Density operators

Density operators

• 1.
  Quantum information andcomputing 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
• 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 operator QM canalso be formulated in terms of density operator or density matrix not just state vector. Suppose system is at state |ψi with probability pi . We call {pi , |ψi } an ensemble of pure states The density operator is defined through ρ= pi |ψi ψi | (13) i Evolution: ρ → U ρU † If the initial state was |ψi then the probability of the outcome 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 operator For theensemble the probability of m is † p(m) = p(m|i)pi = T r(Mm Mm ρ) (15) i After measurement result m: If initially |ψi then m Mm |ψi |ψi = (16) † ψi |Mm Mm |ψi For 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 operator Pure state:Weknow the state exactly and ρ = |ψ ψ|. Also, for pure states T r(ρ2 ) = 1 Otherwise a mixed state. For a mixed state T r(ρ2 ) < 1 Assume that or record for the result m was lost. We could now have state ρm with probability p(m), but we no longer know the value m. Such a system would be described 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: requirements Foran 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
• 10.
  Density operator Postulate 4: If each subsystem is described by a density matrix ρi the joint state of the total system is ρ1 ⊗ ρ2 ⊗ · · · ρn . Density operators shine when, a) describing a quantum system whose state is not known and b) describing subsystems of a composite quantum system. Eigenvalues and eigenvectors of the density matrix do NOT have a special significance with regard too the ensemble of quantum states represented by that density matrix. Department of Physical Sciences, University of Helsinki http://theory.physics.helsinki.fi/˜kvanttilaskenta/ – p. 16/39
• 11.
  Density operator 3 1 For example, ρ = 4 |0 0| + 4 |1 1| probability of being in |1 1/4 and probability of being in |0 3/4 (???) Not necessarily! 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 matrix ρ = 1/2|a a| + 1/2|b b| = 3/4|0 0| + 1/4|1 1| Different ensembles can give rise to the same density matrix Department of Physical Sciences, University of Helsinki http://theory.physics.helsinki.fi/˜kvanttilaskenta/ – p. 17/39
• 12.
  Density operator What classof emsembles gives rise to the same density 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 same density 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 probability 0 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) j so 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 operator Let 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 operator Since A= ˜ ˜ |k k| = ˜ ˜ |ψi ψi | we see that k i ˜ ˜ |k k| = cik c∗ |k ˜ ˜ l|. (25) il k kl i Since operators |k ˜ are linearly independent ˜ l| i cik c∗ = δkl . This ensures that we may append extra il columns 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† is unitary. Department of Physical Sciences, University of Helsinki http://theory.physics.helsinki.fi/˜kvanttilaskenta/ – p. 21/39
• 16.
  Reduced density operator Describesubsystems by using a reduced density operator Suppose the system is composed of A and B , then the reduced 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 respective state spaces and T r(|b1 b2 |) = b2 |b1 as usual. Partial trace 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 operator Reduceddensity matrix gives the correct measurement statistics (this justifies its use physically) √ Take the Bell state (|00 + |11 )/ 2 (pure state): |00 00| + |11 00| + |00 11| + |11 11| ρ= (28) 2 Trace 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 2 This state is a mixed state even though the composite state was pure! Hallmark of entanglement. Department of Physical Sciences, University of Helsinki http://theory.physics.helsinki.fi/˜kvanttilaskenta/ – p. 23/39
• 18.
  Reduced density operator Whythe partial trace? Partial trace turns out to be a unique operation which gives rise to the correct description of observable quantities for subsystems of composite systems. Suppose M is an observable on A and we have a measuring device capable of realizing the ˜ measurement of M ...let M be the corresponding measurement on the composite system AB If the system is in state |m |ψ with |ψ some arbitrary state on B and |m an eigenstate of M , the measuring device 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 operator Therefore,if Pm is the projector onto the m eigenspace of the observable M , then the corresponding projector ˜ on M must be Pm ⊗ IB ...we have ˜ M= mPm ⊗ IB = M ⊗ IB (29) m Then let us show that the partial trace procedure gives the correct measurement statistics for observations on a subsystem. Physical consistency requires that any prescription associating a ’state’ ρA to system A, must have the property that measurement averages be the same as for 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 operator Thisequation is certainly satisfied if ρA = T rB (ρAB ). In fact, partial trace turns out to be a unique function having this property. Let f (·) be any map of density operators on AB to density 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 space of Hermitian operators with respect to Hilbert-Schmidt inner 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) i Therefore, f is uniquely determined by Eq. (30) Moreover, the partial trace satisfies Eq. (30), so it is the unique function having this property. Department of Physical Sciences, University of Helsinki http://theory.physics.helsinki.fi/˜kvanttilaskenta/ – p. 27/39
• 22.
  Schmidt decomposition Suppose |ψis a pure state of the composite system AB . Then there exists orthormal states |iA and |iB such that |ψ = λi |iA |iB , (33) i where λi ≥ 0 are known as Schmidt coefficients and satisfy i λ2 = 1 i Consequence: the reduced density matrices ρA = λ2 |iA iA | i (34) i ρB = λ2 |iB iB | i (35) i have the same eigenvalues. Department of Physical Sciences, University of Helsinki http://theory.physics.helsinki.fi/˜kvanttilaskenta/ – p. 28/39
• 23.
  Schmidt decomposition andPurification 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 Bellinequality What is actually the difference between quantum mechanics and the classical world? What makes quantum mechanics non-classical? In QM an unobserved particle does not possses properties that exist independent of observation. For example, a qubit does not possess definite properties of ’spin in the z-direction’, and ’spin in the x-direction’ each of which can be revealed by performing the appropriate measurement. In the ’EPR-paper’ Einstein, Podolsky and Rosen proposed a thought experiment which they believed demonstrated the incompleteness of QM as a theory of Nature. Department of Physical Sciences, University of Helsinki http://theory.physics.helsinki.fi/˜kvanttilaskenta/ – p. 30/39
• 25.
  EPR and Bellinequality Attempt to find ’elements of reality’ which were not included in QM. Introduced what they claimed was a sufficient condition for a physical property to be an element of reality...namely that it is possible to predict with certainty the value of that property, immediately before measurement. Consider, an entangled pair of qubits (spin singlet) belonging to Alice and Bob |01 − |10 √ (36) 2 Measure 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 Bellinequality No matter what we choose for − the two →v measurements are always opposite to one another. That, if first qubit measurement yields +1 the second will 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 2 But αδ − βγ is a determinant of a unitary matrix and thus just equal to a phase factor eiθ As if the 2nd qubit knows the result of the measurement on the 1st Department of Physical Sciences, University of Helsinki http://theory.physics.helsinki.fi/˜kvanttilaskenta/ – p. 32/39
• 27.
  EPR and Bellinequality Since Alice can predict the measurement result when Bob’s qubit is measured along − , that is an ’element of →v reality’ and should be included in a complete physical theory. Standard QM does not include any fundamental element to represent the value of − · − for all unit → → v σ vectors − . → v EPR hoped for a return to a more classical view: system can be ascribed properties which exist independently of measurements performed. Nature experimentally invalidates EPR, while agrees with QM Key: Bell’s inequality Department of Physical Sciences, University of Helsinki http://theory.physics.helsinki.fi/˜kvanttilaskenta/ – p. 33/39
• 28.
  EPR and Bellinequality Start with a common sense analysis (a’la EPR) and then proceed with QM analysis...find a difference...let nature decide. Perform a measurement outlined in the figure: Charlie prepares two particles and sends one to Alice and one to Bob. Alice can choose to measure two different things which are physical properties labelled by PQ and PR . She doesn’t know in advance what measurement she will conduct...decides by flipping a coin...measurement outcome is (for simplicity) ±1 Alice’s particle has a value Q for the property PQ . This value is assumed to be an objective property of Alice’s particle, which is merely revealed by the measurement. (similarly R is the value revealed by the measurement of PR ) ˜ Department of Physical Sciences, University of Helsinki http://theory.physics.helsinki.fi/ kvanttilaskenta/ – p. 34/39
• 29.
  EPR and Bellinequality ALICE BOB Q=+−1 S=+−1 R=+−1 1 particle 1 particle R=+−1 Department of Physical Sciences, University of Helsinki http://theory.physics.helsinki.fi/˜kvanttilaskenta/ – p. 35/39
• 30.
  EPR and Bellinequality Likewise Bob is capable of measuring values S and T (±1) of properties PS and PT ..also he makes the chooses the measurement randomly Timing is such that Alice and Bob measure at the same time (or at least in causally disconnected manner)...so Alice’s measurement cannot disturb Bob’s measurement 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 before measurements 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 Bellinequality These probabilities may depend on what Charlie does. Let E(·) denote the mean value of a quantity...we have E(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 Bellinequality Combine 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 their outcomes Alice and Bob can determine the left hand side Now lets put QM back in...Let Charlie prepare a quantum system of two qubits |01 − |10 √ (42) 2 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
• 33.
  EPR and Bellinequality 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 with experiments! ...some asumptions going into the derivation 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
• 34.
  EPR and Bellinequality Questionable 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