Lecture 2: Essential quantum mechanics


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1 Department of Physical Sciences, University of Helsinki kvanttilaskenta/ p. 1/46 Quantum information and computing Lecture 2: Essential quantum mechanics JaniPetri Martikainen jamartik Department of Physical Sciences University of Helsinki
2 Department of Physical Sciences, University of Helsinki kvanttilaskenta/ p. 2/46 Disclaimer This lecture is going to extremely boring, for everyone involved. The faster we go through this, the faster we can go on with our lives. I promise to wake you up in the end.
3 Department of Physical Sciences, University of Helsinki kvanttilaskenta/ p. 3/46 Vector spaces Basic objects are vector spaces, we are interested in C n, the space of all ntuples of complex numbers (z 1...z n ) elements of vector space are called vectors, we might use notation z 1. z n (1) Addition takes vectors to another vectors z 1 z 1 z 1 + z (2) z n z n z n + z n
4 Department of Physical Sciences, University of Helsinki kvanttilaskenta/ p. 4/46 Multiplication by a scalar Vector spaces z z 1. zz 1. (3) z n zz n In QM the vector is usually denoted with a ketnotation ψ, where ψ is used to label the vector in question. zero vector: ψ + 0 = ψ, note we do not use 0 notation for the zero vector, since in physics this usually means something different. To make typing easier: (z 1...z n ) implies the column matrix with entries z 1...z n W is a vector subspace of V if it is a vector space and closed under scalar multiplication and addition.
5 Department of Physical Sciences, University of Helsinki kvanttilaskenta/ p. 5/46 Bases and linear independence a vector space is spanned by a set of vectors v 1,..., v n such that any vector in V can be written as a v = i a i v i For example C 2 : v 1 = (1, 0) and v 2 = (0, 1) basis vectors are not unique: v 1 = (1, 1)/ 2 and v 2 = (1, 1)/ 2 would also be OK vectors v 1,..., v n are linearly dependent if a 1 v 1 + a n v n = 0 (4) for some a i, with a i 0 for at least one i Otherwise, vectors are linearly independent and can form a basis of the vector space
6 Department of Physical Sciences, University of Helsinki kvanttilaskenta/ p. 6/46 Bases...operators and matrices Number of basis elements is the dimension of the vector space we will just deal with finite dimensional vector spaces a linear operator between vector spaces V and W is defined as a function A : V W which is linear in its inputs i.e. A( i a i v i ) = i Identity operator I: I v v for all v in V zero operator: 0 v = 0 a i A v i (5) Once the action of linear A on the basis vectors is known, we know the action for all vectors in V
7 Department of Physical Sciences, University of Helsinki kvanttilaskenta/ p. 7/46 Operators and matrices V,W,X are vector spaces and A : V W B : W X are linear operators. BA is the composition of B with A, defined by (BA)( v ) = B(A v ) matrix representations of linear operators: m by n complex matrix with entries A ij sends a vector in C n into C m under matrix multiplication How to find a matrix representation for a linear operator?: Linear operator (from V to W ) defined through its action on the basis vectors A v j = i A ij w i (6)
8 Department of Physical Sciences, University of Helsinki kvanttilaskenta/ p. 8/46 Operators and matrices The matrix entries A ij form the matrix representation of the linear operator A...note: had to define input and output vector space basis. Pauli matrices: σ 0 = I = [ ] σ x = X = [ ] (7) σ y = Y = [ 0 i i 0 ] σ z = Z = [ ] (8)
9 Department of Physical Sciences, University of Helsinki kvanttilaskenta/ p. 9/46 Inner product inner product takes two input vectors and produces a complex number: ( v, w ) In quantum mechanics inner product of v and w is denoted by v w The bravector v is the dual vector to v the matrix representation of dual vectors is just a row vector
10 Department of Physical Sciences, University of Helsinki kvanttilaskenta/ p. 10/46 Inner product requirements 1. (, ) is linear in the second argument: ( v, i λ i w i ) = i λ i ( v, w ) (9) 2. ( v, w ) = ( w, v ) 3. ( v, v ) 0 with equality if only if v = 0 4. For example, C n : ((y 1 y n ), (z 1 z n )) = i y i z i = [y 1 y n] z 1. z n (10)
11 Department of Physical Sciences, University of Helsinki kvanttilaskenta/ p. 11/46 Inner product Hilbert space: vector space with inner product Vectors are orthogonal if their inner product is zero. norm: v v v vector is normalized if its norm is 1 orthonormal basis is a set of normalized orthogonal basis vectors From now on matrix representations of linear operators are assumed to be defined with respect to orthonormal bases.
12 Department of Physical Sciences, University of Helsinki kvanttilaskenta/ p. 12/46 GramSchmidt GramSchmidt procedure to construct an orthonormal basis: w 1,... w n is some basis set. Define v 1 w 1 / w 1 for 1 k n 1 define the new basis vector inductively v k+1 w k+1 k i=1 v i w k+1 v i w k+1 k i=1 v i w k+1 v i (11)
13 Department of Physical Sciences, University of Helsinki kvanttilaskenta/ p. 13/46 Inner product...outer product if w = w i i and v = v j j representation of vectors with respect to orthonormal basis then since i j = δ ij v w = i v i w i Outer product operator w v takes a vector from V to W : ( w v )( v ) = v v w (12) take v = i v i i, then ( i i ) v = i i v = v (13) so i i = I which is the completeness relation
14 Department of Physical Sciences, University of Helsinki kvanttilaskenta/ p. 14/46...CauchySchwartz inequality Can use completeness to represent operators in the outer product notation. Outer product representation of A: A = I W AI V = ij w j w j A v i v i = ij w j A v i w j v i CauchySchwartz inequality: v w 2 v v w w (14) Proof (idea): using GramSchmidt decomposition construct an orthonormal basis i so that w / w w is the first element. Use completeness i i = I and drop some nonnegative terms from v v w w = v i i v w w... (15)
15 Department of Physical Sciences, University of Helsinki kvanttilaskenta/ p. 15/46 Eigenvectors and eigenvalues Diagonal representation or spectral decomposition of A: A = λ i i i (16) where λ i are the eigenvalues and i are the eigenvectors Operator is diagonalizable if it has a diagonal representation [ ] 1 0 For example: Z = = If several eigenvalues are the same...eigenspace is degenerate
16 Department of Physical Sciences, University of Helsinki kvanttilaskenta/ p. 16/46 Hermitian operators Hermitian conjugate or adjoint A of the operator A: ( v,a w ) = (A v, w ) (AB) = B A define v v...(a v ) = v A In matrix representation: A = (A ) T where T indicates transpose An operator which is its own adjoint is Hermitian or selfadjoint
17 Department of Physical Sciences, University of Helsinki kvanttilaskenta/ p. 17/46 Projectors Projectors are an important class of Hermitian operators Suppose W is the kdimensional subspace of ndimensional vector space V. We can construct a basis 1,..., n for V so that 1,..., k is the basis for W Projector into subspace W : P = k i=1 i i Orthogonal complement: Q = I P
18 Department of Physical Sciences, University of Helsinki kvanttilaskenta/ p. 18/46 Unitary operators and tensor products Unitary operator if U U = I = UU Important (among other things) because they preserve inner products between vectors (U v,u w ) = v U U w = v I w = v w (17) tensor product: put vector spaces together to great bigger vector spaces V W... the elements of V W are linear combinations of tensor products v w of elements v in V and w in W. If i and j are orthonormal bases then i j is a basis for V W...we will often abbreviate i j = ij
19 Department of Physical Sciences, University of Helsinki kvanttilaskenta/ p. 19/46 Tensor products properties 1. z( v w ) = (z v ) w = v (z w ) 2. ( v 1 + v 2 ) w = v 1 w + v 2 w 3. v ( w 1 + w 2 ) = v w 1 + v w 2 A and B operators in V and W, define A B( v w ) A v B w (18) ensures linearity
20 Department of Physical Sciences, University of Helsinki kvanttilaskenta/ p. 20/46 Tensor products A is m n matrix and B is p q matrix A 11 B A 12 B A 1n B A A B 21 B A 22 B A 2n B... A m1 B A m2 B A mn B. (19) this is mp nq beast
21 Department of Physical Sciences, University of Helsinki kvanttilaskenta/ p. 21/46 Tensor product example [ 1 2 ] [ 2 3 ] = (20) X Y = [ 0 Y 1 Y 1 Y 0 Y ] = i 0 0 i 0 0 i 0 0 i (21)
22 Department of Physical Sciences, University of Helsinki kvanttilaskenta/ p. 22/46 Operator functions If A = a a a then f(a) = f(a) a a (22) For example, exp(θz) = [ e θ 0 0 e θ ] (23) Trace: Tr(A) = i A ii Trace is invariant under similarity transformation i.e. when A UAU with U unitary Simple example on blackboard...
23 Department of Physical Sciences, University of Helsinki kvanttilaskenta/ p. 23/46 Commutators and anticommutators Commutator: [A,B] = AB BA anticommutator: {A,B} = AB + BA If hermitian A and B commute i.e. [A,B] = 0, then there exists an orthonormal basis in which both A and B are diagonal. for Pauli matrices: [X,Y ] = 2iZ, [Y,Z] = 2iX, and [Z,X] = 2iY
24 Department of Physical Sciences, University of Helsinki kvanttilaskenta/ p. 24/46 Quantum mechanics On its own QM does not tell you what laws of a physical system must obey. It provides a mathematical and conceptual framework for the development of such laws. Postulates of QM were derived after a long process of trial and error...lots of guessing involved so don t be surprised if the motivation is not always clear.
25 Department of Physical Sciences, University of Helsinki kvanttilaskenta/ p. 25/46 Quantum mechanics Postulate 1: A system is completely described by its state vector in the systems Hilbert space (i.e. unit vector in the state space) qubit has a 2dimensional state space with basis 0 and 1. The state vector is generally ψ = a 0 + b 1 What the systems Hilbert space is, is not always clear. That depends...i.e. where do you draw the line where the system ends? Weird classically: we cannot directly observe the state vector.
26 Department of Physical Sciences, University of Helsinki kvanttilaskenta/ p. 26/46 Quantum mechanics Postulate 2: Evolution of a closed system is described by a unitary transformation. ψ(t 2 ) = U(t 2,t 1 ) ψ(t 1 ) (24) QM does not tell what U is...that depends... some examples X is called a bit flip: 0 1 and 1 0, on the other hand Z is called a phase flip: 1 1 Hadamard gate: [ H = ] (25)
27 Department of Physical Sciences, University of Helsinki kvanttilaskenta/ p. 27/46 Quantum mechanics Postulate 2 : Evolution of a closed system is described by the Schrödinger equation i ψ t = H ψ (26) spectral decomposition H = E E E is the system Hamiltonian E are the energy eigenvalues Note: even nonclosed systems can be sometimes described (or well approximated) by a Hamiltonian. For example, under some conditions an atom in a laser field can be described by a Hamiltonian which can be tuned experimentally.
28 Department of Physical Sciences, University of Helsinki kvanttilaskenta/ p. 28/46 Quantum mechanics: Measurement Postulate 3: Measurements described by a measurement operators {M m }, where index m refers to the measurement outcome that may occur. If ψ is the state just before the measurement then the probability of an outcome m is and after the measurement p(m) = ψ M mm m ψ (27) ψ M m ψ ψ M mm m ψ (28) Completeness of measurement operators: m M mm m = I
29 Department of Physical Sciences, University of Helsinki kvanttilaskenta/ p. 29/46 Quantum mechanics: Measurement Completeness simply implies that the probabilities of the measurement outcomes must sum to 1 Measure a qubit in its computational basis: M 0 = 0 0 and M 1 = 1 1 p(0) = ψ M 0 M 0 ψ = a 2, M 0 ψ / a = a/ a 0 Perhaps it is possible to derive the postulate 3 from the 1st and the 2nd postulate. Seems likely, but to prove it is hard. Classical things are easy to distinguish. In QM distinguishing reliably nonorthogonal state vectors is fundamentally impossible ψ 2 = α ψ 1 + β φ
30 Department of Physical Sciences, University of Helsinki kvanttilaskenta/ p. 30/46 Projective measurements Special case of the general measurement Projective measurement is described by an observable, M, a Hermitian operator. This observable can be decomposed as M = m mp m, (29) where P m is a projector onto eigenspace of M with eigenvalue m Measurement outcomes correspond to the eigenvalues m, probability of m is p(m) = ψ P m ψ and just after measurement the state is projected into P m ψ p(m) (30)
31 Department of Physical Sciences, University of Helsinki kvanttilaskenta/ p. 31/46 Projective measurements Easy to calculate average values of measurements: E(M) = mp(m) = m ψ P m ψ = ψ (mp m ) ψ = ψ M ψ (31)
32 Department of Physical Sciences, University of Helsinki kvanttilaskenta/ p. 32/46 Heisenberg s uncertainty principle A and B are Hermitian and suppose ψ AB ψ = x + iy. Then ψ [A,B] ψ = 2iy and ψ {A,B} ψ = 2x This implies: ψ [A,B] ψ 2 + ψ {A,B} ψ 2 = 4 ψ AB ψ 2 (32) By the CauchySchwarz inequality: ψ AB ψ 2 ψ A 2 ψ ψ B 2 ψ...combine with Eq. 32 and drop a nonnegative term ψ [A,B] ψ 2 4 ψ A 2 ψ ψ B 2 ψ (33)
33 Department of Physical Sciences, University of Helsinki kvanttilaskenta/ p. 33/46 Heisenberg s uncertainty principle More normal form: suppose C and D are observables and substitute A = C C and B = D D we obtain the Heisenberg s uncertainty principle (C) (D) [C,D], (34) 2 where (C) = (C C ) 2 is the standard deviation Example, operators X and Y measured for the quantum state 0 : [X,Y ] = 2iZ so (X) (Y ) 0 Z 0 = 1 (35)
34 Department of Physical Sciences, University of Helsinki kvanttilaskenta/ p. 34/46 POVM measurement Measurement postulate has two elements: a) gives a rule to describe measurement statistics b) gives a rule for the postmeasurement state For some applications, postmeasurement state is not of great interest. Probabilities for the measurement outcomes crucial. For example, experiment where the system is measured once at the end of the experiment Analysis in terms of the POVM (positiveoperatorvaluedmeasure) formalism Define E m M mm m whose expectation value gives the probability for the outcome m
35 Department of Physical Sciences, University of Helsinki kvanttilaskenta/ p. 35/46 POVM measurement E m is a positive operator (probabilities are positive) and Em = I E m are known as the POVM elements associated with the experiment and the complete set {E m } is known as the POVM For a projective measurement: P m P m = δ mm P m and the POVM elements are the same as the measurement operators, E m = P mp m = P m Note: Projective measurement is repeatable i.e. if you measure once you project into a state which does not change when you repeat the measurement. I.e. you will always measure the same result.
36 Department of Physical Sciences, University of Helsinki kvanttilaskenta/ p. 36/46 POVM measurement The repeatability suggests than many measurements in QM are NOT projective measurements. (use silvered screen to measure the position of the photon...photon destroyed..cannot repeat the measurement of the photons position) Example: Alice gives Bob a qubit which is either ψ 1 = 0 or ψ 2 = ( )/ 2. Bob can never be absolutely certain of which state he received. However, he can perform a measurement which distinguishes the states some of the time, but NEVER makes an error of misidentification. Consider POVM with elements: E 1 = 2/(1 + 2) 1 1, E 2 = 2/(1 + 2)( 0 1 )( 0 1 )/2, and E 3 = I E 1 E 2
37 Department of Physical Sciences, University of Helsinki kvanttilaskenta/ p. 37/46 POVM measurement These operators are complete and positive...povm OK. Suppose Bob got ψ 1 = 0 and he performs a measurement described by the POVM {E 1,E 2,E 3 }. There is zero probability for outcome E 1 since ψ 1 E 1 ψ 1 = 0...Bob knows that if the outcome is E 1 he must have got ψ 2 from Alice. Also, since ψ 2 is orthogonal with ( 0 1 )/ 2 outcome E 2 must mean he received ψ 1 Sometimes, Bob obtains an outcome E 3 in which case he cannot infer anything about the states identity However! He never makes a mistake in identification.
38 Department of Physical Sciences, University of Helsinki kvanttilaskenta/ p. 38/46 Phase If we have a state exp(iθ) ψ we will find that the measurement statistics of this state and the state ψ are identical. For this reason, from the observational point of view global phase factor plays no role. Relative phase is different: consider states ( )/ 2 and ( 0 1 )/ 2. Magnitude of the amplitudes are the same, but differ in sign (or more generally by a phase factor exp(iθ)) Relative phases may vary from amplitude to amplitude and the concept of a relative phase is basisdependent. This means that states with different relative phases can give rise to different measurement statistics.
39 Department of Physical Sciences, University of Helsinki kvanttilaskenta/ p. 39/46 QM: Composite systems Postulate 4: The state space of a composite system is the tensor product of the component system state spaces. If we have systems 1...n and systems are prepared in ψ i with i = 1...n, then the joint state of the system is ψ 1 ψ 2 ψ n Heuristically: It seems natural that if A is permissible state in A and B is permissible state in B, then A B should be permissible in AB. This combined with the superposition principle (that is if states x and y are OK, the also α x + β y is OK) gives you the tensor product postulate. We will use a subscript notation so that X 2 for example refers to the Pauli σ x acting on the 2nd qubit.
40 Department of Physical Sciences, University of Helsinki kvanttilaskenta/ p. 40/46 General measurement Lets show that we can implement general measurements with unitary evolution and projective measurements. Suppose we have a quantum state in state space Q and we wish to perform measurement described by measurement operators M n on the system Q Introduce an ancilla system, with state space M with an orthonormal basis states m with onetoone correspondence with the possible measurement outcomes. Think of ancilla system as a mathematical device or as an extra physical quantum system which has a state space with the required properties.
41 Department of Physical Sciences, University of Helsinki kvanttilaskenta/ p. 41/46 General measurement Let 0 be any fixed state of M and define U through U ψ 0 = m M m ψ m (36) Using orthonormality of m states and the completeness relation m M mm m = I we see that U preserves inner products: φ 0 U U ψ 0 = m,m φ M mm m ψ m m = m φ M mm m ψ = φ ψ (37)
42 Department of Physical Sciences, University of Helsinki kvanttilaskenta/ p. 42/46 General measurement From this it follows that U can be extended to a unitary operator on the space Q M, which we also denote by U. Suppose we perform a projective measurement on the two systems described by P m = I Q m m...outcome m with probability p(m) = ψ 0 U P m U ψ 0 = m,m ψ M m m (I m m )M m ψ m = ψ M mm m ψ (38) as given by the measurement postulate!
43 Department of Physical Sciences, University of Helsinki kvanttilaskenta/ p. 43/46 General measurement The joint state after the measurement of m is given by P m U ψ 0 ψ U P m U ψ = M m ψ m ψ M mm m ψ (39) just as described by the measurement postulate! Thus unitary dynamics, projective measurements, and the ability to introduce ancilla systems, together allow any general measurement of the form described by the measurement postulate.
44 Department of Physical Sciences, University of Helsinki kvanttilaskenta/ p. 44/46 Entangled states Consider a state ψ = (40) There are NO single qubit states a and b such that ψ = a b...the state is then called entangled Entangled states play a crucial role in quantum computation and information EPR thought experiment based on the possibility of entangled states with parts separated by a large distance. Measurement of, for example, 0 at one end, immediately implies that the one must have a state 0 at the other end.
45 Department of Physical Sciences, University of Helsinki kvanttilaskenta/ p. 45/46 Quantum mechanics We cannot directly observe the state vector. It is as if there is a hidden world in QM which we can access only imperfectly. Observing the state vector, typically changes it. (Play tennis, and each time you look at the ball, its position changes.) Bell s inequalities showed that we are stuck with counterintuitive nature of QM. Which is GREAT!
46 Department of Physical Sciences, University of Helsinki kvanttilaskenta/ p. 46/46 Exercises There are exercises next week! Download from the course web page...return by 1600 on Monday. WAKE UP! WAKE UP! WAKE UP!
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