Exemple of Hilbert spaces with dimension equal to two. *Spin1/2particle

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1 University Paris-Saclay - IQUPS Optical Quantum Engineering: From fundamentals to applications Philippe Grangier, Institut d Optique, CNRS, Ecole Polytechnique. Eemple of Hilbert spaces with dimension equal to two. *Spin/particle *Polarizedphoton:states with linear or circular polarisation; mathematical structure very close to a spin / (factor on angles, see below). 7 Lecture (7 March, 9:5-0:45) : Qubits, entanglement and Bell s inequalities. Lecture (4 March,:00-:30) : From QND measurements to quantum gates and quantum information. Lecture 3 ( March, 9:5-0:45) : Quantum optics with discrete and continuous variables. Lecture 4 (8 March, :0-:30) : Quantum cryptography and optical quantum networks. * Two-levelatom (attention! spontaneous emission). + z z y Spin / Photon Atom *Superconductingcircuit:anharmonic quantum oscillator due to a Josephson junction, allows one to isolate two energy levels These systems are various implementations of a quantum bit (qubit). e f Bloch s sphere. 9 A few considerations on entangled systems. + z θ ϕ u y Bloch's sphere (spin /) Denote the eigenstate along the direction u(θ, ϕ) Normalized vector ~u u cos( )sin( ), u y sin( )sin( ), u z cos( ). ~.~u ~S.~u h ~.~u cos( ) sin( )e i Eigenvalues of ~.~u : ±, eigenstatesof ~ S.~u eigenstatesof~.~u : + ~u i cos( /)e i / + z i +sin( /)e i / zi ~ui sin( /)e i / + z i +cos( /)e i / zi sin( )e i cos( )! * Within classical physics, correlations between measurements carried out on separated subsystems are eplained by attributing to each subsystems some properties which are correlated to properties of the other subsystem. * If one tries to reproduce quantum correlations using such a model, Bell s inequalities show that these properties must be non-local, i.e. must contradict relativistic causality! inacceptable. * Quantum mechanics remains in perfect agreement with relativistic causality, but there is a price : it is impossible to attribute a local physical reality to the state of each subsystem. EPR Parado (Einstein Podolsky Rosen, 935) Quantum non-separability * We will see now that entanglement plays an essential role in quantum mechanics in general, and especially in quantum information...

2 Lecture : Entanglement in a Quantum Measurement Process : from QND measurements to quantum gates. Direct and indirect measurements in Quantum Mechanics How can we measure the state of a qubit? (spin /, photon, atom...). Direct and indirect measurements in Quantum Mechanics. Analysis of a quantum measurement process, no-cloning theorem 3. From quantum measurement to quantum gates. 4. From Shannon to quantum cryptography. *Directmeasurementprocess: Spin / : Stern-Gerlach magnet! spatial splitting as a function of the value of the spin component parallel to the gradient of magnetic field. Photon : Polarizer! spatial splitting as a function of the polarization... The measurement process demolishes the qubit, which is no more available after the measurement. *Indirectmeasurementprocess: One reads the state of the qubit by coupling it to another qubit. One can thus realize an ideal measurement, including the state preparation stage. This is called a Quantum Non Demolition (QND) measurement. Attention : QNDdoes not mean that there is noe ect on the system s state! AQNDmeasurementis aquantummeasurement,sothesystem sstateis changed, unless it is already in an eigenstate of the measured observable. QND measurement of a spin component. 3 QND measurement of a spin component : conclusion. 7 One wants to perform a QND measurement of ˆz on a qubit a : if the qubit is a spin / particle, one gets the spin a to interact with another spin b during a time, andreadouttheresultonspin b. An appropriate interaction Hamiltonian is : H m hg ˆaz ˆb / B eff if az+ z Final state of qubit b if az + B eff if az- y Initial state of qubit b Final state of qubit b if az - Everything happens as if qubit acreatesonqubitbane ective magnetic field, aligned along O, with a sign depending on the state ±i az (see eercise!). +i az +i by! +i az +i bz i az +i by! i i az i bz In the general case the initial state is and from the superposition principle : (0)i ( +i az + i az ) +i by ( )i ( +i az +i bz + i i az i bz ) This is an entangled state like the EPR state seen before : a measurement on qubit b gives + with probability and with probability. For each result, the state of qubit a is perfectly known after the measurement ( reduction of the wave packet ). The quantum measurement of az is done by an indirect measurement, called a QND measurement : qubit a is still there for further action! Attention : the final state is not aduplicationofanarbitraryinitialstate, which would be ( +i az + i az ) ( +i bz + i bz ) : no-cloning!

3 8 9 The no-cloning theorem (). Quantum cloning means to duplicate an arbitrary and unknown initial quantum state onto another system with an initial state 0 i ( white sheet ), keeping the original intact in order to have two identical copies. Considering two di erent initial states to be copied i et i one wants : i 0 i! i i i 0 i! i i The copies are denoted as i et i because they may be made on di erent physical systems : e.g. cloning a polarized photon onto a spin / particle. Theorem : Perfect cloning of an arbitrary unknown quantum state is forbidden both by linearity and by unitarity of quantum mechanics. The no-cloning theorem ().. Demonstration based on linearity : let us consider 3i p ( i+ i) i 0 i! i i i 0 i! i i 3i 0 i! p ( i i + i i) 6 3i 3 i The sum of the copies is not the copy of the sum!. Demonstration based on unitarity : conservation of the scalar product. h i h 0 0 i h ih i h i ( h i)0 h i 0:orthogonalstates h i :identicalcopies. Conclusion : one can clone within a known set of orthogonal states, but it is not possible to clone (perfectly) within a set of non-orthogonal states. Generalization : CNOT quantum gate The interaction between two qubits seen previously implements a logical operation between the qubits : 0, 0i! 0, 0i, et, 0i! i, i. Generalizing, one defines a C-NOT (Controlled-NOT) gate, which does the following (denoting ii a ji b i, ji): 0, 0i! 0, 0i 0, i! 0, i, 0i!, i, i!, 0i The first qubit (control qubit) is unchanged. The second qubit (target qubit) is inverted if the first qubit value is one. This logical gate (applied to many qubits...) is a building block to implement quantum computation. 0 Simple application : preparation and measurement of Bell s states The control qubit is driven in a superposition of the computational states 0i et i ( / rotation of Bloch s vector), then the CNOT gate is applied: 0, 0i! p ( 0i +? i) 0i! p ( 0, 0i +, i) 0, i! p ( 0i + i) i! p ( 0, i +, 0i), 0i! p ( 0i i) 0i! p ( 0, 0i, i), i! p ( 0i i) i! p ( 0, i, 0i) One gets 4 orthogonal entangled states forming a basis of the Hilbert space of the two qubits (with dimension 4) : Bell s states. This transformation is reversible : starting from Bell s states, on can get the four factorized basis states, which are easy to identify from a direct measurement of the two qubits : Bell s states measurement.

4 Simple application : preparation and measurement of Bell s states The control qubit is driven in a superposition of the computational states 0i et i ( / rotation of Bloch s vector), then the CNOT gate is applied: 0, 0i! p ( 0i + i) 0i! p ( 0, 0i +, i) 0, i! p ( 0i + i) i! p ( 0, i +, 0i), 0i! p ( 0i i) 0i! p ( 0, 0i, i), i! p ( 0i i) i! p ( 0, i, 0i) One gets 4 orthogonal entangled states forming a basis of the Hilbert space of the two qubits (with dimension 4) : Bell s states. This transformation is reversible : starting from Bell s states, on can get the four factorized basis states, which are easy to identify from a direct measurement of the two qubits : Bell s states measurement. A few elementary notions about the classical theory of information. Shannon entropy 3 Fundamental results published by Claude Shannon, 948. () By how much is it possible to compress the data of a message, assuming that there is no transmission noise? Answer : the maimum compression rate is the Shannon entropy of the message () What is the maimum transmission rate through a noisy channel? Answer : the maimum transmission rate of information is equal to the channel capacity, tobedefinedbelow Message : (long) string of n letters chosen within an alphabet of k letters : {a,a,...a k } Each letter a is given a probability p(a ),with P k p(a ). Simple eemple : binary alphabet {0, } with p(0) p, p() p. If n>>, themessagewillcontainaboutnp bits and n( p) bits 0. The number of such typical messages is Cn np n! (np)!(n np)!

5 Using Stirling s formula log(n!) n log n n + O(log n) one gets : log (Cnp) n n log (n) np log (np) (n np)log (n np) nh(p) where we define the (binary) Shannon entropy H(p) p log p ( p)log ( p). 4 Remarks : *ShapeofthebinaryentropyH(p) p log p ( p)log ( p): The number of typical messages is then nh(p). 0.6 Main point : When n is very large, it is su cient to assign a codeword to each of these typical messages, because the atypical messages will almost never appear (more rigorously : the error rate due to the atypical messages will be asymptotically negligible). In order to identify these codewords one needs nh(p) bits instead of n, the compression factor is thus H(p) apple *Forp / one gets H(p) : a completely random message (no redundancy) cannot be compressed. 6 This reasoning can be generalized for k letters, by defining H(X) X p()log p() hlog p()i. where H(X) is the Shannon entropy of the message X {, p()} (alphabet and associated probabilities). Amessagewithn letters can then be compressed in nh(x) bits; on says also that each letter carries on average H(X) bits of information. Remarks : *Ifthe6lettersoftheusualalphabetwereperfectlyrandom,eachletter would carry log (6) 4.7 bits. However the probabilities are not equally distributed (esainturlo...) and the letters are strongly correlated, so a letter carries about. bit (see crypto/java/entropy/). *Thereforeitisenoughtousen(H(p) + ) bits to encode a message with n bits, without errors asymptotically. On the other hand, one shows that with n(h(p) ) bits there will necessarily be errors, because there will not be enough codewords to represent all typical sequences.

6 Mutual information. 7 Mutual information. 8 In a message is transmitted through a noisy channel, the received message will be in general di erent from the sent message. So one should answer the question : what do we know about a message drawn in the ensemble X n,if one knows a message drawn in the ensemble Y n? Let us define the conditional probability p(y ) to receive y if was sent. The ensemble of p(y ) caracterizes the channel, and using Bayes formula one can calculate p( y) : p( y) p(, y)/p(y) p(y )p()/ X p(y )p() From the p( y) one defines the conditional entropy H(X Y ) : H(X Y )h log p( y)i h log p(, y)i + hlog p(y)i. One has thus : and also H(X Y )H(X, Y ) H(Y ) H(Y X) H(X, Y ) H(X) One defines then the mutual information I(X; Y ) : I(X; Y )H(X) H(X Y ) H(X)+ H(Y ) H(X, Y ) H(Y ) H(Y X) The mutual information I(X; Y ) is symmetrical with respect to X and Y, and is zero if X and Y are uncorrelated. It is the fundamental quantity measuring the information echanged through the channel. The unit of I(X; Y ) ( bit per symbol ) is the same as the one used for the entropies H(X). Shannon theorem for the transmission through a noisy channel : The maimum number of bits per symbol that can be transmitted through anoisychannelisitscapacityc, definedby: C Ma {p()} I(X; Y ) Taking the maimum over {p()} eliminates the role of the message, and therefore C is a property of the channel itself (in fact, of the p(y )). 3 The density matri () One can (re)formulate quantum mechanics by replacing the state vector (t)i by the projector onto (t)i : (t)i! ˆ (t) (t)ih (t) Then a statistical distribution of states i i with a probability i (classical statistics) willbedescribedbythedensity matri : ˆ X i i i ih i As a result the double average (classical and quantum) of an observable  can be written: hâi stat X i i h i  ii X i i Tr( i ih i Â) Tr(ˆ Â) H(X,Y)

7 4 9 The density matri () It is postulated that any quantum state can be described by a density operator ˆ such that : ˆ is hermitian with trace (but ˆ is not always a projector) All its eigenvalues are positive or zero. The hamiltonian evolution and measurement probabilities are given by : P(a )Tr(ˆ ˆP ) hai Tr(ˆ Â). i h dˆ dt hĥ(t), ˆ (t) i. Remark : One can always diagonalize ˆ and write : ˆ P i i i ih i. One recovers a pure state if there is only one non-zero i,then ˆ (t) (t)ih (t), ˆ ˆ and Tr( ˆ ) Simple eemples of density matrices () Depolarized spin / (silver atom straight from the oven) : ˆ nonpol. +ih+ + ih 0 0 ˆ. to be compared to a pure state ( +i + i)/ p : ˆ pol.selon. Qubit : two-state system { ei, gi} ou { +i, ee ˆ eg ( + nz )/ (n in y )/ ge gg (n + in y )/ ( n z )/ i}: ˆ+~n.~ where we define ~n (n,n y,n z ) with n i real numbers, and where we used the Pauli matrices ~ : 0 0 i 0 y z 0 i 0 0 The eigenvalues of ˆ ˆ+~n.~ are equal to ( ± ~n ) Generalisation of Bloch sphere ( Bloch ball ) z θ n ϕ u y ˆ ( + nz )/ (n in y )/ (n + in y )/ ( n z )/ ˆ+~n.~ with eigenvalues ( ± ~n ). Using Tr( i )0, one gets : 0 i ˆ (i, y, z), h~ i Tr(ˆ ~)~n if ~n one can write ~n ~u and one retrieves a pure state. The operator ˆ is then a projector on the state + ~u i,andonehash~ i ~u if ~n < then ~n is inside the Bloch ball, and one has again h~ i ~n if ~n 0then the spin is depolarized and h~ i ~0. Entangled pair of spin / particles (qubits). In a singlet state for pour spins / : ss i p ( +, i, +i) ˆ ss B A with the ordering of the basis vectors: { +, +i, +, i,, +i,, i} Justify the above epression of ˆ ss. Show that the reduced density operators are given by : ˆ A 0 ˆ 0 B 0 ˆ 0 A ˆ B 4 ˆ 4 Conclude that the two sub-systems are completely depolarized. Show by using the reduced density operators that the reduction of the wave packet does not allow any transfer of information between Alice and Bob when doing an EPR correlation eperiment. 5

8 Von Neumann entropy. 9 Mathematical properties of Von Neumann entropy (). 0 In quantum information the symbols may become quantum states, described by a density matri,andthealphabetisthenanensemble{,p }. The density matri seen by an observer is then X p. One can choose an orthonormal basis { ai} where is diagonal X p a aiha. a The set { aiha,p a } is then équivalent to a classical alphabet, and one has: H(A) X p a log (p a ) Trace( log )S( ) a where S( ) is the Von Neumann entropy of the density matri. In general the symbols are not mutually eclusive (non orthogonal states), and the Shannon and Von Neumann entropies have di erent properties.. The entropy S( ih ) of a pure state is equal to zero.. The entropy S( ) is not modified by a change of basis (isometry). 3. If has D non-zero eignevalues, then S( ) apple log D. 4. For p i 0 and P i p i,onehass( P i p P i i ) i p i S( i ) (S increases when ignoring of the way by which the state was prepared). 5. Measurement : For an observable B with eigenvalues b y one defines Y {b y,p(b y )} and then H(Y ) S( ) (equality if [B, ] 0). 6. Preparation : For a density matri P p ih with X { i, p } then H(X) S( ) (equality if the { } are orthogonal). Mathematical properties of Von Neumann entropy (). Quantum data compression. 7. For a bipartite system AB one has whereas S( A ) S( B ) apple S( AB ) apple S( A )+S( B ) H(X),H(Y ) apple H(X, Y ) apple H(X)+H(Y ). Fundamental di erence between Shannon and Von Neumann! On can get S( A )S( B ) 6 0whereas S( AB )0 (entanglement! impossible classically). 8. Strong subaddidivity : For a tripartite system ABC one has S( ABC )+S( B ) apple S( AB )+S( BC ) (important, not easy to demonstrate!). One considers P p ih,andonedefinesx { i, p }. A message with n symbols is then associated to the density matri n.... What is the minimum number of qubits needed to encode this message? (Reminder : the dimension of the Hilbert space E for N qubits is N ). It can be shown (Schumacher) that log (dime) N ns( ). The entropy S( ) corresponds to the number of qubits per symbol of the message. Principle of the demonstration : For very long messages, the support of the density matri n is included within a subspace of dimension ns( ).ThiscanbeobtainedfromShannon s theorem, by considering a basis where is diagonal.

9 3 Quantum data compression: a few remarks.. Bob received qubits which are e ciently packaged, but in general he cannot simply retrieve the classical information sent by Alice.. In the general case, the symbols are themselves statistical mitures (density matrices), and not pure states. An important question is then to evaluate the maimum amount of classical information, that can be etracted from a quantum message. The answer is given by using the Holevo information : X ({,p })S( ) p S( ). Holevo theorem then states that the maimum accessible classical information (over all possible measurements) is bounded by ({,p }) : Ma Mesures I(X; Y ) apple ({,p }). Quantum cryptography: the characters Eve Alice Bob Demonstration : Not obvious, one has to use strong subadditivity Secret key cryptosystem : one-time pad (G. Vernam, 97) 0 secret channel! 000 " " 000! Eve classical channel ( Demonstrably secure if the key is : random as long as the message used only once (Shannon) 0! Quantum Secret Key Cryptosystem : Bennett-Brassard (984) " 0! # " ÿ quantum channel [ classical channel ( " ! Demonstrably secure if the key is : random as long as the message used only once (Shannon) unknown by Eve : Quantum laws! 0 00