1 Variational calculation of a 1D bound state

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1 TEORETISK FYSIK, KTH TENTAMEN I KVANTMEKANIK FÖRDJUPNINGSKURS EXAMINATION IN ADVANCED QUANTUM MECHAN- ICS Kvantmekanik fördjupningskurs SI38 för F4 Thursday December, 7, Write on each page: Name, study program and year, problem number Motivate in detail! Insufficient motivation leads to reduction of points Allowed material: Formula collection (handed out with the exam), BETA, pocket calculator Grading system: Max 3 points per problem Examiner: Jack Lidmar, tel Variational calculation of a 1D bound state A particle moves in a potential V (x) = { if x < ax if x where a >. Suggest a variational ansatz for the ground state wave function and calculate an estimate of its energy. Calculate also an estimate of the expectation value of the position. Spin in a magnetic field Consider a spin S in a uniform magnetic field B, with a Hamiltonian Ĥ = γb Ŝ. Derive the equation d Ŝ Ŝ = γb. dt Specialize now to a spin- 1 particle and assume that the magnetic field is aligned in the z-direction, B = Be z. Assume further that the state at t = is ψ() = 1 ( + + ), where ± are the eigenstates of Ŝz with eigenvalues ± h/, respectively. What is the state vector at time t = π/γ B? What is the expectation value Ŝ x (t) at the same time? You may need the commutation relations for the components of the spin: [Ŝx, Ŝy] = i hŝz, and cyclic permutations of x,y,z. 1 SEE NEXT PAGE!

Insufficient motivation leads to reduction of points Allowed material: Formula collection (handed out with the exam), BETA, pocket calculator Grading system: Max 3 points per problem Examiner: Jack

2 3 Identical particles Two identical non-interacting spin- 1 particles are placed in a 1D potential { if x L V (x) = otherwise A strong magnetic field is applied so that the projection of the total spin on the z-axis is maximized. Write down the ground state wave function! 4 Perturbed harmonic oscillator A two dimensional harmonic oscillator has a degenerate eigenvalue E = 3 hω. What happens to this energy level due to the perturbation where C is a constant. Ĥ 1 = C ˆxŷ 5 Time-dependent perturbation Assume that a system, described by a time-independent Hamiltonian Ĥ, is perturbed by { Ĥ 1 (t) = Ĥ e t/τ for t < for t >, where Ĥ is small. At time t = the system is in an energy eigenstate n of Ĥ. Calculate, using first order time-dependent perturbation theory, the state of the system at t =. Show that you recover the results of time-independent perturbation theory in the limit τ. 6 Measurement of entangled spins Consider a proton and an electron in a spin singlet state, i.e., a state with total spin zero. Initially they form a bound state, i.e., a hydrogen atom. The separation between the electron and proton is then increased without inducing any transitions of the spin part of the state, until they are a few meters away from each other. Suppose then that the spin component Ŝz of the electron is measured. Suppose further that the outcome of the measurement was h/. What is the state directly after the measurement? Calculate the probabilities of the outcome of a subsequent measurement of Ŝz of the proton! Assume that all measurements can be considered ideal. GOOD LUCK!

What happens to this energy level due to the perturbation where C is a constant.

3 TEORETISK FYSIK, KTH TENTAMEN I KVANTMEKANIK FÖRDJUPNINGSKURS EXAMINATION IN ADVANCED QUANTUM MECHANICS Kvantmekanik fördjupningskurs SI38 för F4 Thursday December, 7, Solutions 1 Variational calculation of a 1D bound state This problem has many solutions! Since the potential is infinite for x < any physical wave function must go to zero at x =. It should also vanish for x. For instance, we can try ψ(x) = x exp( λx/), where λ can be varied: With H = 1 m ˆp + V (x) we get ψ (x)ĥψ(x)dx = 1 m ˆpψ(x) + V (x) ψ(x) dx = = ψ(x) dx = E(λ) = Optimize with respect to λ: So h m (1 λx/) e λx + ax 3 e λx dx = x e λx dx = /λ 3 ψ (x)ĥψ(x)dx ψ(x) dx = h λ 8m + 3a λ de dλ = h λ/4m 3a/λ = λ 3 = 1ma/ h, λ = (1ma/ h ) 1/3 E = 9a λ = 9 ( 1 h ) 1/3 ( ) h 1/3 a a 1 ma ma x = λ3 x 3 e λx = 3 λ h m ψ (x) + ax ψ(x) dx h 4mλ + 6a λ 4 Repeating the calculation using instead a trial wave function ψ(x) = x exp( x /4 ) gives E = 3 ( 6 h ) 1/3 ( ) h 1/3 a a π ma ma ( 8 3 π x = π, = h ) 1/3 16 ma Perhaps you found a better estimate! 3

4 Spin in a magnetic field Using Ehrenfest s theorem, d Ŝ 1 = [ Ŝ, dt i h Ĥ] = iγ h [ Ŝ, B Ŝ] [Ŝ, B Ŝ] = [(Ŝx, Ŝy, Ŝz), B x Ŝ x + B y Ŝ y + B z Ŝ z ] = i h(b y Ŝ z B z Ŝ y,...,...) = i hb Ŝ d Ŝ Ŝ = γb dt Now, for a spin- 1 and B = Be z, the Hamiltonian is diagonal in the eigenstates of Ŝz with energy eigenvalues E ± = hγb/. The time development of the state is then ψ(t) = 1 (e ie +t/ h + + e ie t/ h ) = 1 (e +iγbt/ + + e iγbt/ ) For t = π/γb we get ψ(t) = ψ(), but S x (t) = S x () = h/. 3 Identical particles First we need to calculate the one-particle wave functions: The general solution of the time-independent Schrödinger equation in the region < x < L is φ(x) = A sin(kx)+ B cos(kx). Applying the boundary conditions φ() = φ(l) = and normalizing we find the eigenfunctions φ n (x) = /L sin(nπx/l), n = 1,, 3,.... Since the particles carry spin 1 they are fermions so they cannot occupy the same one-particle state, and the -particle state has to be antisymmetric. In a strong magnetic field the spins of both particles are parallel, so the spin part is symmetric (given by + + ), hence the orbital part is the antisymmetric combination of the two lowest eigenfunctions ψ(x 1, x ) = 1 (φ 1 (x 1 )φ (x ) φ 1 (x )φ (x 1 )) = L (sin πx 1 L sin πx L sin πx L sin πx 1 L ) 4 Perturbed harmonic oscillator The unperturbed Hamiltonian is Ĥ = hω(a xa x + a ya y + 1), with eigenvalues E nxn y = hω(n x + n y + 1), where n x, n y are non-negative integers. Hence the states n x n y =, 11, are degenerate with energy E = 3 hω. We rewrite the perturbation in terms of creation and annihilation operators, Ĥ 1 = C ˆxŷ = D ( a x + a x ) ( a y + a y ), where D = C h/mω. To get the first order corrections E 1 to the energy level we need to calculate the matrix elements n x n y Ĥ1 n x n y for the three degenerate states above and then calculate the eigenvalues of that matrix: E D D E D D E = ( E)(E D ) D (D E) = E(E 4D ) = 4

3 Identical particles First we need to calculate the one-particle wave functions: The general solution of the time-independent Schrödinger equation in the region < x < L is φ(x) = A sin(kx)+ B

5 So E =, ±D. The degenerate energy level splits into three levels with energies E = 3 hω, 3 hω ± C h mω. 5 Time-dependent perturbation The state at time t is ψ(t) = m d m (t)e ie m t/ h m, where m are eigenstates of Ĥ, and d m (t) = δ mn ī h t m Ĥ1 (t ) n e iωmnt dt t when the initial state at t is n. Here ω mn = (E m E n)/ h. Letting t and putting in the explicit form of the time dependence of the perturbation gives d m () = δ mn ī h m Ĥ n e t /τ+iω mnt dt = δ mn ī h m Ĥ n So the state at t = is ψ() = n ī h m m m Ĥ n τ 1 + i(e m E n )/ h 1 τ 1 + iω mn = n n n Ĥ iτ n h + m m Ĥ n m n i hτ 1 (E m E n ). () The first two terms in Eq. () can be combined to (1 iφ) n = e iφ n + O(Ĥ ), where φ = n Ĥ n τ/ h. Since this phase is arbitrary we can just multiply ψ() with e iφ, which gets rid of the second term, with no other change to first order. It is also easy to see that the state is normalized to this order. Taking the limit τ then gives ψ() = n + m m Ĥ n m n E n E m in complete agreement with first order time-independent perturbation theory. (1) 6 Measurement of entangled spins We need only to consider the spin part of the state. A basis for the composite system is given by product states of the form m p m e = m p m e 5

6 where m p = ± and m e = ± refer to the quantum numbers for the spin projection on the z-axis of proton and electron, respectively. The total spin operator is given by S = S p + S e or more exactly S I + I S A general spin state is a linear combination ψ = α ++ + β + + γ + + δ A spin singlet state has total spin zero, so S z ψ = and S ψ =. The first condition gives α = δ =. Instead of applying the second we can equivalently apply S + ψ =, which gives β + γ =. So the initial state before any measurements is ψ = 1 ( + + ). Now we measure S z of the electron. So the observable corresponding to this measurement is really I S z, i.e., it operates only on the electron part of the state. Assuming we get h/ from this measurement the system will end up in a linear combination of the eigenstates of I S z corresponding to h/. There are two such states: ++ and +, and we need to project ψ onto the subspace spanned by these: ψ ( ) ψ = 1 + Renormalizing we get the new state directly after the measurement + Since Sp z + = (S z I)( + ) = S z I + = h + a second measurement of the proton now gives h/ with certainty. 6

and S ψ =. The first condition gives α = δ =. Instead of applying the second we can equivalently apply S + ψ =, which gives β + γ =. So the initial state before any measurements is ψ = 1 ( + + ).

PHY4604 Introduction to Quantum Mechanics Fall 2004 Practice Test 3 November 22, 2004

PHY4604 Introduction to Quantum Mechanics Fall 2004 Practice Test 3 November 22, 2004 PHY464 Introduction to Quantum Mechanics Fall 4 Practice Test 3 November, 4 These problems are similar but not identical to the actual test. One or two parts will actually show up.. Short answer. (a) Recall