Introduction to Quantum Mechanics

Transcription

1 Introduction to Quantum Mechanics R. J. Renka Department of Computer Science & Engineering University of North Texas 03/19/2018

2 Postulates of Quantum Mechanics The postulates (axioms) of quantum mechanics provide a framework for development of the laws governing a system. 1 Every isolated physical system is described by a unit state vector ψ in a complex Hilbert space H (the state space), where two states that differ by a global phase factor (modulus-1 scalar) are equivalent. 2 Physical observables (measurables) are associated with Hermitian (self-adjoint) operators on H. 3 The result of measuring any physical observable is always an eigenvalue of the corresponding operator. 4 Measurement results in a particular eigenvalue/eigenstate (collapse of the wave function). 5 Time evolution of the state of a closed quantum system is governed by the Schrödinger equation i d ψ dt = Ĥ ψ for Hamiltonian operator Ĥ.

3 Time evolution The framework described by the postulates does not tell us, for a given physical system, what the state vector is or even what the state space is, let alone what the Hamiltonian is. These are difficult questions that belong to the theories, such as QED, that are built within the framework of quantum mechanics. A closed quantum system is one that does not interact with other systems. In reality all systems other than the universe as a whole interact somewhat with other systems. Nevertheless, many systems can be described to a good approximation as being closed. The continuous time evolution by Schrödinger s equation corresponds to discrete-time dynamics using unitary operators. The solution to the Schrödinger equation shows that the state at time t 1 is related to the state at time t 2 by a unitary operator U: ψ(t 2 ) = U(t 1, t 2 ) ψ(t 1 ) for U(t 1, t 2 ) exp [ iĥ(t 2 t 1 ) ].

4 Time evolution continued The following exercises show that a Hermitian Hamiltonian operator corresponds to discrete-time dynamics by unitary operators (which preserve the norm of the unit state vector) and, conversely, any unitary operator is of the form U = exp(ia) for some Hermitian operator A. Exercise 1: Suppose A and B are commuting Hermitian operators which are therefore simultaneously diagonalizable. Show that exp(a) exp(b) = exp(a + B). Exercise 2: Prove that U = exp[ iαĥ] is unitary, where α is a real constant and Ĥ is Hermitian. Exercise 3: Use the spectral decomposition to show that A i ln(u) is Hermitian for any unitary operator U, and thus U = exp(ia) for some Hermitian operator A.

5 Quantum measurement The following is a generalization of Postulates 2, 3, and 4. While a closed quantum system evolves without interacting with the rest of the world, interaction with a measuring instrument is necessary at some point. Quantum measurements are described by a set of measurement operators {M m : H H}, where the index m is the measurement outcome which, in state ψ occurs with probability p(m) = ψ M mm m ψ. Following the measurement, the state is M m ψ normalized to a unit vector. Since the probabilities sum to 1 for all states, the measurement operators necessarily satisfy the completeness equation: M mm m = I. m

6 Quantum measurement continued For example, consider a qubit in state ψ = a 0 + b 1. Measurement in the computational basis is defined by Hermitian operators M 0 = 0 0 and M 1 = 1 1. The probabilities are p(0) = a 2 and p(1) = b 2 for amplitudes a = 0 M 0 ψ and b = 1 M 1 ψ, and the state after measurement is either (a/ a ) 0 or (b/ b ) 1, but the modulus-1 scale factors can effectively be ignored. If the system is prepared in state 0 (b = 0) or state 1 (a = 0) then a measurement reveals which state the system was prepared in. If, on the other hand, the initial state is a superposition of 0 and 1 then there is a nonzero probability associated with both outcomes. More generally, it is possible to distinguish among a set of orthonormal states but no quantum measurement is capable of distinguishing among non-orthogonal states. A quantum state contains hidden information not accessible to measurement.

7 Projective measurement A projective measurement is described by a Hermitian operator M on H (an observable) with a spectral decomposition M = m mp m, where P m is the projector onto the eigenspace associated with m. Measurement outcomes are the eigenvalues of M with associated probabilities p(m) = ψ P m ψ = P m ψ 2. Following measurement of outcome m, the state is P m ψ / p(m). Projective measurement is a special case in which the previously defined measurement operators M m are the projectors P m so that M mm m = P mp m = P 2 m = P m.

8 Description of measurement It can be shown that projective measurement augmented by an ancillary system and unitary operations is completely equivalent to the more general description of measurement, and that s the reason for the choice of postulates. This raises the question of what use is the more general description. The measurement operators M m do not have to satisfy the condition P i P j = δ ij P i for projective measurements. The simpler structure gives rise to useful properties not possessed by P m. Projective measurements are repeatable in that a second measurement results in the same outcome as that of the first measurement. However, many important measurements in quantum mechanics are not repeatable and are therefore not projective. Measuring the position of a photon, for example, may destroy the photon.

9 Expectation The mean or expectation value of a projective measurement is E(M) = mp(m) = m ψ P m ψ m m ( ) = ψ mp m ψ = ψ M ψ. m The average value of the observable M is often written M. The variance associated with M is ( M) 2 = (M M I ) 2 = M 2 M 2. Exercise 4: Suppose a quantum system is prepared in an eigenstate ψ of some observable M, with corresponding eigenvalue m. What is the average observed value of M and the standard deviation?

10 Heisenberg Uncertainty Principle Suppose Hermitian operators A, B, C satisfy the commutator relation [A, B] AB BA = ic. Examples include Position and momentum: [X, P] = i I Pauli matrices: [σ 1, σ 2 ] = 2iσ 3 Spin or angular momentum: [S x, S y ] = i S z Theorem 1: ( A)( B) 1 2 ψ C ψ = 1 ψ [A, B] ψ, 2 where ψ is an arbitrary quantum state, and ( A) is the uncertainty (standard deviation) in A: ( A) 2 = ψ (A A) 2 ψ = ψ A 2 ψ A 2 for expected value operator A = ψ A ψ I, identity I.

11 Proof of Heisenberg Uncertainty Principle proof: Let α = A A and β = B B. Then ψ α 2 ψ = ψ (A A) 2 ψ = ( A) 2, ψ β 2 ψ = ψ (B B) 2 ψ = ( B) 2, and [α, β] = [A A, B B] = [A, B] = ic. By the Cauchy-Schwarz inequality, ( A) 2 ( B) 2 = ψ α 2 ψ ψ β 2 ψ = αψ αψ βψ βψ αψ βψ 2 = αψ βψ βψ αψ = ψ αβ ψ ψ βα ψ.

12 Proof continued Also, αβ = w [α, β] = w ic, for Thus βα = w 1 2 [α, β] = w 1 2 ic w = 1 (αβ + βα). 2 ( A) 2 ( B) 2 ψ w + (1/2)iC ψ ψ w (1/2)iC ψ = ψ w ψ 2 (i/2) ψ w ψ ψ C ψ + (i/2) ψ C ψ ψ w ψ + (1/4) ψ C ψ 2 = ψ w ψ 2 + (1/4) ψ C ψ 2 (1/4) ψ C ψ 2.

13 Significance of the uncertainty principle A common misconception of the principle is that measuring observable A to some accuracy causes the value of B to be disturbed by an amount ( B) such that the inequality is satisfied. While it is true that measurement causes disturbance to the system, this is not the content of the principle. Rather it reflects an intrinsic property of nature and describes what would happen in a large number of experiments. The most important consequence is the fact that position and momentum of a particle cannot be determined simultaneously with arbitrary accuracy: X P /2. This sets quantum mechanics apart from classical mechanics.

14 Measurement exercises Exercise 5: Suppose we have a qubit in state 0, and we measure the observable associated with Pauli matrix X. What is the average value X and standard deviation X? Exercise 6: Let v σ v 1 σ 1 + v 2 σ 2 + v 3 σ 3 for unit vector v and Pauli matrices σ i. Show that v σ has eigenvalues ±1, and that the projectors onto the corresponding eigenspaces are P ± = (I ± v σ)/2. Exercise 7: Calculate the probability of obtaining the result ±1 for a measurement of v σ, given that the state prior to measurement is 0. What is the state after the measurement if +1 is obtained?

15 Positive Operator-Valued Measure If we are given only the positive operators E m M mm m instead of M m then we can compute probabilities of measurement outcomes p(m) = ψ E m ψ but not the post-measurement state which is often of little interest. The complete set of POVM elements E m is a POVM (Positive Operator-Valued Measure). The POVM elements coincide with the measurement operators (only) in the case of a projective measurement with E m P mp m = P m = M m. Given a positive operator E m, we can compute M m E m. We can therefore define a POVM to be any set of positive operators E m such that m E m = I. This describes a measurement defined by operators E m.

16 POVM example To demonstrate the utility of having only measurement statistics, suppose Alice gives Bob a qubit prepared in either state ψ 1 = 0 or ψ 2 = + ( )/ 2. Since these states are not orthogonal, it is not possible for Bob to distinguish between them with perfect reliability, but he can perform a measurement which may distinguish between them and never makes an error in identification. Consider the POVM with three elements E , E , E 3 I E 1 E 2. Then if the measurement outcome is the eigenvalue associated with E 1, Bob knows that he was not given ψ 1 since p(1) = ψ 1 E 1 ψ 1 = 0, and he concludes that he was given state ψ 2. Similarly, if outcome E 2 occurs, then it must have been ψ 1 that Bob received. Outcome E 3 reveals no information.

17 Phase Since states ψ and e iθ ψ have the same measurement statistics, the global phase factor e iθ for real θ is irrelevant. Two amplitudes a 1 and a 2 differ by a relative phase if a 1 = e iθ a 2 for some real θ. More generally, two states differ by a relative phase in some basis if all amplitudes of one have the same magnitudes as the corresponding amplitudes of the other in that basis. For example + (1/ 2) 0 + (1/ 2) 1 and (1/ 2) 0 (1/ 2) 1 are the same up to relative phase shifts in the computational basis but not in the +, basis. Measurement in the computational basis results in p(0) = p(1) = 1/2 for both states, but the states give rise to different statistics when measured in other bases and are therefore not physically equivalent.

18 Composite systems Another postulate is that the state space of a composite physical system is the tensor product of the component state spaces. If system i is prepared in state ψ i then the joint state of the total system is ψ 1 ψ 2 ψ n. A state which cannot be expressed as the product of states of its component systems is entangled. Entangled states play a crucial role in quantum computation, leading to strange effects such as superdense coding and the violation of Bell s inequality. Exercise 8: Prove that the two qubit state ( )/ 2 a b for all single qubit states a and b. Notation: Subscripts sometimes denote states and operators on different systems as in the following exercise. Exercise 9: Show that the average value of the observable X 1 Z 2 for a two qubit system measured in the state ( )/ 2 is zero.

19 No-cloning theorem Consider a quantum machine with a data slot in pure state ψ to be copied into a target slot which starts in some standard pure state s so that the initial state of the copying machine is ψ s. Suppose the copy is to be effected by some unitary operator U, and the copying procedure works for two particular states ψ and φ so that U( ψ s ) = ψ ψ, U( φ s ) = φ φ. Equating inner products gives ψ φ = ( ψ φ ) 2. Hence either ψ = φ or ψ and φ are orthogonal. It is impossible to perfectly clone an unknown quantum state using unitary evolution. It has been shown that, even allowing non-unitary cloning devices, cloning of non-orthogonal states (pure or mixed) necessarily incurs loss of fidelity.

20 Superdense coding Alice can transmit two classical bits to Bob by sending a single qubit. Suppose they share the pair of qubits of the entangled state ψ = ( )/ 2. This is a fixed known state which could be distributed to Alice and Bob by a third party. Depending on which of the four two-bit values 00, 01, 10 or 11 Alice wishes to transmit, she applies operator I, Z, X, or iy to her qubit, resulting in one of the Bell states: 1 00 : I ( ψ ) = ( )/ 2 = (1, 0, 0, 1)/ : Z( ψ ) = ( )/ 2 = (1, 0, 0, 1)/ : X ( ψ ) = ( )/ 2 = (0, 1, 1, 0)/ : iy ( ψ ) = ( )/ 2 = (0, 1, 1, 0)/ 2. The Bell states form an orthonormal basis and can therefore be distinguished by measurement in the Bell basis (projective measurement with projectors P m = m m ). Information is physical, and surprising physical theories lead to surprising information processing capability.

21 Superdense coding continued Exercise 10: Verify that the Bell states form an orthonormal basis for the two qubit state space. Exercise 11: Show that ψ E I ψ takes the same value for all four Bell states ψ and any operator E. Hint: (E I ) ab = E a b. It follows from the previous exercise that if a malevolent third party Eve intercepts Alice s qubit on the way to Bob, she can only measure the first qubit and can infer nothing about the message being transmitted.

22 Density operator The density operator formulation of quantum mechanics is an alternative that is mathematically equivalent to the state vector approach and is convenient when the state of a system is not completely known. Given an ensemble of pure states {p i, ψ i } with probabilities p i, the density operator is ρ i p i ψ i ψ i. In order to formulate the postulates of quantum mechanics in terms of density, suppose the evolution of a closed system is described by unitary operator U. If the system starts in state ψ i with probability p i, then the evolved system will be in state U ψ i with probability p i and the evolved density will be p i U ψ i ψ i U = UρU. i

23 Measurements in terms of density Suppose we perform a measurement described by operators M m. Then the probability of getting result m given initial state ψ i is p(m i) = ψ i M mm m ψ i = tr(m mm m ψ i ψ i ). Hence the total probability of measuring result m is p(m) = i p(m i)p i = i p i tr(m mm m ψ i ψ i ) = tr(m mm m ρ). If the initial state was ψ i then the state after measuring m is ψi m = (M m ψ i )/ ψ i M mm m ψ i, and we have the ensemble of states ψi m with probabilities p(i m) so that the density is ρ m = i p(i m) ψ m i ψ m i = i p(i m) M m ψ i ψ i M m ψ i M mm m ψ i.

24 Measurements continued Using the equations above, along with Bayes rule p(i m) = p(i, m)/p(m) = p(m i)p i /p(m), we obtain ρ m = i p i M m ψ i ψ i M m tr(m mm m ρ) = M mρm m tr(m mm m ρ). Suppose we have lost the record of the result m of a measurement. Then we have a system in state ρ m with probability p(m) described by the density operator ρ = m p(m)ρ m = m tr(m M m ρm m M m ρ) tr(m mm m ρ) = m M m ρm m. A quantum system whose state ψ is known exactly is in a pure state with density ρ = ψ ψ. Otherwise ρ is in a mixed state.

25 General properties of the density operator Theorem 2: An operator ρ is the density operator associated with some (non-unique) ensemble {p i, ψ i } iff it satisfies 1 (Trace condition): tr(ρ) = 1. 2 (Positivity condition): ρ is a positive operator. proof: A density operator ρ = i p i ψ i ψ i is easity shown to satisfy the properties. Conversely, suppose ρ satisfies the properties. Since ρ is positive, it has a spectral decomposition: ρ = i λ i φ i φ i, where the eigenvectors { φ i } are orthonormal and the eigenvalues λ i are real, nonnegative, and add to 1 (by the trace condition). Hence {λ i, φ i } is an ensemble of states giving rise to the density operator ρ. Density operators and quantum mechanics can be described without reference to state vectors. We define a density operator to be a positive operator with trace equal to one.

26 Postulates reformulated The following reformulation of the postulates is mathematically equivalent to the description in terms of state vectors but is more convenient when the state is not known and for description of subsystems of a composite system. Postulate 1: Associated with any isolated physical system is a complex Hilbert space (state space). The system is completely described by a positive (semidefinite) density operator ρ with trace 1. If the system is in state ρ i with probability p i then ρ = i p iρ i. Postulate 2: The evolution of a closed quantum system is described by a unitary transformation U = U(t 1, t 2 ): ρ(t 2 ) = Uρ(t 1 )U.

27 Postulates reformulated continued Postulate 3: Quantum measurements are described by a collection of measurement operators M m for measurement outcome m with associated probability p(m) = tr(m mm m ρ). After measurement the state is M m ρm m tr(m mm m ρ). The operators satisfy the completeness equation M mm m = I. m Postulate 4:: The state space of a composite physical system is the tensor product of the component state spaces. For system i prepared in state ρ i, the joint state is ρ = ρ 1 ρ 2 ρ n.

28 Mixed vs. pure states Theorem 3: A density operator ρ satisfies tr(ρ 2 ) 1, with equality if and only if the system described by ρ is in a pure state. proof: Let ρ = i p i ψ i ψ i, where p i 0 and i p i = 1 so that 0 pi 2 p i 1 for all i. Hence tr(ρ 2 ) = tr(( i p i ψ i ψ i ) 2 ) = tr( i p2 i ψ i ψ i ) = i p2 i i p i = 1. Also, if ρ = ψ ψ then tr(ρ 2 ) = tr(ρ) = 1. Conversely, suppose the system is in a mixed state. Then 0 < p i < 1 for some i, and therefore pi 2 < p i and tr(ρ 2 ) = i p2 i < i p i = 1. 3 Example: For a , b , ρ = 1 2 a a b b = Hence two different ensembles give rise to the same density operator. The eigenvectors and eigenvalues (3/4 and 1/4) of ρ indicate only one of many possible ensembles.

29 Unitary freedom in the ensemble An arbitrary set of vectors ψ i generates the operator i ψ i ψ i. Theorem 4: The sets ψ i and φ i generate the same density matrix if and only if ψ i = j u ij φ j, where U is a unitary matrix, and we pad the smaller set of vectors with zero vectors so that the two sets have the same number of elements. As a consequence of the theorem, ρ = i p i ψ i ψ i = j q j φ j φ j for normalized states ψ i, φ j and probability distributions p i and q j if and only if pi ψ i = j u ij qj φ j for some unitary matrix U, where we may pad the smaller ensemble with zero probabilities. The theorem thus characterizes the freedom in ensembles giving rise to a given density matrix ρ. In the previous example U is defined by u 11 = u 12 = u 21 = u 22 = 1/ 2.

30 Reduced density operator The reduced density operator for subsystem A of a composite system described by density operator ρ AB is ρ A tr B (ρ AB ) for partial trace over B defined by tr B ( a 1 a 2 b 1 b 2 ) a 1 a 2 tr( b 1 b 2 ) = a 1 a 2 b 2 b 1 with extension by linearity. It can be shown that the reduced density operator provides the correct statistics for measurements made on system A. For example, for a system in the product state ρ AB = ρ A ρ B, and ρ A = tr B (ρ A ρ B ) = ρ A tr(ρ B ) = ρ A ρ B = tr A (ρ A ρ B ) = ρ B.

31 Reduced density operator example The (pure) Bell state ( )/ 2 has density operator ( ) ( ) ρ = 2 2 = 1 [ ]. 2 Tracing out the second qubit gives the reduced density operator for the first qubit ρ = tr 2 (ρ) = 1 2 [tr 2( ) + + tr 2 ( )] = 1 2 [ 0 0 ) ] = 1 2 I since tr( ab cd ) = tr( a c b d ) = a c d b. Note that, while the joint system is in a known pure state, the first qubit is in a mixed state since tr((i /2) 2 ) = 1/2 < 1. This is a hallmark of quantum entanglement.

32 EPR argument In 1935, Albert Einstein, Boris Podolsky, and Nathan Rosen published a paper titled Can quantum-mechanical description of physical reality be considered complete? in Phys. Rev., 47: This introduced the famous EPR paradox in which Einstein referred to spukhafte Fernwirkung which translates to spooky action at a distance. The EPR argument is as follows: A physical property is an element of reality if it is possible to predict with certainty the value of the property immediately before a measurement. A complete theory in physics must include a representation of every element of reality. If Alice and Bob share an entangled pair of qubits such as the spin singlet ψ = ( )/ 2 then Alice can make a measurement of spin in the direction v and predict what Bob will measure for the value of v σ with certainty. Therefore the value is an element of reality, but

33 EPR argument continued Standard quantum mechanics does not include any fundamental element intended to represent the value of v σ for all unit vectors v; it can only give probabilities of measurement outcomes and is therefore not complete. Refer to the theorem on the next page. The EPR attempt to force a return to classical physics was not widely accepted and was proved invalid 30 years later by CERN physicist John Bell s inequalities.

34 EPR argument continued Theorem: For ψ = ( )/ 2 a measurement of spin along any axis v, v = 1, on both qubits results in opposite values, +1 for one and 1 for the other regardless of the choice of v. proof: Suppose a and b are the eigenstates of v σ. Then there are complex numbers α, β, γ, and δ such that Hence 0 = α a + β b 1 = γ a + δ b. ( )/ 2 = (αδ βγ) ( ab ba )/ 2. But αδ βγ is the determinant of the unitary matrix with rows [α β] and [γ δ] and is thus equal to e θ for θ R. Hence ab ba = 2 2 up to an unobservable global phase factor.

35 Bell inequality The following thought experiment is due to Clauser, Horne, Shimony, and Holt (CHSH) in Suppose Charlie prepares two particles and sends one to Alice and the other to Bob. Alice has two measuring apparatuses which can measure physical properties P Q and P R with outcomes 1 or 1 for the values Q of P Q and R of P R. Similarly, Bob can choose to measure either P S with value S or P T with value T. Measurements are simultaneous and independent of each other. QS + RS + RT QT = (Q + R)S + (R Q)T, where (Q + R)S = 0 or (R Q)T = 0 since R, Q = ±1. In either case, QS + RS + RT QT = ±2.

36 Bell inequality continued Let p(q, r, s, t) be the probability that, before the measurements, Q = q, R = r, S = s, and T = t. Then we have the mean value E(QS + RS + RT QT ) = p(q, r, s, t)(qs + rs + rt qt) Also, by linearity q,r,s,t q,r,s,t p(q, r, s, t) 2 = 2. E(QS +RS +RT QT ) = E(QS)+E(RS)+E(RT ) E(QT ) and hence E(QS) + E(RS) + E(RT ) E(QT ) 2. Now suppose Charlie sends Alice and Bob the two halves of the entangled pair ψ = ( )/ 2 and let Q = Z 1, S = ( Z 2 X 2 )/ 2 R = X 1, T = (Z 2 X 2 )/ 2.

37 Bell inequality continued Then QS = RS = RT = 1/ 2, and QT = 1/ 2 so that QS + RS + RT QT = 2 2 > 2. We thus have an experimental test that will resolve the paradox. The results of experiments are resoundingly in favor of the quantum mechanical prediction. The Bell inequality is violated. So what are the assumptions behind the Bell inequality? 1 The physical properties P Q, P R, P S, P T have definite values which exist independent of observation realism. 2 Alice s measurement does not influence Bob s result locality. Together, these assumptions are known as local realism and one or both are wrong. Most physicists say drop realism.