Quantum Mechanics Class Notes Week 5 The Postulates of QM Postulate 1. The state (at time t) of a particle in a Quantum Mechanical system is represented by a wave function (t) in Hilbert Space (HS). Any multiple α with α from the set of complex numbers refers to the same physical state. The state is uniquely defined if we know the expectation values of any possible observable (see below). Postulate 2. The observables Ω of the system are Hermitian operators in the same HS. Only the eigenstates j!i of an operator have sharp, well defined values ! of the observable. Applying the operator to the eigenstate gives a real number ! which can be used to label the basis vectors j!i Ω j!i = ! j!i The eigen vectors of any observable form a basis for the HS. In general, we may have more than one operator with mutual eigenvectors Ω; Λ; with eigenvectors j!; λi that can be labelled by the eigenvalues to form a complete orthonormal basis set. Ω j!; λi = ! j!; λi ;Λ j!; λi = λ j!; λi Thus, any state can be written as X X j i = a!i,λi j!i; λii = h!i; λij j i j!i; λii !i,λi i;j Postulate 3. The observables and their eigenvalues determine what can be measured for any given system. Measuring the observable Ω on a (normalized) state yields one of the eigenvalues ! with a probability P (!) given by jPj > j2, where P is the projection operator on the sub-space of all vectors in HS with the same eigenvalue ! under Ω. If ! is a non-degenerate eigenvalue with only one eigen vector j! >, then Pj >= j! >< !j > and P (!) = j < !j > j2. To calculate the expectation value of the observable for any wave function using any hermitian operators, we can write X X h j Ω j i = h j Ω j!ii h!ij i = h j !ii !i h!ij i !i !i X 2 X = !i jh!ij ij = !iP (!i) !i !i The function itself, Ψ, may be completely specified (100% certain) but the outcome of a measurement still cannot be predicted and could in principle yield any !i. Example Given HS basis vectors 1 0 j1 >= ; j2 >= 0 1 and the operator as a 0 Ω = 0 b c1 2 2 the most general form of the wavefunction is jΨi = normalised by c1 + c2 = 1 where c1; c2 are c2 complex numbers.. Any measurement of Ω on the state of Ψ can only give a or b (eigenvalues of Ω) and the state will then collapse into the corresponding eigenstate with this eigenvalue: 2 c1 Ω a; P = jc1j ! j new >= j1 > −! 2 c2 b; P = jc2j ! j new >= j2 > Second Example Assume the operator Ω has several eigenvalues !1;!2;!3; :::. For each eigenvalue, you can have several eigenvectors that span a subspace of the full HS. E.g., assume !3 has several eigenvectors, j!3;ji (j = 1:::n). A measurement of Ω that yields the value !3 \collapses" the initial state j > by projecting it onto P the vector space V!3 with P!3 = j!3;ji h!3;jj. j This means that the final state is not uniquely specified by the measurement alone - we also need to know the a priori state. This can be avoided by doing a complete set of measurements with commuting observ- ables. Continuous set of eigenvalues In this case, we cannot interpret j h!j j i j2 as a probability, but rather as a probability density. For instance, if we measure the continuous observable X with a continuous set of eigenstates jxi, X jxi = x jxi, then the probability of finding the particle at exactly the position x0 is infinitely small; instead, one can define the probability for finding the particle at a position x within the interval x0 : : : x0 + ∆x as 2 P (x0 : : : x0 + ∆x) = jhx0j Ψij ∆x in the limit ∆x ! 0. Further Points on QM measurements: • In practice, it is not easy to devise a \perfect quantum measurement" that can give any possible value of an observable. Instead, it is easier to make a "projection measurement" that answers a simple yes-no question, e.g., \is the particle within a box of length L?". In that case, the final \collapsed" wave function is an eigenfunction of the projection operator, or perhaps a statistical ensemble of such eigenfunctions. • Even if you know the initial state, you cannot predict the result of a measurement unless the initial state is an eigenstate to the observable being measured. • If you make the same measurements immediately following one another, you keep getting the same answer and the system remains in the same eigen state. However, if you wait awhile, the system evolves with time and may no longer be in an eigenstate of the observable you are measuring. • You can \prepare" a system in a definite state of HS by making measurements of a complete set of observables (mutually compatible operators). Quantum Mechanics then tells you what the probabil- ity of any outcome for any measurement of any observable will be at any later time. Postulate 4. Given a generator g, the Poisson bracket of it with any observable gives δO δO = fO; gg x ! x + fx; gg p ! p + fp; gg δH = 0; fH; gg = 0 δO(δt) = fO;Hgδt Replacing the Poisson brackets with commutators and substituting in the observables Ω; G, the change in Ω is given by i~δhΩi = hΨj [Ω; G] jΨi hΩi = hΨj Ω jΨi Example With H as the generator acting on Ω i~δhΩi = hΨj [Ω;H] jΨi δt @ hΨj Ω jΨi i = hΨj [Ω;H] jΨi ~ @t Assuming jΨi ! jΨi (t), the state depends on time @Ψ @Ψ i~ Ω jΨi + hΨj Ω = hΨj ΩH jΨi − hΨj ΩH jΨi @t @t @Ψ @Ψ −i~ Ω jΨi + hΨj Ω i~ = hΨj Ω jHΨi − hHΨj Ω jΨi @t @t This implies @Ψ i = HΨ ~ @t Schr¨odinger'sEquation @ jΨi (t) i = H jΨi (t) ~ @t jΨi is a vector in HS and describes the state of the system at any given time t. From the Schr¨odinger Equation, the change in state over time can be predicted if the Hamiltonian H is known. A formal solution of the equation, ignoring the HS and pretending H is \like a number" is given as −iHt jΨi (t) = e ~ jΨi (0) For any observable, the expectation value after ∆t can be written ∆t hOi(t + ∆t) = hOi(t) + hΨj [O; H] jΨi i~ The Hamiltonian in V space depends on the representation of classical observables in the Hilbert space and can be as simple as a 2x2 matrix or as complicated as a differential operator. Example The Hamiltonian from HW problem set 4.5 in matrix form is given by E -V H = i;j −V E The HS has only 2 dimensions, corresponding to the two possible states of the N atom relative to the plane of 3 N atoms (above or below). Applying this to the Schrodinger Equation: a jΨi = b a_ E -V a Ea − V b i = = ~ b_ −V E b −V a + Eb Adding up the two equations i~(_a + b_) = (Ea + Eb − V a − V b) = (E − V )(a + b) c_ = (_a + b_) (E − V )c i~c_ = (E − V )c;_c = i~ (E−V )t (E−V )t c(t) = c0e i~ ;(a + b)(t) = (a + b)(t0)e i~ Subtracting the two i~(_a − b_) = (Ea − V b + V a − Eb) = (E + V )(a − b) (E+V )t (a − b)(t) = (a − b)(t0)e i~ 1 h (E−V )t (E+V )t i a(t) = (a + b)(t )e i~ + (a − b)(t )e i~ 2 0 0 1 h (E−V )t (E+V )t i b(t) = (a + b)(t )e i~ − (a − b)(t )e i~ 2 0 0 Et 1 h −V t V t −V t V t i a(t) = e i~ a e i~ + e i~ + b e i~ − e i~ 2 0 0 −iEt V t V t a(t) = e ~ a0 cos + ib0 sin ~ ~ −iEt V t V t b(t) = e ~ b0 cos + ia0 sin ~ ~ Over all time V t V t −iEt a0 cos + ib0 sin jΨi (t) = e ~ ~ ~ b cos V t + ia sin V t 0 ~ 0 ~ Given initial conditions, you can determine the state of the system at any time. At 1 t = 0; jΨi (0) = the atom in on top 0 π 0 t = T = ~ ; jΨi (T ) = the system will have flipped back 2V −1 The system moves up and down periodically. By making a measurement we are actually looking for the 1 eigenvectors of the observable. At t = 0 there is a 100% chance of getting . Even though the system 0 is perfect, after t = 0, it changes to a superposition of the two eigenvectors. The outcome still has to be 1 0 either or , but the probability of each state has to be calculated. 0 −1 2 1 P rob(+1) = jΨ(t)i 0 −iEt V t 2 −cos = e ~ ~ V t cos2 ~ V t P rob(−1) = sin2 ~ The sum of the 2 probabilities always gives 1. If the probability of either is 100%, then that is the state of the system. E -V We can use the Hamiltonian matrix, H = by replacing it in the formal solution jΨi (t) = −V E iHt e ~ jΨi (0) 0 1 it E -V @ A ~ −V E it E -V e = 1 + ::: ~ −V E Using e(a+b) = eaeb 0 1 0 1 −it E 0 it 0 -V @ A @ A ~ 0 E ~ −V 0 = e e −1 2 2 −2 −iEt 1 0 0 iV t~ 1 iV t ~ 0 = e ~ 1 + −1 + 2 2 −2 ::: 0 1 iV t~ 0 2 0 iV t ~ V t V t −iEt cos i sin the propagator U(t) = e ~ ~ ~ i sin V t cos V t ~ ~ If H is Hermitian, then U is Unitary and tells what the state of the system is at any time t jΨi (t) = U(t) jΨi (0) 1 jΨi (0) = 0 One more way of looking at the Schrodinger equation is to look first for the eigenvalues H jφni = En jφni Any state X X H jφni hφnj Ψi (t) = En hφnj Ψi jφni n n @ X i jΨi = H jφ i hφ j Ψi ~@t n n n @ X φ i jΨi = H jφ i hφ j Ψi > n ~@t n n n If Ψ is an eigenfunction @ i jΨ i = E jφ i ~@t n n n En becomes a number so we can write iEnt jφni (t) = e ~ jφni (0) This means that the initial state at t = 0 becomes a multiple of a complex number at all t and doesn't change the eigenstates (stationary and time-independent) solutions to the Schrodinger equation.