Density operators
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Density operators Presentation Transcript

  • 1. Quantum information and computing lecture 3: Density operators Jani-Petri Martikainen Jani-Petri.Martikainen@helsinki.fi http://www.helsinki.fi/˜jamartik Department of Physical Sciences University of HelsinkiDepartment of Physical Sciences, University of Helsinki http://theory.physics.helsinki.fi/˜kvanttilaskenta/ – p. 1/39
  • 2. Entanglement application:superdense coding Simple example demonstrating the application of basic quantum mechanics. Alice is in possesion of two classical bits of information and want’s to send these bits to Bob. However, she can only send a single qubit. Is her task possible? Answer: Yes! Assume that Alice and Bob share a pair of qubits in an entangled state |00 + |11 |ψ = √ (7) 2 Department of Physical Sciences, University of Helsinki http://theory.physics.helsinki.fi/˜kvanttilaskenta/ – p. 8/39
  • 3. Entanglement application:superdense coding Note: someone else might have prepapared the state |ψ and just sending the qubits to Alice and Bob before hand. If Alice wishes to transmit 00, she does nothing to her qubit. If Alice wishes to transmit 01, she applies phase flip Z to the her qubit. If Alice wishes to transmit 10, she applies quantum NOT gate X to the her qubit. If Alice wishes to transmit 11, she applies iY to the her qubit. 2 3 0 1 iY = 4 5 (8) −1 0 Department of Physical Sciences, University of Helsinki http://theory.physics.helsinki.fi/˜kvanttilaskenta/ – p. 9/39
  • 4. Entanglement application:superdense coding The states map according to: |00 + |11 00 : |ψ → √ (9) 2 |00 − |11 01 : |ψ → √ (10) 2 |10 + |01 10 : |ψ → √ (11) 2 |01 − |10 11 : |ψ → √ (12) 2 Department of Physical Sciences, University of Helsinki http://theory.physics.helsinki.fi/˜kvanttilaskenta/ – p. 10/39
  • 5. Entanglement application:superdense coding Alice then sends her qubit to Bob The four states above are an orthonormal basis of the 2-qubit Hilbert space. (known as the Bell basis, Bell states, and EPR pairs) Orthogonal states can be distinguished by making an appropriate quantum measurement From the measured state Bob can then identify which of the four alternatives Alice send him. In some sense, this is delayed communication. The qubits where entangled and in order for them to get entangled they must have interacted in the past. The channel capacity was already “waiting” as a resource in the entangled state. Department of Physical Sciences, University of Helsinki http://theory.physics.helsinki.fi/˜kvanttilaskenta/ – p. 11/39
  • 6. Density operatorQM 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 statesThe density operator is defined through ρ= pi |ψi ψi | (13) iEvolution: ρ → U ρU †If the initial state was |ψi then the probability of theoutcome m is † † p(m|i) = ψi |Mm Mm |ψi = T r(Mm Mm |ψi ψi |) (14) Department of Physical Sciences, University of Helsinki http://theory.physics.helsinki.fi/˜kvanttilaskenta/ – p. 12/39
  • 7. Density operatorFor the ensemble the probability of m is † p(m) = p(m|i)pi = T r(Mm Mm ρ) (15) iAfter measurement result m: If initially |ψi then m Mm |ψi |ψi = (16) † ψi |Mm Mm |ψiFor the ensemble † m m Mm ρMm ρm = p(i|m)|ψi ψi | = † (17) i T r(Mm Mm ρ)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
  • 8. Density operatorPure state:We know the state exactly and ρ = |ψ ψ|.Also, for pure states T r(ρ2 ) = 1Otherwise a mixed state. For a mixed state T r(ρ2 ) < 1Assume 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 † † Mm ρMm ρ = p(m)ρm = T r(Mm Mm ρ) † m m T r(Mm Mm ρ) † = Mm ρMm (18) m Department of Physical Sciences, University of Helsinki http://theory.physics.helsinki.fi/˜kvanttilaskenta/ – p. 14/39
  • 9. Density operator: requirementsFor an operator ρ to be a density operator:1. ρ must have a trace equal to one2. ρ must be a positive operator Department of Physical Sciences, University of Helsinki http://theory.physics.helsinki.fi/˜kvanttilaskenta/ – p. 15/39
  • 10. Density operatorPostulate 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
  • 11. Density operator 3 1For example, ρ = 4 |0 0| + 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
  • 12. Density operatorWhat class of emsembles gives rise to the samedensity matrix? ˜Vectors (not necessarily normalized) |ψi generate the ˜ ˜operator ρ ≡ |ψi ψi |... connection to usual ensemble ˜i = √pi |ψi .picture of density operators : |ψ ˜ ˜Answer:The sets |ψi and |φi generate the samedensity matrix if and only if ˜ |ψi = ˜ uij |φj , (21) j ˜where uij is a unitary matrix. (Pad which ever set |ψi or ˜|φi 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
  • 13. Density operator ˜Proof: suppose |ψi = ˜ uij |φj . Then j ˜ ˜ |ψi ψi | = ˜ ˜ uij u∗ |φj φk | (22) ik i ijk = u∗ uij ˜ ˜ |φj φk | = ˜ ˜ δjk |φj φk | ki jk i jk = ˜ ˜ |φj φj | (23) jso they generate the same operator. Conversely, suppose ˜ ˜ ˜ ˜A = i |ψi ψi | = j |φj φj |. Department of Physical Sciences, University of Helsinki http://theory.physics.helsinki.fi/˜kvanttilaskenta/ – p. 19/39
  • 14. Density operatorLet A = k λk |k k| be a decomposition of A into ˜orthonormal states |k with λk > 0. We wish to relate |ψi to √states |k ˜ ˜ ˜ = λk |k and similarly |φi to states |k . Let |ψ be ˜any vector orthonormal to space spanned by |ki , so ˜ ψ|k = 0 for all k . Therefore 0 = ψ|A|ψ = ˜ ˜ ψ|ψi ψi |ψ = ˜ | ψ|ψi |2 (24) i i ˜ ˜and thus ψ|ψi = 0 for all i. It follows that each |ψi is a ˜linear combination |ψi = k cik |k .˜ Department of Physical Sciences, University of Helsinki http://theory.physics.helsinki.fi/˜kvanttilaskenta/ – p. 20/39
  • 15. Density operatorSince A = ˜ ˜ |k k| = ˜ ˜ |ψi ψi | we see that k i ˜ ˜ |k k| = cik c∗ |k ˜ ˜ l|. (25) il k kl iSince operators |k ˜ are linearly independent ˜ l| i cik c∗ = δkl . This ensures that we may append extra ilcolumns to c to obtain a unitary matrix v such that ˜ ˜|ψi = k vik |k where we have appended zero vectors to ˜the list |k . Similarly we can find w, such that ˜ ˜ ˜ ˜|φi = k wik |k . Thus |ψi = j uij |φi where u = vw† isunitary. Department of Physical Sciences, University of Helsinki http://theory.physics.helsinki.fi/˜kvanttilaskenta/ – p. 21/39
  • 16. Reduced density operatorDescribe subsystems by using a reduced density operatorSuppose the system is composed of A and B , then thereduced density operator for system A is ρA ≡ T rB (ρAB ). (26)Above the partial trace is defined by T rB (|a1 a2 | ⊗ |b1 b2 |) ≡ |a1 a2 |T r(|b1 b2 |) (27)where |ai and |bi are any vectors in the respectivestate spaces and T r(|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
  • 17. Reduced density operatorReduced density matrix gives the correct measurement statistics (this justifies its usephysically) √Take the Bell state (|00 + |11 )/ 2 (pure state): |00 00| + |11 00| + |00 11| + |11 11| ρ= (28) 2Trace out the second qubit: T r2 (|00 00|) + T r2 (|11 00|) + T r2 (|00 11|) + T r2 (|11 11|)ρ1 = T r2 (ρ) = 2 |0 0| 0|0 + |1 0| 1|0 + |0 1| 0|1 + |1 1| 1|1 |0 0| + |1 1| = = = I/2 2 2This 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
  • 18. Reduced density operatorWhy 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 the ˜measurement of M ...let M be the correspondingmeasurement on the composite system ABIf 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
  • 19. Reduced density operatorTherefore, if Pm is the projector onto the m eigenspaceof the observable M , then the corresponding projector ˜on M must be Pm ⊗ IB ...we have ˜ M= mPm ⊗ IB = M ⊗ IB (29) mThen 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. ˜ T r(M ρA ) = T r(M ρAB ) = T r((M ⊗ IB )ρAB ) (30) Department of Physical Sciences, University of Helsinki http://theory.physics.helsinki.fi/˜kvanttilaskenta/ – p. 25/39
  • 20. Reduced density operatorThis equation is certainly satisfied if ρA = T rB (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 T r(M f (ρAB )) = T r((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 ) = T r(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
  • 21. Reduced density operator... f (ρAB ) = Mi T r(Mi f (ρAB )) i = Mi T r((M ⊗ IB )ρAB ) (32) iTherefore, 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
  • 22. Schmidt decompositionSuppose |ψ is a pure state of the composite systemAB . Then there exists orthormal states |iA and |iBsuch that |ψ = λi |iA |iB , (33) iwhere λi ≥ 0 are known as Schmidt coefficients andsatisfy i λ2 = 1 iConsequence: the reduced density matrices ρA = λ2 |iA iA | i (34) i ρB = λ2 |iB iB | i (35) ihave the same eigenvalues. Department of Physical Sciences, University of Helsinki http://theory.physics.helsinki.fi/˜kvanttilaskenta/ – p. 28/39
  • 23. Schmidt decomposition and Purification Schmidt number is the number of non-zero λi and in some sense it quantifies the amount of entanglement between systems A and B . a state of the composite system is a product state (and thus not entangled) if and only if it has a Schmidt number 1. Purification: suppose we are given a state ρA of a quantum system A. We can introduce another system R and a pure state |AR there, such that ρA = T rR (|AR AR|) This mathematical procedure is known as purification and enables us to associate pure states with mixed states. Department of Physical Sciences, University of Helsinki http://theory.physics.helsinki.fi/˜kvanttilaskenta/ – p. 29/39
  • 24. 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
  • 25. EPR and Bell inequalityAttempt 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 √ (36) 2Measure spin along − axis i.e. − · − for both spins → v → → v σ Department of Physical Sciences, University of Helsinki http://theory.physics.helsinki.fi/˜kvanttilaskenta/ – p. 31/39
  • 26. EPR and Bell inequalityNo matter what we choose for − the two →vmeasurements 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 − · − then → → v σ |0 = α|a + β|b , |1 = γ|a + δ|b (37)and |01 − |10 |ab − |ba √ = (αδ − βγ) √ (38) 2 2But αδ − βγ is a determinant of a unitary matrix andthus just equal to a phase factor eiθAs 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
  • 27. EPR and Bell inequalitySince Alice can predict the measurement result whenBob’s qubit is measured along − , that is an ’element of →vreality’ and should be included in a complete physicaltheory.Standard QM does not include any fundamentalelement to represent the value of − · − for all unit → → v σvectors − . → vEPR 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 withQMKey: Bell’s inequality Department of Physical Sciences, University of Helsinki http://theory.physics.helsinki.fi/˜kvanttilaskenta/ – p. 33/39
  • 28. 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) ±1Alice’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
  • 29. EPR and Bell inequalityALICE BOBQ=+−1 S=+−1R=+−1 1 particle 1 particle R=+−1Department of Physical Sciences, University of Helsinki http://theory.physics.helsinki.fi/˜kvanttilaskenta/ – p. 35/39
  • 30. EPR and Bell inequalityLikewise Bob is capable of measuring values S and T(±1) of properties PS and PT ..also he makes thechooses the measurement randomlyTiming is such that Alice and Bob measure at the sametime (or at least in causally disconnected manner)...soAlice’s measurement cannot disturb Bob’smeasurementLook at QS + RS + RT − QT = (Q + R)S + (R − Q)Tsince R, Q = ±1 it follows that either (Q + R)S = 0 or(R − Q)T = 0In either case QS + RS + RT − QT = ±2suppose 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
  • 31. EPR and Bell inequalityThese probabilities may depend on what Charlie does.Let E(·) denote the mean value of a quantity...we haveE(QS + RS + RT − QT ) = p(q, r, s, t)(qs + rs + rt − qt) qrst ≤ p(q, r, s, t) × 2 qrst = 2 (39)and X E (QS + RS + RT − QT ) = p(q, r, s, t)qs qrst X X + p(q, r, s, t)rs + p(q, r, s, t)rt qrst qrst X − p(q, r, s, t)qt qrst = 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
  • 32. 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 handsideNow lets put QM back in...Let Charlie prepare aquantum system of two qubits |01 − |10 √ (42) 2Alice 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
  • 33. EPR and Bell inequalitySimple 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 wrongHowever, 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
  • 34. EPR and Bell inequalityQuestionable assumptions:1. Physical properties PQ , PR ,PS , and PT have definite values which exist independent of observation (assumption of realism)2. Alice’s measurement does not influence Bob’s measurement (locality) World is not locally realistic! entanglement can be a fundamentally new resource which goes beyond classical resources...How to exploit it, is the key question of quantum information and computing. Department of Physical Sciences, University of Helsinki http://theory.physics.helsinki.fi/˜kvanttilaskenta/ – p. 40/39