Quantum Mechanics II: Examples Michael A. Nielsen University of Queensland Goals: 1. To apply the principles introduced in the last lecture to some illustrative examples: superdense coding, and quantum teleportation. 2. Revised form of postulates 2 (dynamics) and 3 (measurement). 3. Introduce more elements of the Dirac notation. 4. Discuss the philosophy underlying quantum information science.
Alice Superdense coding Bob ab ab Theorist s impression of a measuring device
Alice Superdense coding Bob ab ab
Alice Superdense coding Bob X Z 0 = 1 ; X 1 = 0 0 = 0 ; Z 1 = 1 ab 00 : Apply I 01 : Apply Z 10 : Apply X 11 : Apply XZ 00 + 11 00 + 11 2 2 00 + 11 00 11 2 2 00 + 11 10 + 01 2 2 00 + 11 10 01 2 2 00 + 11 2 ab
Superdense coding can be viewed as a statement about the interchangeability of physical resources. 1 ebit + 1 qubit of communication 2 bits of classical communication
Worked exercise: Could Alice and Bob still communicate two bits using the superdense coding protocol if the initial 01 10 state shared by Alice and Bob was? 2
1 Revised measurement postulate Recall postulate 3: If we measure ψ in an orthonormal basis e,..., e, then we obtain the result with probability d P( ) = e ψ. 2 The measurement disturbs the system, leaving it in a state determined by the outcome. e Problem: Imagine we measure a quantum system, A, in the orthonormal basis e,..., e. 1 d A Suppose system A is part of a larger system, consisting of two components, A and B. How should we describe the effect of the measurement on the larger system?
1 Revised measurement postulate Recall postulate 3: If we measure ψ in an orthonormal basis e,..., e, then we obtain the result with probability d P( ) = e ψ. 2 The measurement disturbs the system, leaving it in a state determined by the outcome. e The revised postulate replaces the orthogonal states e1,..., with a complete set of orthogonal subspaces V,..., V. V = V1 V2... V m. 1 m e d = ( 1 + 2 ) + 3 ( e1 e2 e3 ) = ( e1 e2 ) ( e3 ) Example: ψ α e β e γ e sp,, sp, sp A general measurement can be thought of as asking the question " which of the subspaces V,..., V m are we in? " 1
1 Revised measurement postulate Recall postulate 3: If we measure ψ in an orthonormal basis e,..., e, then we obtain the result with probability d P( ) = e ψ. 2 The measurement disturbs the system, leaving it in a state determined by the outcome. e Mathematically, it is convenient to describe the subspaces V,..., V m in terms of their corresp onding proectors, P,..., P m. 1 1 ( ) ( 1 2 ) Example: The proector P onto sp e, e acts as P α e + β e + γ e = α e + β e 1 2 3 1 2 In general, the proector P onto a subspace V acts as the identity on that subspace, and annihilates everything orthogonal to V.
Revised measurement postulate Let P1,..., P m be a set of proectors onto a complete set of orthogonal subspaces of state space. P = I ; P P =δ P k k This set of proectors defines a measurement. If we measure ψ then we get outcome with probability Pr( ) = ψ P ψ. The measurement unavoidably disturbs the system, leaving P ψ it in the post-measurement state ψ P ψ
Example: A two-outcome measurement on a qutrit A general state of a qutri t may be written ψ = α 0 + β 1 + γ 2. 1 ( ) ( ) P proects onto sp 0, 1 ; and P2 proects onto sp 2. α α 2 Pr(1) = ψ P1 ψ = β β = α + β γ 0 Pr(2) = ψ P 2 ψ = γ 2 2 ψ = P ψ α 0 + β 1 = ' 1 1 2 2 ψ P1 ψ α + β ψ P ψ γ 2 = ~ 2 ψ P ψ = γ ' 2 2 2
1 2 Example: Measuring the first of two qubits Suppose we want to perform a measurement in the basis e, e for the first of two qubits. The rule is to first form the corresponding proectors PP 1, 2 onto the state space of that qubit, and then to tensor them with the identity on the second qubit, obtaining P I and P I. 1 ( ) = ψ ( P I ) 0 2 Example: If the state of two qubits is ψ = α00 00 + α01 01 + α10 10 + α11 11 then measuring the first qubit in the computational basis gives the result 0 with probability Pr 0 ψ ( α00 00 α01 01 α10 10 α11 11 ) ( α00 00 α01 01 ) = + + + + 2 2 = α + α 00 01
Example: Measuring the first two of three qubits Suppose we have three qubits in the state α e a + β e b + γ e c + δ e d. 1 2 3 4 e1, e2, e3, e4 is an orthonormal basis for the state space of the first two qubits. a, b, c, d are normalized states of the third qubit. Measuring the first two qubits in the basis e1, e2, e3, e4 gives the result 1 with probability Pr(1) = ψ ψ ( P I ) 1 ( α 1 ) = ψ e a 2 = α Post-measurement state is e a. 1
Alice Teleportation Bob
Alice Teleportation Bob 01 01
Alice Teleportation Bob α 0 + β 1 00 + 11 2 00 + 11 + 2 α 000 + α 011 + β 100 + β 111 = 2 ( α 0 β 1 ) 00 1 00 + 11 1 00 11 = + 2 2 2 2 10 1 01 + 10 1 01 10 = 2 2 2 2 01 1 01 + 10 1 01 10 = + 2 2 2 2 11 1 00 + 11 1 00 11 = 2 2 2 2
Alice Teleportation Bob 01 1 00 + 11 = + 2 2 ( α 0 β 1 ) 1 00 11 + 2 2 ( α 0 β 1 ) 1 01 + 10 + + 2 2 ( α 1 β 0 ) 1 01 10 + 2 2 ( α 1 β 0 ) I ( α 0 β 1 ) + ( 0 1 ) Z α + β ( 0 1 ) X α + β ( 0 1 ) ZX α + β
Teleportation can be viewed as a statement about the interchangeability of physical resources. 1 ebit + 2 classical bits of communication 1 qubit of communication Compare with superdense coding: 1 ebit + 1 qubit of communication 2 bits of classical communication 1 qubit of communication = 2 bits of communication (Mod 1 ebit)
The fundamental question of information science 1. Given a physical resource energy, time, bits, space, entanglement; and 2. Given an information processing task data compression, information transmission, teleportation; and 3. Given a criterion for success; We ask the question: How much of 1 do I need to achieve 2, while satisfying 3? Pursuing this question in the quantum case has led to, and presumably will continue to lead to, interesting new information processing capabilities. How to write a quant-ph Are there any fundamental scientific questions that can be addressed by this program?
What fundamental problems are addressed by quantum information science? You Your challenger Knowing the rules Understanding the game
Knowing the rules of quantum mechanics Understanding quantum mechanics What high-level principles are implied by quantum mechanics?
Robert B. Laughlin, 1998 Nobel Lecture I give my class of extremely bright graduate students, who have mastered quantum mechanics but are otherwise unsuspecting and innocent, a take-home exam in which they are asked to deduce superfluidity from first principles. There is no doubt a special place in hell being reserved for me at this very moment for this mean trick, for the task is impossible. Superfluidity, like the fractional quantum Hall effect, is an emergent phenomenon a low-energy collective effect of huge numbers of particles that cannot be deduced from the microscopic equations of motion in a rigorous way and that disappears completely when the system is taken apart (Anderson, 1972)
Quantum information science as an approach to the study of complex quantum systems Quantum processes Shor s algorithm teleportation communication theory of entanglement quantum phase transitions cryptography quantum error-correction Complexity
A few quanta of miscellanea The outer product notation The spectral theorem diagonalizing Hermitian matrices Historical digression on measurement The trace operation Quantum dynamics in continuous time: an alternative form of the second postulate
Let ψ and φ be vectors. ( ) Outer product notation Define a linear operation (matrix) ψ φ by ψ φ γ ψ φ γ ( ) Example: 1 0 α 0 + β 1 1 α = α 1 Connection to matrices: If a = a b = b a b k = b a *, and then k. a1 * * * But a 2 b1 b 2 1 = bk a. a1 * * * Thus a b = a 2 b1 b2 b3.
Outer product notation Example: Example: 1 0 0 = 1 0 0 0 1 1 = 0 1 1 1 0 = 0 0 0 0 = 0 1 1 0 Example: Z = 0 1 = 0 0 1 1 1 0 1 Example: 0 1 = 0 1 0 = 0 0 0 0 0 Example: 1 0 = 1 0 1 = 1 0 0 1 Exa mp le: X = = 0 1 + 1 0 1 0 Exercise: Find an outer product representation for Y.
Outer product notation One of the advantages of the outer product notation is that it provides a convenient tool with which to describe proectors, and thus quantum measurements. ( ) ( 1 2 ) Recall: The proector P onto sp e, e acts as P α e + β e + γ e = α e + β e 1 2 3 1 2 This gives us a simple explicit formula for P, since e e + e e α e + β e + γ e = α e + β e ( )( ) 1 1 2 2 1 2 3 1 2 More generally, the proector onto a subspace spanned by orthonormal vectors e1,..., em is given by P = e e. Exercise: Suppose e1,..., e is an orthonormal basis for state d space. Prove that I = e e. Exer ci se: Prove that a b = b a.
The spectral theorem Theorem: A is diagonalizable, A = Udiag ( λ1,, λ ) U, d where U is unitary, and λ,, λ are the eigenvalues Suppose A is a Hermitian matrix, A = A. Then ( ) But diag λ1,, λ = λ. d Thus A = λ e e, where e U is the λ eigenvector of A, A e = λ e. 1 d of A. A = λkpk, where Pk is the proector onto the k λ eigenspace of A. k
Examples of the spectral theorem 1 0 Example: Z = = 0 0 1 1 0 1 0 1 0 ± 1 Example: X = has eigenvectors 1 0 ± 2 corresponding eigenvlaues ± 1., with + + - 1 1 1 1 = 1 1-1 2 1 2-1 -1 1 1 1 1 1 1 = 2 1 1 2 1 1 0 1 = 1 0
Historical digression: the measurement postulate formulated in terms of observables Our form: A complete set of proectors P onto orthogonal subspaces. Outcome occurs with probability Pr( ) = ψ P ψ. The corresponding post-measurement state is P ψ. ψ P ψ Old form: A measurement is described by an observable, a Hermitian operator M, with spectral decomposition M = λp. The possible measurement outcomes correspond to the eigenvalues λ, and the outcome λ occurs with probability Pr( ) = ψ P ψ. The corresponding post-measurement state is P ψ. ψ P ψ
An example of observables in action Example: Suppose we "measure Z". Z has spectral decomposition Z = 0 0-1 1, so this is ust like measuring in the computational basis, and calling the outcomes "1" and "-1", respectively, for 0 and 1. Exercise: Find the spectral decomposition of Z Z. Show that measuring Z Z corresponds to measuring the parity of two qubits, with the result +1 corresponding to even parity, and the result -1 corresponding to odd parity. Exercise: Suppose we measure the observable M for a state ψ which is an eigenstate of that observable. Show that, with certainty, the outcome of the measurement is the corresponding eigenvalue of the observable.
tr ( A) A The trace operation 0 1 1 0 Examples: X = tr( X ) 0; I tr( I ) 2. 1 0 = = = 0 1 Cyclicity proper t tr( AB ) =t ( BA) tr ( AB ) ( AB ) y: r. = = A B k k k ( a b ) Exercise: Prove that tr =a b. = = ( BA) kk = tr( BA) B A k k k k
An alternative form of postulate 2 Postulate 2: The evolution of a closed quantum system is described by a unitary transformation. ψ ' = U ψ But quantum dynamics occurs in continuous time!
An alternate form of postulate 2 The evolution of a closed quantum system is described by Schroedinger's equation: d ψ i = H ψ dt where H is a constant Hermitian matrix known as the Hamiltonian of the system. The eigenvectors of H are known as the energy eigenstates of the system, and the corresponding eigenvalues are known as the energies. ( ) ( ) Example: H = ωx has energy eigenstates 0 + 1 / 2 and 0 1 / 2, with corresponding energies ± ω
Connection to old form of postulate 2 The solution of Schroedinger's equation is ψ( t) = exp( iht) ψ(0) U exp( iht) ψ ' = U ψ