vectors in the two spaces, their superposition is not

A direct product vector would have the form

|ϕ⟩ =

(∑

m

c_m|m₁⟩

) ⊗

(∑

n

d_n|n₂⟩ )

But not all the states

will have this form

A state that cannot be written in a factorised form is said to be entangle (in German verschränkt)

Entangled states reveal the non-local nature of the wave function

Suppose that we have a system which is described by two parts and

We have seen that the formalism of the density matrix allows us to characterise the behaviour of a quantum system when a measurement is performed

A particularly interesting situation is when we carry out a measurement only on a part of the complete system

ρ = |ψ⟩⟨ψ| = ∑

m,n,m′,n′

c*_m′n′c_mn|m₁⟩|n₂⟩⟨m′₁|⟨n₂′|

where we have used the shorthand notation

For the state

m,n

c_mn|m₁⟩ ⊗ |n₂⟩

The density matrix is

Let us consider an observable acting only on the second subspace: its action

is diagonal on the subspace and therefore its expectation value is

The result for

can also be expressed as

⟨A₂⟩ = Tr₂ (A₂ρ₂)

where we have defined a density matrix reduced to the second subspace

ρ₂ = Tr₁ρ = Tr₁ ∑

m,n,m′,n′

c*_m′n′c_mn|m₁⟩|n₂⟩⟨m′₁|⟨n′₂| = ∑

m,n,n′

c*_mn′c_mn|n₂⟩⟨n₂′|

⟨ψ|A₂|ψ⟩ = ∑

m′,n′∑

m,n

c*_m′n′c_mn⟨m′₁, n′₂|A₂|m₁, n₂⟩

= ∑

m,n,n′

c*_mn′c_mn⟨n′₂|A₂|n₂⟩

This is of course a rather common situation: the entire universe is in principle

described by a unique wave function, but in practice a measurement is always

done on small subsystem of it

As an explicit example let us consider a system of non-identical spin-1/2 particles in the pure state

|ψ⟩ = 1

2 (| + ⟩₁| − ⟩₂ + | − ⟩₁ + ⟩₂)

Its density matrix is

ρ = |ψ⟩⟨ψ | = 12 (| + ⟩₁| − ⟩₂ + | − ⟩₁ + ⟩₂) (⟨₁ + |⟨₂ − | + ⟨₁ − |⟨₂ + |)

Suppose that we want to measure the spin of the second particle: we need to consider the density matrix obtained by taking the trace with respect to 1

ρ₂ = Tr₁ρ = 1

2 [⟨ + |₁(| + ⟩₁| − ⟩₂ + | − ⟩₁ + ⟩₂) (⟨₁ + |⟨₂ − | + ⟨₁ − |⟨₂ + |)| + ⟩₁] + 12 [⟨ − |₁(| + ⟩₁| − ⟩₂ + | − ⟩₁ + ⟩₂) (⟨₁ + |⟨₂ − | + ⟨₁ − |⟨₂ + |)| − ⟩₁]

= 12 [| − ⟩₂⟨ − |₂ + | + ⟩₂⟨ + |₂]

But this is the density matrix for a completely unpolarised mixed state (recall

the discussion at page )

We can thus conclude that, even if the system is in a pure state, under a partial measurement the system may behave as a mixed state

In other words, a quantum superposition of states can become a classical mixture if we average over degrees of freedom that are not measured

One can show that a necessary and sufficient condition under which the density matrix of a pure state still corresponds to a pure state when a

measurement is performed over a subsystem is that the state can be written in a factorised form

If we can write

1⟩|ψ₂⟩

then

ρ₂² = ρ₂

This explains why in many cases it is possible to describe a quantum system in terms of the Hamiltonian of a part of it

Even if the entire universe is described by a single wave function, in practice, if

factorisation holds, one can focus on a subsystem of it !

Einstein-Podolsky-Rosen experiment

The EPR Gedankenexperiment considers an entangled system formed by two subsystems

A simplified version due to Bell can be stated as follows: we have a spin-zero particle which decays in two spin

particles, say an electron and a positron Since the total angular momentum is null, the wave function is

|ψ₀⟩ = 1

2 (| + ⟩| − ⟩ − | − ⟩| + ⟩)

The two particles move away from the source in such a way that time evolution does not modify the wave function, and when the particles are sufficiently distant spin measurements are carried out that are causally disconnected

Local realism states that, since the particles are causally disconnected, there is nothing that can happen to one of them as a consequence of what we do on the other, or, using the words of the original paper

..since at the time of measurement the two systems no longer interact, no real

change can take place in the second system in consequence of anything that may be

done in the first system.

Quantum mechanics contradicts these statements

Indeed, if we measure the spin of the first particle in the state and we find spin up, after the measurement the wave function of the system is projected onto

|ψ₀′⟩ = | + ⟩| − ⟩

and the spin of the second particle is down

So, it seems that there is transfer of information at a speed larger than the speed of light and a violation of causality

Such conclusion is, however, not correct

We can understand it by using the example of Bertlmann socks Suppose that there are two brothers who choose the socks in a drawer that contains only two pairs: one is blue and the

other is not

Local realism states that a property of the system which is measurable must

necessarily be a property of the system before the measurement is done

If, during the day, when the brothers are far away, we observe that the socks of one of them are not blue, we can immediately conclude that the socks of the other are blue

Of course this does not violate local realism, even if the brothers are very far away:

this is because the colour of the socks was determined before being measured When they dress up we don’t know who will wear the blue socks and there is

probability that the socks of each brother are of one colour or of the other

Let’s now go back to our state

: the problem comes when we realise that

|ψ₀⟩ = 1

2 (| + ⟩| − ⟩ − | − ⟩| + ⟩)

can also be written as a superposition of spin states along an arbitrary axis For example, since

|ψ₀⟩ = 1 (| + ⟩_x| − ⟩_x − | − ⟩_x| + ⟩_x)

we have

| ± ⟩_x = 1

2 (| + ⟩ ± | − ⟩)

Therefore, if we measure the spin of the first particle along the axis, then the second particle has opposite spin along the axis. But if we do the measurement along the axis, then the second particle will have opposite spin along the axis

z z

x x

But the spin operators along the two directions do not commute

The second particle cannot be before the measurement in a state with given spin both along the and the direction

The explanation used for Bertlmann socks does not work here, and thus quantum mechanics contradicts the hypothesis of local realism

Since for Einstein and collaborators the local realism is principle they don’t want to abandon, they conclude that Quantum Mechanics cannot be complete

Einstein expressed his rejection of Quantum Mechanics with the famous sentence:

“God does not play dice”

This expresses the non-local character of quantum theory: the non locality is a

consequence of the existence of correlated many-particle states and the fact that one

can linearly superimpose them

Hidden variables and the Bell theorem

By insisting on local realism, one can assume that a more complete theory exists and that a result of a measurement is determined by a property that the system possesses before the measurement

We now show that such description leads to results different from those of quantum mechanics

We can therefore introduce the hypothesis that the complete state of the system is described not only by the variables we use in the quantum-mechanical

description but also by some additional hidden variables unknown to us

If we knew such hidden variables we could predict with certainty the result of a measurement of any operator

We go back to the EPR experiment: we assume that the electron and the positron travel far apart in the state

|ψ₀⟩ = 1

2 (| + ⟩| − ⟩ − | − ⟩| + ⟩)

α β

Alice Bob

ℏ/2

−ℏ/2 ℏ/2

−ℏ/2

Source

The density matrix for this state is

ρ = |ψ₀⟩⟨ψ₀| = 1

4 (I − σ₁ ⋅ σ₂)

where the and are Pauli matrices acting on the spin states of the two particles

This result can be easily checked by showing that on an arbitrary state

|ψ⟩ = ∑

σ,σ′=±

c_σσ′|σ⟩|σ′⟩

The operator

projects onto

4 (I − σ₁ ⋅ σ₂) |ψ₀⟩ = 1

2 (| + ⟩| − ⟩ − | − ⟩| + ⟩)

Electron and positrons move away from the source and reach two detectors

operated by two physicists, Alice and Bob

The detectors measure the angular momentum along the directions

n₁ = (0, sin α, cosα) n₂ = (0, sin β, cosβ)

We ask ourselves what is the joint probability that Alice measures the spin of particle 1 up along the direction and Bob measures the spin of particle 2 down along the

In quantum mechanics we can compute this probability We have

This probability can be computed explicitly or by using the density operator

P₊₋(α, β) = |⟨ψ₀| + ⟩_n1| − ⟩_n2|² = ⟨ψ₀| + ⟩_n1| − ⟩_n2⟨_n1 + |⟨_n2 − |ψ₀⟩

where, by recalling the result in Eq. the projector

is

𝒫_n1,+n2,− = I + n₁ ⋅ σ₁

2 ⊗ I − n₂ ⋅ σ₂ 2

= ∑

k

⟨ψ₀| + ⟩_n1| − ⟩_n2⟨_n1 + |⟨_n2 − |e_k⟩⟨e_k|ψ₀⟩ = Tr (ρ𝒫_n1,+n2,−)

More generally the four correlated probabilities read

P₊₊(α, β) = P₋₋(α, β) = 14 (1 − n₁ ⋅ n₂) P₋₊(α, β) = P₊₋(α, β) = 14 (1 + n₁ ⋅ n₂)

In the case

we recover the previous result

and

Then the required probability is

P₊₋(α, β) = Tr (ρP_n1,+n2,−) = Tr ( 1

4 (I − σ₁ ⋅ σ₂) I + n₁ ⋅ σ₁ 2

I − n₂ ⋅ σ₂

2 )

Since the Pauli matrices are traceless the only non trivial contribution comes from the term

and we get

P₊₋(α, β) = 1 4 1

4 × 4 + 1

4 × 1

4 × Tr ((σ₁ ⋅ σ₂)(n₁ ⋅ σ₁)(n₂ ⋅ σ₂))

= 14 + 14 × 1

4 × Tr ((σ₁)_i(σ₂)_i(n₁)_j(σ₁)_j(n₂)_k(σ₂)_k) = 14 (1 + n₁ ⋅ n₂) 2δ_ij

2δ_ik

We now introduce the correlation function defined as the probability of equal outcomes minus the probability of different outcomes

C(α, β) = ∑

σ=± (P_σσ − P_σ,−σ)

We find

C(α, β) = − n₁ ⋅ n₂

We now assume that the probabilistic nature of quantum mechanics is simply a result of our lack of knowledge of some hidden variable

We assume that the first and the second particle are characterised by

response functions for a measurement of the spin (in units of ) along the

directions and

σ₁(α, λ) = ± 1 σ₂(β, λ) = ± 1

To explain the observed anticorrelation we have to assume

This is what quantum mechanics predicts

The parameter is unknown and measurements correspond to averages over with

some probability density

such that

We can now make predictions for the measurements

For example

1(α, ± ) = ∫ dλρ(λ) 1 ± σ₁(α, λ) 2

And this should be

to agree with the quantum mechanical result Correlated probabilities can be computed analogously

P_σσ′ = ∫ dλρ(λ) 1 + σσ₁(α, λ) 2

1 + σ′σ₂(β, λ) 2

C(α, β) = P₊₊ + P₋₋ − P₊₋ − P₋₊ = ∫ dλρ(λ)σ₁(α, λ)σ₂(β, λ)

The correlation coefficient becomes

∫ dλρ(λ) = 1

The model is completely defined by giving

We can thus define the quantity

s(λ, α,α′, β, β′) = σ₁(α,λ)[σ₂(β, λ) − σ₂(β′, λ)] + σ₁(α′,λ)[σ₂(β,λ) + σ₂(β′,λ)]

Since and can only take the values we have

s(λ, α, α′, β, β′) = ± 2

Therefore

−2 ≤ ∫ dλρ(λ)s(λ, α, α′, β, β′) ≤ 2

and introducing the correlations

we find

|C(α, β) − C(α, β′) + C(α′, β) + C(α′, β′)| ≤ 2

This is the Clauser-Horne-Shimony-Holt (CHSH) inequality

According to local realism we can simultaneously measure the spin along different

directions , and , corresponding to angles , and ,

It is easy to see that the quantum mechanical result

C(α, β) = − n₁ ⋅ n₂ = − cos(α − β)

violates such inequality

For example if we choose the angles as in the figure we obtain

π/8 C(α, β) − C(α, β′) + C(α′, β) + C(α′, β′)

= sin(π/8) − 3 cos(π/8) = − 2.38896

n₁

n₂

n′₁ n′₂

We must therefore conclude that quantum mechanics is indeed not compatible with local realism and with theories with hidden variables (Bell theorem)

Identities analogous to the CHSH identity had been derived earlier by Bell in 1964

For example, in the case

one can arrive to the alternative inequality

C(α, β) − 1 ≤ C(α, γ) + C(γ, β)

which is one of the Bell’s inequalities

Since measurements are found to be in agreement with QM predictions, this