Physics 215 Solution Set 3 Winter 2018 1. Consider the One

Physics 215 Solution Set 3 Winter 2018

1. Consider the one-dimensional problem of a particle moving in a delta-function potential:

V (x)= Aδ(x) . −

(a) Solve for the bound state energies and wave functions. Consider the cases A> 0 and A< 0 separately.

The time-independent Schrödinger equation for this problem is ~2 d2ψ Aδ(x)ψ(x)= Eψ(x) . (1) −2m dx2 − Integrating this equation from x = ǫ to x = ǫ, where ǫ > 0 is a positive infinitesimal − quantity, ~2 ǫ ψ′(ǫ) ψ′( ǫ) Aψ(0) = E ψ(x) dx . −2m − − − Z−ǫ where ψ′ dψ/dx. We require that ψ(x) is continuous at x = 0. If this were not true ≡ then dψ/dx would behave like a delta function near x = 0 and d2ψ/dx2 would behave lie the derivative of a delta function.1 Such behavior is not compatible with eq. (1). Hence, it follows that ǫ ǫ ψ(x) dx ψ(0) dx =2ǫψ(0) , ≃ Z−ǫ Z−ǫ which vanishes in the limit of ǫ 0. We conclude that → 2mA lim ψ′(ǫ) ψ′( ǫ) = ψ(0) . (2) ǫ→0 − − − ~2 That is, dψ/dx is not continuous at x = 0; its discontinuity is determined by eq. (2). To solve for the bound state energies, we solve eq. (1) for x = 0. In this case, we can set δ(x) = 0. Normalizable energy eigenstates exist if and only if E6 < 0 and A> 0. That is, we solve the differential equation, ~2 d2ψ + E ψ(x)=0 , for x = 0, −2m dx2 | | 6 where E = E , under the assumption that limx→±∞ ψ(x) = 0. The solution to this equation after−| imposing| the boundary conditions at x = is ±∞ Ne−αx , for x> 0, ψ(x)= αx (Ne , for x< 0,

1Recall that dΘ(x)/dx = δ(x), where Θ(x) is the Heavyside step function.

1 where N is a normalization constant, and

2m E α | | . (3) ~2 ≡ r

Note that the sign of α (which is positive) is determined since limx→±∞ ψ(x) = 0. The energy eigenvalues are determined by imposing eq. (2). Noting that N = ψ(0), eq. (2) yields 2mAN 2αN = , − − ~2 which implies that mA α = . (4) ~2 Note that we must require that A> 0 in order that bound state solutions exist, since A< 0 is incompatible with the requirement that α> 0. Finally, inserting eq. (3) into eq. (4) yields

mA2 E = E = . −| | − 2~2 That is, there is precisely one bound state. The corresponding energy eigenfunction is

1/2 mA 2 ψ (x)= e−mA|x|/~ , (5) E ~2 where we have determined that the normalization constant is N = mA/~2 by requiring that ∞ p ψ (x) 2dx =1 . | E | Z−∞ The normalization condition yields

1 ∞ ∞ 1 ∞ 1 ~2 = e−2α|x| dx =2 e−2α|x| dx = e−2α|x| = = , N 2 −α α mA Z−∞ Z0 0 | | after using eq. (4). The phase of N is conventionally taken to be unity, in which case N =(mA/~2)1/2, as indicated in eq. (5).

(b) In the case of A > 0, consider a scattering process where the incident wave enters from the left with E = ~2k2/(2m) > 0 (where E is the energy eigenvalue of the Hamiltonian). Determine the corresponding reflection coefficient R and the transmission coefficient T as a function of k. Write the coefficients in terms of the dimensionless parameter b E/Eg where E is the ground state energy obtained in part (a), in the case of A > 0.≡ What is − g the behavior of T (b) as b 1? → −

2 Consider a scattering process where the incident wave enters from the left. Then, the solution to ~2 d2ψ Eψ(x)=0 , for x = 0, −2m dx2 − 6 where E > 0, is given by

Beikx + Ce−ikx , for x< 0. ψ(x)= ikx (6) (De , for x> 0, and 2mE k = , (7) ~2 r Note that k > 0, under the assumption that no incident wave enters from the right. As discussed in part (a), we demand that the wave function ψ(x) is continuous at x = 0, whereas dψ/dx is discontinuous at x = 0 according to eq. (2). Noting that ψ(0) = B + C = D and lim ψ′( ǫ)= ik(B C) , lim ψ′(ǫ)= ikD , ǫ→0 − − ǫ→0 which we make use of in eq. (2), we end up with two equations, B + C = D, 2mA ik D B + C = D, − − ~2 which can be rewritten in the following form, D C =1 , (8) B − B D 2imA C 1 + =1 . (9) B − ~2k B Adding eqs. (8) and (9), we immediately solve for D/B, D ~2k = . (10) B ~2k imA − Plugging back in to eq. (8) yields, C imA = . (11) B ~2k imA − The corresponding reﬂection and transmission coeﬃcients are C 2 m2A2 R = = , (12) B ~4k2 + m2A2

2 4 2 D ~ k T = = . (13) B ~4k2 + m2A2

3 It is convenient to rewrite eqs. (12) and (13) in terms of the the energy eigenvalue for the scattering problem, E = ~2k2/(2m) [cf. eq. (7)], and the parameter E mA2/(2~2), 0 ≡ − which we recognize from part (a) is the ground state energy of the attractive delta function potential. In terms of E and E0, eqs. (12) and (13) yield E E R = 0 , T = . −E E E E − 0 − 0

Note that the transmission coeﬃcient T (E) exhibits a pole at the ground state energy E0 of the attractive delta function potential.

(c) In the case of A< 0, consider a scattering process where the incident wave enters from the left with E = ~2k2/(2m) > 0. Investigate the location of any poles of the transmission amplitude in the complex k plane and the complex E plane, respectively. Explain your result in light of the fact that the repulsive delta function potential possesses no bound states.

In part (b), the transmission amplitude for one-dimensional scattering was obtained in eq. (10). In the case under consideration here, the only diﬀerence from part (b) is that the sign of A is negative. From eq. (10), we see that the poles of the transmission amplitude correspond to ~2k imA = 0. Solving for k yields − imA k = . (14) ~2 where k (2mE/~2)1/2 > 0. Since A < 0, it follows that Im k < 0, which means that ≡ in the complex k-plane, the pole is located on the negative imaginary axis. Hence, in the complex E-plane, the pole is located on the negative Re E-axis on the second Riemann sheet. In contrast, poles corresponding to bound states reside on the negative Re E-axis on the ﬁrst Riemann sheet of the complex E-plane. We conclude that the repulsive delta function potential supports no bound states, as expected. Indeed, if we rewrite eq. (6) in the form,

1 R eikx + e−ikx , for x< 0. ψ(x) T T (15) ∼  eikx , for x> 0, and substitute eq. (14) into eq. (15), we see that the resulting wave function diverges expo- nentially as x . This implies that the resulting wave function cannot correspond to a bound state. → ±∞

4 2. A one-dimensional potential has the following form:

+ , for x< 0, ∞ V (x)= V , for 0 < x < b, − 0  0 , for x > b, where V0 and b are positive constants.

(a) Find V0 as a function of b such that there is just one bound state, of about zero binding energy, for a particle of mass M.

To solve the bound state problem, we require that V

A sin(kx) , for 0 x < b , ψ(x)= −κx ≤ (Be for x > b, where E = E and −| | 2M(V E ) 2M E k = 0 −| | , κ = | | , (16) ~2 ~2 r r with k > 0 and κ> 0.2 Continuity of ψ and dψ/dx at x = b yields two equations,

A sin(kb)= Be−κb , Ak cos(kb)= Bκe−κb . − Diving these two equations, we obtain 1 1 tan(kb)= . (17) k −κ Note that eq. (16) implies that 2MV k2 + κ2 = 0 . (18) ~2 We can also rewrite eqs. (17) and (18) as

kb cot(kb)= κb , (19) − 2MV b2 (kb)2 +(κb)2 = 0 . (20) ~2

2We require κ> 0 so that ψ(x) 0 as x . The requirement of k> 0 is conventional, since the sign of k can be absorbed into the deﬁnition→ of the→ constant ∞ A.

5 Eqs. (19) and (20) can be solved graphically by looking for intersection of the function y = x cot x with the circle of radius r = (2MV b2/~2)1/2 in the ﬁrst quadrant of the x–y − 0 plane (where x = kb and y = κb).

κb

bound state

kb π 2

It is clear from the ﬁgure above that if V0 is less than some critical value, then the radius 1 of the circle, r < 2 π, in which case there are no intersections in the ﬁrst quadrant, and 1 therefore no bound states. The critical value of of V0 is obtained by setting r = 2 π. That is, 2MV b2 π2 0 = , ~2 4 which yields π2~2 V = . (21) 0 8Mb2

(b) Applying this crude model to the deuteron (a bound state of a proton and a neutron), −13 1 evaluate V0 in MeV, assuming b = 1.3 10 cm and M = 2 mp (where mp is the proton mass). ×

2 The rest mass of the proton is mpc = 938.272 MeV. Therefore, eq. (21) yields π2~2 π2(6.5817 10−22 MeV sec)2(3 1010 cm/sec)2 V = = × × 60.7 MeV . 0 8Mb2 4(1.3 10−13 cm)2(938.272 MeV) ≃ × 6 1 (c) Why did I set M = 2 mp rather than M = mp in part (b)?

Applying the Schr¨odinger equation to the bound state of a proton and a neutron, we proceed as in classical mechanics by converting the two-body problem into a one-body problem by separating oﬀ the center-of-mass motion. The corresponding mass employed in the one- body problem is the reduced mass, M = mpmn/(mp + mn). Since mp mn, it is a good 1 ≃ approximation to take M = 2 mp.

REMARK: We have obtained here an estimate of the mean value of the nuclear potential in light of the fact that the deuteron is a weakly bound system with a bound state energy of Eb 2.2 MeV. Since Eb V0 (which was estimated in part (b) to be 60 MeV), it follows≃ that − our approximation≪ of E 0 [corresponding to the critical value∼ of V obtained b ≃ 0 in eq. (21)] is justiﬁed.

3. Consider a one-dimensional quantum mechanical problem with a time-independent Hamil- tonian, H. The time evolution operator, evaluated in the coordinate basis, also known as the propagator, is given by, ∞ G(x, t ; x′, 0) = x e−iHt/~ x′ = dp x p p e−iHt/~ x′ , (22) h | | i h | ih | | i Z−∞ where we have taken the initial time to be t0 = 0.

(a) The free particle Hamiltonian is given by H = P 2/(2m). Evaluate the free particle propagator by explicitly performing the p-integration using eq. (22).

Plugging H = P 2/(2m) into eq. (22), and using the fact that P p = p p ,3 | i | i ∞ ip2t G(x, t ; x′, 0) = dp x p p x′ exp . h | ih | i −2m~ Z−∞ Using x p = (2π~)−1/2eipx/~ and p x′ = x p ∗, it follows that h | i h | i h | i 1 ∞ i p2t G(x, t ; x′, 0) = exp p(x x′) dp . (23) 2π~ ~ − − 2m Z−∞ In order to perform the integral given in eq. (23), we “complete the square,” by rewriting the argument of the exponent in eq. (23) as,

p2t p(x x′) it m(x x′) 2 im(x x′)2 i − = p − + − . − 2m~ − ~ −2m~ − t 2~t 3Since P is self-adjoint, we also have p P = p p (since the eigenvalues of a self-adjoint operator are real), and p f(P )= p f(p), for any functionh | f(Ph)| expressed as a Taylor series in P . h | h | 7 Plugging this last result back into eq. (23) yields,

1 im(x x′)2 ∞ it m(x x′) 2 G(x, t ; x′, 0) = exp − exp p − dp . 2π~ 2~t −2m~ − t Z−∞ ( ) We now change integration variables by deﬁning p′ p m(x x′)/t. The limits of integration ≡ − − do not change and we are left with, 1 im(x x′)2 ∞ itp′ 2 G(x, t ; x′, 0) = exp − exp dp′ . 2π~ 2~t −2m~ Z−∞ The integral can now be evaluated by making use of, ∞ ∞ 2 2 π e−iap dp =2 e−iap dp = e−iπ/4 , for a> 0, (24) a Z−∞ Z0 r which was derived in the class handout entitled, A Gaussian integral with a purely imaginary argument. Identifying a t/(2m~), and noting that a> 0 if we take t> 0 corresponding to propagation forward in time,≡ we end up with, m 1/2 im(x x′)2 G(x, t ; x′, 0) = e−iπ/4 exp − . (25) 2π~t 2~t Finally, if we interpret e−iπ/4 = i−1/2, we can rewrite eq. (25) as, m 1/2 im(x x′)2 G(x, t ; x′, 0) = exp − . (26) 2πi~t 2~t

2 1 2 2 (b) For the one-dimensional harmonic oscillator, where H = P /(2m)+ 2 mω X , can you evaluate the propagator by explicitly performing the p-integration in eq. (22)? Why is this calculation doomed to failure?

If we try to compute the propagator of the one-dimensional harmonic oscillator using the same technique as the one employed in part (a), then we would have to evaluate it P 2 p exp + 1 mω2X2 x′ . (27) h | − ~ 2m 2 | i If we could write it P 2 itP 2 itmω2X2 exp + 1 mω2X2 =? exp exp , (28) − ~ 2m 2 −2m~ − 2~ then it would follow that it P 2 it p2 p exp + 1 mω2X2 x′ =? exp + 1 mω2x′ 2 p x′ , (29) h | − ~ 2m 2 | i − ~ 2m 2 h | i after employing the eigenvalue equations for P and X.

8 However, eqs. (28) and (29) are both false. In particular, eq. (28) would be correct if and only if P 2 and X2 were commuting operators. But, we know that X,P = i~I, so that 2 2 2 2 X ,P = X X,P + X,P X = XP X,P +X X,P P + P X,P X + X,P PX =2i~(XP +PX). (30) Thus, X2,P 2 = 0 (nor is it a c-number). Hence, to evaluate the matrix exponential on the 6 left hand side of eq. (28), one must employ the Zassenhaus formula for matrix exponentials,4 exp t(A + B) = etAetB exp 1 t2 A, B exp 1 t3 2 B, [A, B] + A, [A, B] , − 2 6 ··· where the exponents of higher order in t involve nested commutators. Consequently, the right hand side of eq. (28) must be corrected by multiplying by a complicated product of exponentials. Hence, it is not possible to simply replace the operators P and X with their corresponding eigenvalues in eq. (27). In fact, one can surmount these difficulties by a clever manipulation of eq. (27). The first step is to divide up the interval from x to x′ into N equal segments. In the limit of infinitely large N, each segment is infinitesimally small. By inserting N complete sets of position eigenstates and corresponding momentum eigenstates and then formally taking N , one can develop the so-called path integral representation of the propagator. → ∞ The resulting path integral can then be evaluated exactly in the case of the harmonic oscillator. Since we do not have time in this course to pursue this formalism, we shall seek another technique for computing the propagator of the one-dimensional harmonic oscillator. This is the goal of the next two parts of this problem.

NOTE ADDED:

Perhaps I am being a little too hasty in stating that a direct evaluation of eq. (27) is doomed to failure. Indeed, a direct evaluation has been carried out in F. A. Barone, H. Boschi-Filho and C. Farina, Three methods for calculating the Feynman propagator, Amer- ican Journal of Physics 71, 483 (2003). In this paper, the authors provide three different methods of evaluation. The first method corresponds to the method you will employ in parts (c) and (d) of this problem. The third method employs the path integral technique alluded to above. But the second method makes use of the remarkable fact that the exponential in eq. (27) can be expressed as the following product of exponentials, it P 2 exp + 1 mω2X2 = eiωt/2 exp( αX2) exp( βP 2) exp(ωtPX/~) exp(βP 2) exp(αX2) , − ~ 2m 2 − − (31) 4The Zassenhaus formula for matrix exponentials is sometimes referred to as the dual of the Baker- Campbell Hausdorff formula. The latter provides a formula for eAeB in terms of a single matrix exponential, 1 whose argument is an infinite series of terms, A+B + 2 A, B + , where succeeding terms consist of nested commutators. For further details, see, e.g., Fernando Casas, An···der Nurua and Mladen Nadinic, Efficient computation of the Zassenhaus formula, Comp. Phys. Commun. 183, 2386 (2012).

9 where mω 1 α , β . ≡ 2~ ≡ −4m~ω Using eq. (31), it is now possible to evaluate eq. (27), with the help of the relation, p′ exp(ωtPX/~ p = e−iωtδ(p′ e−iωtp) . (32) h | | i − (Can you prove the above result?) As a challenge to the reader, try evaluating eq. (27) using the results of eqs. (31) and (32) and then derive the propagator for the harmonic oscillator using eq. (22).

(c) Show that G(x, t; x′, 0) = x, t x′, 0 , where the x, t are basis states in the Heisenberg h | i | i representation. Deduce the following diﬀerential equation for G, ∂G i~ = x, t H x′, 0 , ∂t h | | i where the boundary condition at t = 0 is G(x, 0; x′, 0) = δ(x x′). −

Recall that for basis states in the Heisenberg representation, x , t = eiHt/~ x , (33) | i | i assuming that the Hamiltonian H is time-independent. It immediately follows that x , t | i satisﬁes the wrong-sign Schr¨odinger equation, ∂ i~ x , t = H x , t . (34) ∂t | i − | i For the corresponding bras, we take the adjoint of eqs. (33) and (34) to obtain, x , t = x e−iHt/~ , (35) h | h | and ∂ i~ x , t = x , t H, (36) ∂t h | h | where we have used the fact that H is self-adjoint. Eq. (50) then yields, G(x, t; x′, 0) x e−iHt/~ x′ = x , t x′ , 0 , ≡h | | i h | i after using eqs. (33) and (35). Finally, using eq. (36), it immediately follows that ∂G i~ = x, t H x′, 0 . (37) ∂t h | | i The relevant boundary condition at t = 0 is G(x, 0; x′, 0) = x , 0 x′ , 0 = x x′ = δ(x x′) . (38) h | i h | i −

10 (d) Evaluate the propagator for the one-dimensional harmonic oscillator by employing the following steps. First, by using the Heisenberg equations of motion, express P P (0) ≡ in terms of X(t) and X X(0). Then solve the diﬀerential equation obtained in part (c), subject to the boundary≡ condition at t = 0.

The one-dimensional harmonic oscillator Hamiltonian is given by P 2 H = + 1 mω2X2 . (39) 2m 2 To solve for the harmonic oscillator propagator, we shall solve the diﬀerential equation given by eq. (37). Since H is time-independent, H(t)= H(0), and we may employ eq. (39), where P P (0) and X X(0). Our strategy is to rewrite P (0) in terms of the Heisenberg picture ≡ ≡ operators X(t) and X X(0). To accomplish this, we make use of the Heisenberg equations of motion, ≡ dP (t) i dX(t) i = P (t) , H(t) , = X(t) , H(t) . dt −~ dt −~ The commutators are easily evaluated: P 2(t) P (t) , H(t) = P (t) , + 1 mω2X(t) = i~mω2X(t) , 2m 2 − P 2(t) P (t) X(t) , H(t) = X(t) , + 1 mω2X(t) = i~ . 2m 2 m That is, the Heisenberg equations of motion are given by, dP (t) dX(t) P (t) = mω2X(t) , = . dt − dt m Taking the derivative of dX(t)/dt with respect to time and using the equation for dP (t)/dt yields d2X(t) = ω2X(t) . dt − The solution to this equation is P X(t)= sin ωt + X cos ωt , (40) mω where P P (0) and X X(0). ≡ ≡ We can now express P in terms of X(t) and X, mω P = X(t) X cos ωt . (41) sin ωt − We would like to insert this into eq. (39). In computing P 2, it is important to remember that X(t) does not commute in general with X(0). In particular, in light of eq. (40), P sin ωt i~ sin ωt X,X(t) = X, sin ωt + X cos ωt = X,P ]= . mω mω mω 11 Thus, squaring eq. (41) yields

m2ω2 P 2 = X2 + X2 cos2 ωt cos ωt X(t)X + XX(t) sin2 ωt − m2ω2 i~ sin ωt cos ωt = X2 + X2 cos2 ωt 2X(t)X cos ωt . sin2 ωt − − mω Inserting this result into eq. (39), we end up with

mω2 i~ sin ωt cos ωt H = X2(t)+ X2 cos2 ωt 2X(t)X cos ωt + 1 mω2X2 . (42) 2 sin2 ωt − − mω 2 We can now evaluate x , t H x′ , 0 by employing the eigenvalue equations, h | | i X(t) x , t = x x , t , X x , 0 = X(0) x , 0 = x x , 0 . | i | i | i | i | i It then follows that mω2 i~ sin ωt cos ωt x , t H x′ , 0 = x2 + x′ 2 cos2 ωt 2xx′ cos ωt + 1 mω2x′ 2 x , t x , 0 h | | i 2 sin2 ωt − − mω 2 h | i mω2 i~ω cos ωt = x2 + x′ 2 2xx′ cos ωt x , t x , 0 2 sin2 ωt − − 2 sin ωt h | i Plugging this result into eq. (37) yields the following diﬀerential equation,

∂G imω2 ω cos ωt = − x2 + x′ 2 2xx′ cos ωt G, (43) ∂t 2~ sin2 ωt − − 2 sin ωt subject to the boundary condition speciﬁed by eq. (38). Obtaining the solution to eq. (43) is straightforward. Integrating over t, and making use of dt 1 cos ωt 1 cos ωt 1 dt = cot ωt , dt = , dt = ln sin ωt , sin2 ωt −ω sin2 ωt −ω sin ωt sin ωt ω | | Z Z Z we end up with

imω (x2 + x′ 2) cos ωt 2xx′ G(x, t ; x′, 0) = g(x, x′) exp − 1 ln sin ωt , ~ 2 ( 2 sin ωt − | |) where g(x, x′) is a function of x and x′ to be determined. We can rewrite G(x, t ; x′, 0) in the following form,

2 ′ 2 ′ ′ ′ −1/2 imω (x + x ) cos ωt 2xx G(x, t ; x , 0) = g(x, x ) sin ωt exp ~ − . (44) | | ( 2 sin ωt )

12 We now impose the boundary condition specified by eq. (38). However, we cannot just set t = 0 in eq. (44), since the resulting expression is not well defined. Instead, we shall consider the limit as t 0+ and impose the condition that5 → lim G(x, t ; x′, 0) = δ(x x′) . t→0+ − Since cos ωt 1 and sin ωt ωt as t 0, it follows that ≃ ≃ → im(x x′)2 δ(x x′) = lim g(x, x′)(ωt)−1/2 exp − . (45) − t→0+ 2~t Eq. (45) should be compared with the following representation of the delta function6 1 (x x′)2 δ(x x′) = lim exp − , (46) − ∆→0 (π∆2)1/2 − ∆2 If we define ∆2 2i~t/m, then we can therefore identify7 ≡ mω 1/2 g(x, x′)= . (47) 2πi~ Plugging this result back into eq. (44), we end up with,

−1/2 2 ′ 2 ′ ′ mω imω (x + x ) cos ωt 2xx G(x, t ; x , 0) = ~ exp ~ − , (48) 2πi sin ωt ( 2 sin ωt ) assuming that 0 < t < π/ω, in which case we can drop the absolute value signs from the factor sin ωt that appears in the prefactor of the exponential in eq. (44).8 | |

(e) Check that in the limit of ω 0, the result of part (d) reduces to the free particle propagator obtained in part (a). →

If one evaluates eq. (48) in the limit of ω 0, the end result is, → m 1/2 im(x x′)2 G (x, t ; x′, 0) = exp − , (49) 0 2πi~t 2~t which is the expression for the free-particle propagator previously obtained in eq. (26). 5The t 0 limit should be taken such that t = 0 is approached from the positive side (this is the meaning of the symbol→ t 0+), since one is typically interested in propagation forward in time, i.e. t> 0. 6See e.g., eq.→ (1.10.19) of Ramamurti Shankar, Principles of Quantum Mechanics, 2nd edition, p. 61. 7Strictly speaking, these steps are valid only for Re ∆2 > 0. When ambiguities of this nature arise in quantum mechanics, the safest procedure is to add an infinitesimal real part to ∆2 in such a way that convergence is assured. At the end of the calculation, this infinitesimal quantity can be taken to zero. 8The result given in eq. (48) is ambiguous, since it is not clear how to interpret i−1/2 in the prefactor. One can show that the naive choice, i−1/2 = e−iπ/4, is correct only for 0 < t < π/ω. This ambiguity is associated with the fact that sin ωt =0 at t = π/ω which means that the propagator diverges at this point (called the first caustic). For values of t > π/ω, an additional phase factor arises that cannot be fixed by the limiting procedure of eqs. (45)–(47). A simple way of deriving this phase factor is given in Nora S. Thornber and Edwin F. Taylor, American Journal of Physics, 66, 1022 (1998). You can also find references there to the original literature and references to the few textbooks that address this phase ambiguity.

13 4. The partition function is deﬁned by,

Z(β) Tr e−βH , ≡ where H is a time-independent Hamiltonian operator and β is a real positive parameter.

(a) In the case of a one-dimensional quantum mechanical problem, show that

Z(β)= G(x, i~β ; x, 0) dx , − Z where the propagator is deﬁned by,

G(x, t; x′, 0) x e−iHt/~ x′ . (50) ≡h | | i

Evaluating the trace in a coordinate basis,

Z(β) Tr e−βH = d3x ~x e−βH ~x = d3xG(~x, t; ~x, 0) , ≡ h | | i Z Z t=−i~β

which yields the result quoted in eq. (50).

(b) Show that the ground state energy E0 is given by: 1 ∂Z E0 = lim . β→∞ −Z ∂β

One can also compute Z(β) by evaluating the trace with respect to a basis of energy eigenstates. Assuming that the basis states are normalized to unity, E E = δ , where h m| ni mn H E = E E , it follows that | ni n | ni Z(β) Tr e−βH = E e−βH E = e−βEn E E = e−βEn , (51) ≡ h n| | ni h n| ni n n n X X X where I have adopted a convention in which E E for m < n. Henceforth, we will assume m ≤ n that the ground state energy E0 is non-degenerate. Hence, eq. (51) yields

−βEn En e βE0 βE1 βE2 1 ∂Z n E0 e + E1 e + E2 e + = X = ··· −Z ∂β −βEn eβE0 + eβE1 + eβE2 + e ··· n X E + E eβ(E1−E0) + E eβ(E2−E0) + = 0 1 2 ··· , 1+ eβ(E1−E0) + eβ(E2−E0) + ··· 14 where E0 is the (non-degenerate) ground state energy as previously noted. This means that E E > 0 for n =1, 2, 3,.... Taking the limit as β , it follows that e−β(En−E0) 0 n − 0 →∞ → for all n =1, 2, 3,.... Consequently, 1 ∂Z E0 = lim . (52) β→∞ −Z ∂β

REMARK: Using eq. (51), one can immediately establish the following result, ∞ 1 e−βEZ(β)dβ = . E + E 0 n n Z X However, in the case of the harmonic oscillator, both sides of this relation are divergent. Thus, this relation does not provide a method for computing the energy levels of the harmonic oscillator. Nevertheless, all is not lost as we now examine in parts (c) and (d) of this problem.

(c) Using the results of part (b) of this problem and part (d) of Problem 3, compute the ground state energy of the one-dimensional harmonic oscillator.

One can now evaluate Z(β) as follows. First, we set x′ = x and t = i~β in the expression for G(x, t ; x′, 0) given in eq. (48). Using sin(iz)= i sinh z and cos(iz−) = cosh z, we obtain,9

mω −1/2 mωx2(cosh β~ω 1) G(x, i~β ; x, 0) = exp − . (53) − 2π~ sinh β~ω − ~ sinh β~ω It follows that ∞ 1 2 1/2 Z(β)= dxG(x, i~β ; x, 0) = . − 2 cosh β~ω 1 Z−∞ − In light of the identity, sinh2(z/2) = 1 (cosh z 1), we can rewrite Z(β) as 2 − 1 Z(β)= , (54) 1 ~ 2 sinh 2 β ω which yields, 1 ∂Z = 1 ~ω coth 1 β~ω . −Z ∂β 2 2 Finally, we use eq. (52) to compute the ground state energy.

1 ∂Z 1 E0 = lim = ~ω , β→∞ −Z ∂β 2 as expected for the ground state energy of the harmonic oscillator. 9In contrast to eq. (48), no phase ambiguities arise in the derivation of eq. (53). Indeed, since β > 0, the prefactor is equal to the positive inverse square root of a positive quantity.

15 (d) Consider the the propagator for a one-dimensional quantum system governed by a time-independent Hamiltonian with only discrete (bound state) energy levels, E , for { n} n =0, 1, 2, 3,.... Using eq. (50), show that the full energy spectrum of H can be determined from10

∞ ∞ Tr e−iHt/~ = dxG(x, t ; x, 0) = E e−iHt/~ E = e−iEnt/~ . (55) h n| | ni −∞ n n=0 Z X X Strictly speaking, the sums on the right-hand side of eq. (57) are not convergent. However, one can give mathematical meaning to these sums by extending the time parameter to the complex plane. In particular, let t = i~β (where β is a positive real parameter). Using eq. (57), obtain all the energy eigenvalues− of the one-dimensional harmonic oscillator.

In analogy with part (a) of this problem, we can write Tr e−iHt/~ in two diﬀerent ways. With respect to the coordinate basis,

∞ ∞ Tr e−iHt/~ = dx x e−iHt/~ x = G(x, t ; x, 0) . h | | i Z−∞ Z−∞ where we have made use of eq. (50). Alternatively, we can compute the trace with respect to the energy eigenstate basis,

Tr e−iHt/~ = E e−iHt/~ E = e−iEnt/~ , h n| | ni n n X X under the assumption that the energy eigenstates are normalized to unity. Hence, eq. (55) is established. If we now put t = i~β, then we obtain − Z(β) Tr e−βH = e−βEn , (56) ≡ n X which is a result that was previously obtained in part (b) [cf. eq. (51)]. In part (c) we derived an explicit formula for Z(β) [cf. eq. (54)]. We can rewrite this result in the following way:

−β~ω/2 ∞ ∞ 1 1 e ~ ~ 1 ~ Z(β)= = = = e−β ω/2 e−βn ω = e−β(n+ 2 ) ω . 1 ~ eβ~ω/2 e−β~ω/2 1 e−β~ω 2 sinh 2 β ω n=0 n=0 − − X X Comparing with eq. (56), we conclude that the energy levels of the harmonic oscillator are

~ 1 En = ω(n + 2 ) . 10In eq. (57), the trace is expressed as a diagonal sum of matrix elements by employing two diﬀerent basis choices (the coordinate basis and the energy basis, respectively). Of course, the trace is a basis-independent quantity, so one may choose any orthonormal basis to compute it.

16 5. Consider a particle in one dimension trapped between two impenetrable walls at x = 0 and x = L.

(a) Determine the bound state energy levels, En, of the particle. (Here, n labels the possible energy eigenvalues: n = 1 is the ground state, n = 2 is the ﬁrst excited state, etc.).

We solve the time-independent Schr¨odinger equation, ~2 d2ψ(x) Hψ(x)= = Eψ(x) , (57) −2m dx2 subject to the boundary conditions, ψ(0) = ψ(L) = 0, which are a consequence of the two impenetrable walls at x = 0 and x = L. As usual, we deﬁne

2mE k = , ~2 r where k > 0. Then, the solution to eq. (57) is

A sin kx , for 0 x L, ψ(x)= ≤ ≤ 0 , for x > L and x< L, ( − after imposing the boundary condition ψ(0) = 0. We next impose the boundary condition ψ(L) = 0, which yields,

kL = nπ , for any positive integer n. (58)

Note that since k > 0, n cannot be negative, and n = 0 is not allowed since this would correspond to ψ(x) = 0, which is not an eigenfunction of H. Eq. (58) then yields, ~2k2 π2~2n2 E = = . (59) n 2m 2mL2

The normalization constant A is easily determined by setting

L 1 2 nπx 1 2 = sin dx = 2 L, A 0 L | | Z which yields A = 2/L after setting the arbitrary phase factor to unity. It is convenient to write the expressions for the energy eigenstates with the help of the Heavyside step function. p Then, the eigenfunction corresponding the the nth energy level is given by

2 nπx ψ (x)= sin [Θ(x) Θ(x L)] , (60) n L L − − r where Θ(x) is the Heavyside step function, and the quantity Θ(x) Θ(x L) in eq. (60) − − ensures that the wave function is zero beyond the two impenetrable walls at x = 0 and x = L.

17 (b) Suppose that at time t = 0, the state of the particle is given by the wave function

ψ(x, t =0)= Ax(L x) [Θ(x) Θ(x L)] , − − − where Θ(x) is the Heavyside step function and A is a normalization constant. If an energy measurement is performed at time t = 0, what is the probability that the particle will be observed to be in the ground state? Find an exact expression for the probability that the particle will be observed to be in a state of energy En (for any positive integer n).

The normalization constant is determined in the usual way,

1 L L5 = x2(L x)2dx = L2 1 1 + 1 = . A 2 − 3 − 2 5 30 | | Z0 Hence, after setting the the arbitrary phase factor to unity,

30 A = 2 . rL Thus, 30 ψ(x, t =0)= 2 x(L x) [Θ(x) Θ(x L)] , (61) rL − − − Next, we expand ψ(x, t = 0) in terms of energy eigenstates,

∞

ψ(x, 0) = cnψn(x) , (62) n=1 X where the energy eigenstates (which are a complete set of states), are given by eq. (60). Using the fact that the ψ (x) are orthonormal, { n } L ∗ ψn(x)ψm(x) dx = δnm , Z0 ∗ we can multiply eq. (62) on the left by ψn(x) and integrate to project out the coeﬃcients cn. Thus,

L 1/2 1/2 L ∗ 30 2 πnx cn = ψn(x)ψ(x, 0) dx = 5 x(L x) sin dx . 0 L L 0 − L Z Z after the change of variables, y = πx/L,

2√15 π y cn = 2 y 1 sin(ny)dy . π 0 − π Z 18 Employing the integrals, π π y sin(ny)dy = ( 1)n , − n − Z0 π π2 2 y2 sin(ny)dy = ( 1)n + ( 1)n 1 , − n − n3 − − Z0 we end up with 4√15 c = 1 ( 1)n . n π3n3 − − We conclude that the probability of ﬁnding the particle in the state with energy En, denoted by Pn, is given by 240 P = c 2 = 1 ( 1)n]2 . n | n| π6n6 − − The ground state corresponds to n = 1. Thus, the probability that the particle will be observed to be in the ground state is 960 P = =0.9986 . 1 π6 This means that ψ(x, 0) is a pretty good approximation to the ground state energy eigenfunction!

(c) Evaluate the expectation value of the Hamiltonian with respect to the wave function given in eq. (61). What is the average value of the energy at time t = 0?

The average energy is given by the expectation value of the Hamiltonian. Thus, in light of eq. (57), ~2 L d2ψ(x) E = ψ H ψ = ψ∗(x) dx . h i h | | i −2m dx2 Z0 Plugging in eq. (61), 30 L ~2 d2 30~2 L 5~2 E = x(L x) x(L x)dx = x(L x)dx = . (63) h i L5 − −2m dx2 − mL5 − mL2 Z0 Z0

WARNING : The calculation presented above is too naive, although the end result turns out to be correct. Note that starting from eq. (61), it follows that

dψ 30 = (L 2x)[Θ(x) Θ(x L)] + x(L x)[δ(x) δ(x L)] , (64) dx L2 − − − − − − r 19 after using δ(x) = dΘ(x)/dx. Luckily, the second term on the right hand side of eq. (64) does not contribute, since xδ(x)=0 and(L x)δ(x L) = 0. Hence, − − dψ 30 = (L 2x) [Θ(x) Θ(x L)] . dx L2 − − − r However, when we take the second derivative,

d2ψ 30 = 2[Θ(x) Θ(x L)]+(L 2x)[δ(x) δ(x L)] , dx2 L2 − − − − − − r we cannot immediately discard the terms proportional to the delta functions. Indeed, (L − 2x)δ(x)= Lδ(x) and (L 2x)δ(x L)= Lδ(x L), Hence, − − − − d2ψ 30 = 2 [Θ(x) Θ(x L)] + L [δ(x)+ δ(x L)] , dx2 L2 − − − − r When we compute E more carefully, we obtain h i ~2 L d2ψ(x) E = ψ∗(x) dx h i −2m dx2 Z0 15~2 ∞ = x(L x) [Θ(x) Θ(x L)] 2 [Θ(x) Θ(x L)] + L [δ(x)+ δ(x L)] . −mL5 − − − − − − − Z−∞ Since an overall factor of x(L x) appears in the integrand, we can use it to kill the delta function terms, i.e., x(L x)[δ−(x)+ δ(x L)] = 0. It then follows that − − 30~2 L 5~2 E = x(L x)dx = . h i mL5 − mL2 Z0 The reason I mention all this is that keeping track of the terms that arise by taking multiple derivatives of the Heavyside step function will sometimes matter. In particular, if you were to calculate E2 = ψ H2 ψ , the naive calculation of the integrand would give zero, but the correcth calculationi h | would| i yield terms proportional to the derivative of the delta function, which contributes to the ﬁnal result. Indeed, the correct calculation yields E2 = 30~4/(m2L4). (Try it!) h i

(d) The expectation value of H can be computed by a diﬀerent method than the one used in part (c). First, expand ψ(x, 0) as a linear combination of energy eigenstates, and then show that the expectation value of H can be expressed as an inﬁnite sum. Using this technique, obtain an expression for the average value of the energy at time t = 0, and then employ the result obtained in part (c) to determine the value of the sum,

∞ 1 . (2n + 1)4 n=0 X 20 We can also compute E by inserting a complete set of energy eigenstates, h i ∞ E = ψ H ψ = ψ H E E ψ = E ψ E E ψ = E c 2 , (65) h i h | | i h | | nih n| i n h | nih n| i n| n| n n n=1 X X X after making use of eq. (62) and identifying c = E ψ . n h n| i In part (b), we computed

960 240 , for n odd, c 2 = 1 ( 1)n]2 = π6n6 | n| π6n6 − −   0 , for n even.  Inserting this result into eq. (65) and making use of eq. (59), we end up with

480~2 1 E = 4 2 4 , h i π mL n>0 n nXodd which we can rewrite as 480~2 ∞ 1 E = . (66) h i π4mL2 (2n + 1)4 n=0 X Comparing eqs. (63) and (66), we conclude that

∞ 1 π4 = . (67) (2n + 1)4 96 n=0 X

ADDED NOTE:

Sums such as the one given in eq. (67) are related to the famous Riemann zeta function,11

∞ 1 1 ∞ 1 ζ(z)= = ( 1)n+1 , nz 1 21−z − nz n=1 n=1 X − X where the ﬁrst sum is valid for Re z > 1 and the second sum is valid for Re z > 0 (assuming that z = 1 where the Riemann zeta function diverges). It then follows that 6 1 1 ∞ 1 1 ∞ 1 = + ( 1)n+1 = (1 2−z)ζ(z) , for Re z > 1. nz 2 nz 2 − nz − n>0 n=1 n=1 nXodd X X 11See e.g., I.S. Gradshteyn and I.M. Ryzhik, Table of Integral, Series and Products, 8th edition (Academic Press, Waltham, MA, 2015) pp. 1046–1049.

21 It is well known that ζ(4) = π2/90. Hence, 1 π4 4 = . n>0 n 96 nXodd in agreement with eq. (67).

(e) After preparing the state given by eq. (61) at time t = 0, suppose that instead of performing an energy measurement at time t = 0, I wait a while and then make the ﬁrst energy measurement at a later time t> 0. Do any of the results obtained in parts (b) and (c) change? Explain.

The results of parts (b) and (c) do not change as a function of time. To prove this, we ﬁrst note that eq. (60) is equivalent to

ψ(0) = c E , | i n | i n X when evaluated with respect to the coordinate basis. Multiplying on the left by E and h m| using E E = δ , it immediately follows that c = E ψ(0) . Next, we recall that h m| ni mn n h n| i ψ(t) = U(t) ψ(0) = U(t) E E ψ(0) = c e−iEnt/~ E (68) | i | i | nih n| i n | ni n n X X where we have used the fact that the time evolution is given by U(t) = exp iHt/~ for a time-independent Hamiltonian, and H E = E E . It immediately follows− that | ni n | ni ∞ −iEnt/~ ψ(x, t)= cne ψn(x) , (69) n=1 X after evaluating eq. (68) with respect to the coordinate basis. From eq. (69), we conclude that the probability of a state with wave function ψ(x, t) to be observed in an energy eigenstate with energy En at time t is given by c e−iEnt/~ 2 = c 2 , | n | | n| which is indeed time-independent. To see that E is also time-independent, we make use of Ehrenfest’s equation, h i d i ∂Ω Ω = [H, Ω] + . dt h i ~ ∂t

Setting ω = H, it follows that d E =0 , dt h i for any time-independent Hamiltonian. This is the quantum mechanical statement of energy conservation.

22 6. Consider a periodic potential in one-dimension which satisﬁes V (x + ℓ)= V (x).

(a) Show that the translation operator T = exp( iℓP/~) commutes with the Hamilto- − nian: P 2 H = + V (x) . (70) 2m

We ﬁrst note that [P 2 , T ] = 0, since P commutes with any function of P . Thus, we focus on [T,V (X)]. Consider the action of this commutator on an arbitrary basis state x , | i [T,V (X)] x = TV (X) V (X)T x = V (x) V (X) T x , (71) | i − | i − | i where we have used V (X) x = V (x) x , where X x =x x . Using the result of problem 5(c) on Problem Set 1 (with| iP = ~K),| iti follows that| i | i

T x = exp( iℓP/~) x = x + ℓ . (72) | i − | i | i Since V (X) x + ℓ = V (x + ℓ) x + ℓ , it then follows from eqs. (71) and (72) that | i | i [T,V (X)] x = V (x) V (x + ℓ) x + ℓ =0 , (73) | i − | i where we have used the periodicity of V(x) in the ﬁnal step above. Since eq. (73) is true for an arbitrary basis state x , it follows that [T,V (X)] = 0. | i

(b) We may choose the energy eigenstates to be simultaneous eigenstates of the translation operator T . Show that the general form of such eigenfunctions is:

ψ(x) = exp(ipx/~)up(x) . where up(x+ℓ)= up(x). That is, the eigenfunctions are plane waves modulated by a function with the periodicity of the potential. Consider the eigenvalue problem for T ,

T ψ = t ψ . (74) | i | i To determine the possible values of t, examine eq. (74) in the p-representation. Since

p T ψ = p exp( iℓP/~) ψ = e−iℓp/~ p ψ , h | | i h | − | i h | i we can conclude that t = e−iℓp/~ . (75)

Next, we consider the eigenvalue equation for T in the x-representation. Using eqs. (74) and (75), x T ψ = e−iℓp/~ x ψ . (76) h | | i h | i 23 In light of eq. (72), T x = x + ℓ . It follows that x T = x ℓ . To verify this last assertion,12 consider | i | i h | h − | x T x′ = x x′ + ℓ = δ(x x′ ℓ) , h | | i h | i − − when we act with T to the right. If we act with T to the left we will get the same result if x T = x ℓ , namely, h | h − | x T x′ = x ℓ x′ = δ(x x′ ℓ) , h | | i h − | i − − Consequently, eq. (76) yields

x ℓ ψ = e−iℓp/~ x ψ . h − | i h | i which we can rewrite as ψ(x ℓ)= e−iℓp/~ψ(x) . (77) − Although ψ(x) is not a periodic function, it is straightforward to check that e−ipx/~ψ(x) is periodic, since the equation,

e−i(x−ℓ)p/~ψ(x ℓ)= e−ipx/~ψ(x) , − is equivalent to eq. (77). Thus, we introduce the notation,

u (x) e−ipx/~ψ(x) . p ≡ Then, ipx/~ ψ(x)= e up(x) , where up(x + ℓ)= up(x). This result is known as Bloch’s Theorem. The description of ψ(x) as a plane wave modulated by a function that possesses the periodicity of the potential V (x) is called a Bloch wave.

12Another way to derive x T = x ℓ is as follows. The action of the operator T on the ket x is T x T x . Likewise,h the| actionh − of the| operator T on the bra x is x T T †x . Since P| isi | i ≡ | i h | h | ≡ hermitian, it follows that T † = exp(iℓP/~). Thus T †x = T † x = x ℓ and the adjoint of T †x is | i | − i T †x = x T = x ℓ .

h | h − |

24 APPENDIX: An alternative solution to Problem 6

Since V (x + ℓ)= V (x), we can expand V (X) in a Fourier series, ∞

V (X)= cn exp(2πinX/ℓ) . n=−∞ X To compute TV (X), where T = exp( iℓP/~), we need to calculate the product of two − exponentials. Here we make use of the identities, A B 1 e e = exp A + B + 2 [A, B] , (78) eBeA = expA + B 1 [A, B] , − 2 under the assumption that A, [A, B] = B, [A, B] = 0. Using [X,P ]= i~I, iℓP 2πinX TV (X)= c exp + iπn n − ~ ℓ − n X iℓP 2πin(X ℓ) = c exp + − + iπn n − ~ ℓ n X = V (X ℓ)T = V (X)T. − Hence, we have demonstrated that [T,V (X)] = 0. It then follows that [H , T ] = 0. Thus, we can choose ψ = E, t to be a simultaneous eigenstate of H and T . That is, H E, t = E E, t and T |E,i t =| t E,i t . Consequently, | i | i | i | i x HT E, t = tE x E, t , x TH E, t = E x ℓ E, t , h | | i h | i h | | i h − | i after noting [cf. footnote 12] that x T = x ℓ . Since HT = TH, it follows that h | h − | ψ(x ℓ)= tψ(x) , (79) − where ψ(x) x E, t . Finally, we compute t by evaluating p T E, t in two diﬀerent ways, ≡h | i h | | i p T E, t = t p E, t , h | | i h | i and p T E, t = p exp( iℓP/~) E, t = exp( iℓp/~) p E, t . h | | i h | − | i − h | i Subtracting the last two equations yields t = e−iℓp/~. Plugging this back into eq. (79) yields ψ(x ℓ)= e−iℓp/~ψ(x) , (80) − which reproduces eq. (77). The ﬁnal step is the same as in our previous solution. Namely, we introduce, u (x) e−ipx/~ψ(x) . p ≡ and show that eq. (80) implies that u (x ℓ)= u (x). Hence, we conclude as before that p − p ipx/~ ψ(x)= e up(x) , where up(x + ℓ)= up(x).