density operators

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Quantum information and computing lecture 3: Density operators Jani-Petri Martikainen [email protected] http://www.helsinki.fi/˜jamartik Department of Physical Sciences University of Helsinki Department of Physical Sciences, University of Helsinki http://theory.physics.helsinki.fi/˜kvanttilaskenta/ – p. 1/39

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Page 1: Density operators

Quantum information and computing

lecture 3: Density operators

Jani-Petri [email protected]

http://www.helsinki.fi/˜jamartik

Department of Physical Sciences

University of Helsinki

Department of Physical Sciences, University of Helsinki http://theory.physics.helsinki.fi/˜kvanttilaskenta/ – p. 1/39

Page 2: Density operators

Entanglement application:superdense coding

Simple example demonstrating the application of basicquantum mechanics.

Alice is in possesion of two classical bits of informationand want’s to send these bits to Bob. However, she canonly send a single qubit. Is her task possible?

Answer: Yes!

Assume that Alice and Bob share a pair of qubits in anentangled state

|ψ〉 =|00〉 + |11〉√

2(7)

Department of Physical Sciences, University of Helsinki http://theory.physics.helsinki.fi/˜kvanttilaskenta/ – p. 8/39

Page 3: Density operators

Entanglement application:superdense coding

Note: someone else might have prepapared the state|ψ〉 and just sending the qubits to Alice and Bob beforehand.

If Alice wishes to transmit 00, she does nothing to herqubit.

If Alice wishes to transmit 01, she applies phase flip Z tothe her qubit.

If Alice wishes to transmit 10, she applies quantum NOTgate X to the her qubit.

If Alice wishes to transmit 11, she applies iY to the herqubit.

iY =

2

4

0 1

−1 0

3

5 (8)

Department of Physical Sciences, University of Helsinki http://theory.physics.helsinki.fi/˜kvanttilaskenta/ – p. 9/39

Page 4: Density operators

Entanglement application:superdense coding

The states map according to:

00 : |ψ〉 → |00〉 + |11〉√2

(9)

01 : |ψ〉 → |00〉 − |11〉√2

(10)

10 : |ψ〉 → |10〉 + |01〉√2

(11)

11 : |ψ〉 → |01〉 − |10〉√2

(12)

Department of Physical Sciences, University of Helsinki http://theory.physics.helsinki.fi/˜kvanttilaskenta/ – p. 10/39

Page 5: Density operators

Entanglement application:superdense coding

Alice then sends her qubit to Bob

The four states above are an orthonormal basis of the2-qubit Hilbert space. (known as the Bell basis , Bellstates , and EPR pairs )

Orthogonal states can be distinguished by making anappropriate quantum measurement

From the measured state Bob can then identify which ofthe four alternatives Alice send him.

In some sense, this is delayed communication. Thequbits where entangled and in order for them to getentangled they must have interacted in the past. Thechannel capacity was already “waiting” as a resource inthe entangled state.

Department of Physical Sciences, University of Helsinki http://theory.physics.helsinki.fi/˜kvanttilaskenta/ – p. 11/39

Page 6: Density operators

Density operator

QM can also be formulated in terms of density operatoror density matrix not just state vector.

Suppose system is at state |ψi〉 with probability pi. Wecall {pi, |ψi〉} an ensemble of pure states

The density operator is defined through

ρ =∑

i

pi|ψi〉〈ψi| (13)

Evolution: ρ→ UρU †

If the initial state was |ψi〉 then the probability of theoutcome m is

p(m|i) = 〈ψi|M †mMm|ψi〉 = Tr(M †

mMm|ψi〉〈ψi|) (14)

Department of Physical Sciences, University of Helsinki http://theory.physics.helsinki.fi/˜kvanttilaskenta/ – p. 12/39

Page 7: Density operators

Density operatorFor the ensemble the probability of m is

p(m) =∑

i

p(m|i)pi = Tr(M †mMmρ) (15)

After measurement result m: If initially |ψi〉 then

|ψmi 〉 =

Mm|ψi〉√

〈ψi|M †mMm|ψi〉

(16)

For the ensemble

ρm =∑

i

p(i|m)|ψmi 〉〈ψm

i | =MmρM

†m

Tr(M †mMmρ)

(17)

since p(i|m) = p(m|i)pi/p(m)

Department of Physical Sciences, University of Helsinki http://theory.physics.helsinki.fi/˜kvanttilaskenta/ – p. 13/39

Page 8: Density operators

Density operatorPure state: We know the state exactly and ρ = |ψ〉〈ψ|.Also, for pure states Tr(ρ2) = 1

Otherwise a mixed state . For a mixed state Tr(ρ2) < 1

Assume that or record for the result m was lost. Wecould now have state ρm with probability p(m), but weno longer know the value m. Such a system would bedescribed by a density operator

ρ =∑

m

p(m)ρm =∑

m

Tr(M †mMmρ)

MmρM†m

Tr(M †mMmρ)

=∑

m

MmρM†m (18)

Department of Physical Sciences, University of Helsinki http://theory.physics.helsinki.fi/˜kvanttilaskenta/ – p. 14/39

Page 9: Density operators

Density operator: requirements

For an operator ρ to be a density operator:

1. ρ must have a trace equal to one

2. ρ must be a positive operator

Department of Physical Sciences, University of Helsinki http://theory.physics.helsinki.fi/˜kvanttilaskenta/ – p. 15/39

Page 10: Density operators

Density operator

Postulate 4 : If each subsystem is described by a densitymatrix ρi the joint state of the total system isρ1 ⊗ ρ2 ⊗ · · · ρn.

Density operators shine when, a) describing a quantumsystem whose state is not known and b) describingsubsystems of a composite quantum system.

Eigenvalues and eigenvectors of the density matrix doNOT have a special significance with regard too theensemble of quantum states represented by thatdensity matrix.

Department of Physical Sciences, University of Helsinki http://theory.physics.helsinki.fi/˜kvanttilaskenta/ – p. 16/39

Page 11: Density operators

Density operator

For example, ρ = 3

4|0〉〈0| + 1

4|1〉〈1| probability of being in

|1〉 1/4 and probability of being in |0〉 3/4 (???) Notnecessarily!

suppose 1/2 prob. for both

|a〉 =√

3/4|0〉 +√

1/4|1〉 (19)

and|b〉 =

3/4|0〉 −√

1/4|1〉 (20)

The density matrixρ = 1/2|a〉〈a| + 1/2|b〉〈b| = 3/4|0〉〈0| + 1/4|1〉〈1|Different ensembles can give rise to the same densitymatrix

Department of Physical Sciences, University of Helsinki http://theory.physics.helsinki.fi/˜kvanttilaskenta/ – p. 17/39

Page 12: Density operators

Density operatorWhat class of emsembles gives rise to the samedensity matrix?

Vectors (not necessarily normalized) |ψ̃i〉 generate theoperator ρ ≡∑ |ψ̃i〉〈ψ̃i|... connection to usual ensemblepicture of density operators : |ψ̃i〉 =

√pi|ψi〉.

Answer:The sets |ψ̃i〉 and |φ̃i〉 generate the samedensity matrix if and only if

|ψ̃i〉 =∑

j

uij |φ̃j〉, (21)

where uij is a unitary matrix. (Pad which ever set |ψ̃i〉 or|φ̃i〉 is shorter with additional vectors having probability0 so that both sets have the same number of elements.)

Department of Physical Sciences, University of Helsinki http://theory.physics.helsinki.fi/˜kvanttilaskenta/ – p. 18/39

Page 13: Density operators

Density operator

Proof: suppose |ψ̃i〉 =∑

j uij |φ̃j〉. Then

i

|ψ̃i〉〈ψ̃i| =∑

ijk

uiju∗ik|φ̃j〉〈φ̃k| (22)

=∑

jk

(

i

u∗kiuij

)

|φ̃j〉〈φ̃k| =∑

jk

δjk|φ̃j〉〈φ̃k|

=∑

j

|φ̃j〉〈φ̃j | (23)

so they generate the same operator. Conversely, supposeA =

i |ψ̃i〉〈ψ̃i| =∑

j |φ̃j〉〈φ̃j |.

Department of Physical Sciences, University of Helsinki http://theory.physics.helsinki.fi/˜kvanttilaskenta/ – p. 19/39

Page 14: Density operators

Density operator

Let A =∑

k λk|k〉〈k| be a decomposition of A intoorthonormal states |k〉 with λk > 0. We wish to relate |ψ̃i〉 tostates |k̃〉 =

√λk|k〉 and similarly |φ̃i〉 to states |k̃〉. Let |ψ〉 be

any vector orthonormal to space spanned by |k̃i〉, so〈ψ|k̃〉 = 0 for all k. Therefore

0 = 〈ψ|A|ψ〉 =∑

i

〈ψ|ψ̃i〉〈ψ̃i|ψ〉 =∑

i

|〈ψ|ψ̃i〉|2 (24)

and thus 〈ψ|ψ̃i〉 = 0 for all i. It follows that each |ψ̃i〉 is alinear combination |ψ̃i〉 =

k cik|k̃〉.

Department of Physical Sciences, University of Helsinki http://theory.physics.helsinki.fi/˜kvanttilaskenta/ – p. 20/39

Page 15: Density operators

Density operator

Since A =∑

k |k̃〉〈k̃| =∑

i |ψ̃i〉〈ψ̃i| we see that

k

|k̃〉〈k̃| =∑

kl

(

i

cikc∗il

)

|k̃〉〈l̃|. (25)

Since operators |k̃〉〈l̃| are linearly independent∑

i cikc∗il = δkl. This ensures that we may append extra

columns to c to obtain a unitary matrix v such that|ψ̃i〉 =

k vik|k̃〉 where we have appended zero vectors tothe list |k̃〉. Similarly we can find w, such that|φ̃i〉 =

k wik|k̃〉. Thus |ψ̃i〉 =∑

j uij |φ̃i〉 where u = vw† isunitary.

Department of Physical Sciences, University of Helsinki http://theory.physics.helsinki.fi/˜kvanttilaskenta/ – p. 21/39

Page 16: Density operators

Reduced density operator

Describe subsystems by using a reduced density operator

Suppose the system is composed of A and B, then thereduced density operator for system A is

ρA ≡ TrB(ρAB). (26)

Above the partial trace is defined by

TrB(|a1〉〈a2| ⊗ |b1〉〈b2|) ≡ |a1〉〈a2|Tr(|b1〉〈b2|) (27)

where |ai〉 and |bi〉 are any vectors in the respectivestate spaces and Tr(|b1〉〈b2|) = 〈b2|b1〉 as usual. Partialtrace must also be linear in its input.

Department of Physical Sciences, University of Helsinki http://theory.physics.helsinki.fi/˜kvanttilaskenta/ – p. 22/39

Page 17: Density operators

Reduced density operator

Reduced density matrix gives the correct measurement statistics (this justifies its usephysically)

Take the Bell state (|00〉 + |11〉)/√

2 (pure state):

ρ =|00〉〈00| + |11〉〈00| + |00〉〈11| + |11〉〈11|

2(28)

Trace out the second qubit:

ρ1 = Tr2(ρ) =Tr2(|00〉〈00|) + Tr2(|11〉〈00|) + Tr2(|00〉〈11|) + Tr2(|11〉〈11|)

2

=|0〉〈0|〈0|0〉 + |1〉〈0|〈1|0〉 + |0〉〈1|〈0|1〉 + |1〉〈1|〈1|1〉

2=

|0〉〈0| + |1〉〈1|2

= I/2

This state is a mixed state even though the composite state was pure! Hallmark ofentanglement.

Department of Physical Sciences, University of Helsinki http://theory.physics.helsinki.fi/˜kvanttilaskenta/ – p. 23/39

Page 18: Density operators

Reduced density operator

Why the partial trace?

Partial trace turns out to be a unique operation whichgives rise to the correct description of observablequantities for subsystems of composite systems.

Suppose M is an observable on A and we have ameasuring device capable of realizing themeasurement of M ...let M̃ be the correspondingmeasurement on the composite system AB

If the system is in state |m〉|ψ〉 with |ψ〉 some arbitrarystate on B and |m〉 an eigenstate of M , the measuringdevice must give the result m with a probability of 1.

Department of Physical Sciences, University of Helsinki http://theory.physics.helsinki.fi/˜kvanttilaskenta/ – p. 24/39

Page 19: Density operators

Reduced density operatorTherefore, if Pm is the projector onto the m eigenspaceof the observable M , then the corresponding projectoron M̃ must be Pm ⊗ IB...we have

M̃ =∑

m

mPm ⊗ IB = M ⊗ IB (29)

Then let us show that the partial trace procedure givesthe correct measurement statistics for observations ona subsystem.

Physical consistency requires that any prescriptionassociating a ’state’ ρA to system A, must have theproperty that measurement averages be the same asfor the whole system i.e.

Tr(MρA) = Tr(M̃ρAB) = Tr((M ⊗ IB)ρAB) (30)

Department of Physical Sciences, University of Helsinki http://theory.physics.helsinki.fi/˜kvanttilaskenta/ – p. 25/39

Page 20: Density operators

Reduced density operator

This equation is certainly satisfied if ρA = TrB(ρAB).

In fact, partial trace turns out to be a unique functionhaving this property.

Let f(·) be any map of density operators on AB todensity operators on A such that

Tr(Mf(ρAB)) = Tr((M ⊗ IB)ρAB) (31)

Let Mi be orthonormal basis of operators for the spaceof Hermitian operators with respect to Hilbert-Schmidtinner product (see text book page 76) (X,Y ) = Tr(XY ),then we can expand f(ρAB) in this basis...

Department of Physical Sciences, University of Helsinki http://theory.physics.helsinki.fi/˜kvanttilaskenta/ – p. 26/39

Page 21: Density operators

Reduced density operator

...

f(ρAB) =∑

i

MiTr(Mif(ρAB))

=∑

i

MiTr((M ⊗ IB)ρAB) (32)

Therefore, f is uniquely determined by Eq. (30)

Moreover, the partial trace satisfies Eq. (30), so it is theunique function having this property.

Department of Physical Sciences, University of Helsinki http://theory.physics.helsinki.fi/˜kvanttilaskenta/ – p. 27/39

Page 22: Density operators

Schmidt decompositionSuppose |ψ〉 is a pure state of the composite systemAB. Then there exists orthormal states |iA〉 and |iB〉such that

|ψ〉 =∑

i

λi|iA〉|iB〉, (33)

where λi ≥ 0 are known as Schmidt coefficients andsatisfy

i λ2i = 1

Consequence: the reduced density matrices

ρA =∑

i

λ2

i |iA〉〈iA| (34)

ρB =∑

i

λ2

i |iB〉〈iB| (35)

have the same eigenvalues.Department of Physical Sciences, University of Helsinki http://theory.physics.helsinki.fi/˜kvanttilaskenta/ – p. 28/39

Page 23: Density operators

Schmidt decomposition and PurificationSchmidt number is the number of non-zero λi and insome sense it quantifies the amount of entanglementbetween systems A and B.

a state of the composite system is a product state (andthus not entangled) if and only if it has a Schmidtnumber 1.

Purification: suppose we are given a state ρA of aquantum system A. We can introduce another systemR and a pure state |AR〉 there, such thatρA = TrR(|AR〉〈AR|)This mathematical procedure is known as purificationand enables us to associate pure states with mixedstates.

Department of Physical Sciences, University of Helsinki http://theory.physics.helsinki.fi/˜kvanttilaskenta/ – p. 29/39

Page 24: Density operators

EPR and Bell inequalityWhat is actually the difference between quantummechanics and the classical world? What makesquantum mechanics non-classical?

In QM an unobserved particle does not posssesproperties that exist independent of observation. Forexample, a qubit does not possess definite properties of’spin in the z-direction’, and ’spin in the x-direction’ eachof which can be revealed by performing the appropriatemeasurement.

In the ’EPR-paper’ Einstein, Podolsky and Rosenproposed a thought experiment which they believeddemonstrated the incompleteness of QM as a theory ofNature.

Department of Physical Sciences, University of Helsinki http://theory.physics.helsinki.fi/˜kvanttilaskenta/ – p. 30/39

Page 25: Density operators

EPR and Bell inequality

Attempt to find ’elements of reality’ which were notincluded in QM. Introduced what they claimed was asufficient condition for a physical property to be anelement of reality...namely that it is possible to predictwith certainty the value of that property, immediatelybefore measurement.

Consider, an entangled pair of qubits (spin singlet)belonging to Alice and Bob

|01〉 − |10〉√2

(36)

Measure spin along −→v axis i.e. −→v · −→σ for both spins

Department of Physical Sciences, University of Helsinki http://theory.physics.helsinki.fi/˜kvanttilaskenta/ – p. 31/39

Page 26: Density operators

EPR and Bell inequalityNo matter what we choose for −→v the twomeasurements are always opposite to one another.That, if first qubit measurement yields +1 the secondwill give −1 and vice versa.

Suppose |a〉 and |b〉 are the eigenstates of −→v · −→σ then

|0〉 = α|a〉 + β|b〉, |1〉 = γ|a〉 + δ|b〉 (37)

and|01〉 − |10〉√

2= (αδ − βγ)

|ab〉 − |ba〉√2

(38)

But αδ − βγ is a determinant of a unitary matrix andthus just equal to a phase factor eiθ

As if the 2nd qubit knows the result of the measurementon the 1st

Department of Physical Sciences, University of Helsinki http://theory.physics.helsinki.fi/˜kvanttilaskenta/ – p. 32/39

Page 27: Density operators

EPR and Bell inequality

Since Alice can predict the measurement result whenBob’s qubit is measured along −→v , that is an ’element ofreality’ and should be included in a complete physicaltheory.

Standard QM does not include any fundamentalelement to represent the value of −→v · −→σ for all unitvectors −→v .

EPR hoped for a return to a more classical view:system can be ascribed properties which existindependently of measurements performed.

Nature experimentally invalidates EPR , while agrees withQM

Key: Bell’s inequality

Department of Physical Sciences, University of Helsinki http://theory.physics.helsinki.fi/˜kvanttilaskenta/ – p. 33/39

Page 28: Density operators

EPR and Bell inequalityStart with a common sense analysis (a’la EPR) andthen proceed with QM analysis...find a difference...letnature decide.

Perform a measurement outlined in the figure: Charlieprepares two particles and sends one to Alice and oneto Bob.

Alice can choose to measure two different things whichare physical properties labelled by PQ and PR. Shedoesn’t know in advance what measurement she willconduct...decides by flipping a coin...measurementoutcome is (for simplicity) ±1

Alice’s particle has a value Q for the property PQ. Thisvalue is assumed to be an objective property of Alice’sparticle, which is merely revealed by the measurement.(similarly R is the value revealed by the measurementof PR)

Department of Physical Sciences, University of Helsinki http://theory.physics.helsinki.fi/˜kvanttilaskenta/ – p. 34/39

Page 29: Density operators

EPR and Bell inequality

1 particle 1 particle

ALICE BOBQ=+−1

R=+−1

S=+−1R=+−1

Department of Physical Sciences, University of Helsinki http://theory.physics.helsinki.fi/˜kvanttilaskenta/ – p. 35/39

Page 30: Density operators

EPR and Bell inequality

Likewise Bob is capable of measuring values S and T(±1) of properties PS and PT ..also he makes thechooses the measurement randomly

Timing is such that Alice and Bob measure at the sametime (or at least in causally disconnected manner)...soAlice’s measurement cannot disturb Bob’smeasurement

Look at QS +RS +RT −QT = (Q+R)S + (R−Q)T

since R,Q = ±1 it follows that either (Q+R)S = 0 or(R−Q)T = 0

In either case QS +RS +RT −QT = ±2

suppose next that p(q, r, s, t) is the prob. that beforemeasurements the system is in a state with Q = q,R = r, S = s, and T = t.

Department of Physical Sciences, University of Helsinki http://theory.physics.helsinki.fi/˜kvanttilaskenta/ – p. 36/39

Page 31: Density operators

EPR and Bell inequalityThese probabilities may depend on what Charlie does.Let E(·) denote the mean value of a quantity...we have

E(QS +RS +RT −QT ) =∑

qrst

p(q, r, s, t)(qs+ rs+ rt− qt)

≤∑

qrst

p(q, r, s, t) × 2

= 2 (39)

and

E (QS + RS + RT − QT ) =X

qrst

p(q, r, s, t)qs

+X

qrst

p(q, r, s, t)rs +X

qrst

p(q, r, s, t)rt

−X

qrst

p(q, r, s, t)qt

= E(QS) + E(RS) + E(RT ) − E(QT ) (40)

Department of Physical Sciences, University of Helsinki http://theory.physics.helsinki.fi/˜kvanttilaskenta/ – p. 37/39

Page 32: Density operators

EPR and Bell inequalityCombine and you have an example of a Bell inequality

E(QS) + E(RS) + E(RT ) − E(QT ) ≤ 2 (41)

By repeating the measurements and comparing theiroutcomes Alice and Bob can determine the left handside

Now lets put QM back in...Let Charlie prepare aquantum system of two qubits

|01〉 − |10〉√2

(42)

Alice and Bob perform measurements Q = Z1,S = (−Z2 −X2)/

√2, R = X1, and T = (Z2 −X2)/

√2

Department of Physical Sciences, University of Helsinki http://theory.physics.helsinki.fi/˜kvanttilaskenta/ – p. 38/39

Page 33: Density operators

EPR and Bell inequality

Simple calculation (show this...) shows that〈QS〉 = 1/

√2, 〈RS〉 = 1/

√2, 〈RT 〉 = 1/

√2, and

〈QT 〉 = −1/√

2 and thus

〈QS〉 + 〈RS〉 + 〈RT 〉 − 〈QT 〉 = 2√

2 (43)

Violates the inequality and is consistent withexperiments! ...some asumptions going into thederivation of the Bell inequality must have been wrong

However, if Alice and Bob choose (for example),Q = Z1, S = −Z2, R = X1, and T = −X2 then

E(QS) + E(RS) + E(RT ) − E(QT ) = 2 (44)

and there is no violation.

Department of Physical Sciences, University of Helsinki http://theory.physics.helsinki.fi/˜kvanttilaskenta/ – p. 39/39

Page 34: Density operators

EPR and Bell inequality

Questionable assumptions:

1. Physical properties PQ, PR,PS, and PT have definitevalues which exist independent of observation(assumption of realism)

2. Alice’s measurement does not influence Bob’smeasurement (locality)

World is not locally realistic!

entanglement can be a fundamentally new resourcewhich goes beyond classical resources...How to exploitit, is the key question of quantum information andcomputing.

Department of Physical Sciences, University of Helsinki http://theory.physics.helsinki.fi/˜kvanttilaskenta/ – p. 40/39