xp ih) px pxih = xppx xppx + pxih pxih = 0

Transcription

1 QM A1 Method (1) Use [ x, p] = xp px = i xp = px + i and px = xp i Then [ xp, px] = xppx px( xp) = xppx px( px + ih) = xppx ( px) px pxih = = xppx ( xp ih) px pxih = xppx xppx + pxih pxih = Method () ( xppx) f = x( ( xf ) ) = x( f + xf ) = x( f + f + xf ) = xf + x f ( pxxp) f = ( ( x f ) ) = xf + x f We see that ( xppx) f = ( pxxp) f xppx = pxxp [ xp, px ] = Method (3) Use px = xp i, and then: [ xp, px] = [ xp, xp i ] = [ xp, xp] [ xp, i ] = =

2 QM A Part a. One requirement for linearity is A( f + g) = Af + A g. Here, we have A( f + g) = f + g Af + Ag= f + g but these two are not equal for all functions f and g, so the operator is NOT linear. Counterexample: consider the constant functions f( x ) = 1 and gx ( ) = 1. A( f + g) = 1 + ( 1) = Af + Ag= = Part b. The adjoint O of operator O is defined by f O g = O f g. Therefore, if some operator O is selfadjoint ( O = O), it must satisfy f Og = O f g. Checking this requirement for operator B we note it doesn t satisfy this requirement: f Bg = f g*, but Bf g = f* g = g f** = g* f = f Bg * f Bg The two are not equal for all functions f and g, so the operator is NOT self-adjoint.

3 QM A3 A krypton fluoride (KrF) laser beam (wavelength λ = 48 nm ) hits a polished metal surface. As 5 a result, electrons leave the metal, the fastest ones having a speed of m/s. a. Find the energy per photon in the KrF laser beam, in ev. b. Find the work function Φ of the metal, in ev. ANSWERS Part a. E hc = = 5. ev e λe Part b. K e 1 melv = = 1.8 ev e Now K = E Φ Φ= E K = = 3. ev

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5 QM B1 Part a. First, we note S = S and S = S, because spin components are observables. From this we x x y y obtain S = ( S + is ) = S + ( is ) = S is = S ; hence, also S = S. Next, we note that + x y x y x y + ( ) SS = S S = SS, so the operator SS is self-adjoint. Alternative solution: S S = ( S + is )( S is ) = S ( S is ) + is ( S is ) = S S is S + is S + S S = + x y x y x x y y x y x x x y y x y y = S + S i[ S, S ] = S + S i( i S ) = S + S + S x y x y x y z x y z which, being a sum of self-adjoint operators (observables), is self-adjoint. Part b. From S = S ± is we find ± x y 1 1 S = 1 ( S + S ) = 1 + = + 1 x i S = is ( S) = i = = + i y 1 1 i Some students may double-check these matrices for self-adjointness: S S * T * x = S 1 = = = 1 1 * T * y Part c. i i i = S i = = = i i x y 1 a S ψ = ( + ) ψ ; take ψ = y b i i a = b a ib a a ib i, normalized: i b = = b b i 1 i 1 1 i i 1 1 i Check: = = i 1 i 1 1 Part d.

6 1 1 1 This probability is ( ) i Part e. The spin ket collapsed to,1 = 1 = = 5%. 1 1, so this probability is ( ) 1 1, =.

7 QM B Part a. We have ψ( x,) = c ϕ, with (ignoring the common factor A) unnormalized expansion n n n 3 coefficients c =, c =, c =, c =. 1 n> n Now c = + + = 4+ + = n E = ( n + 1 ) ω are: n The probabilities P to find energy n c 4 8 P = = = c + c + c 1 c 9 1 P = = = c + c + c 9 1 c 4 P = = = c + c + c (check: P + P1 + P = + + = 1; correct) and so E = PE + PE + PE = ( ω) + ( ω) + ( ω) = + + ( ω) = ω( 1.31 ω 1 1 ) Part b. The stationary states are also parity eigenstates: ˆ Pϕ = ηϕ = ( 1) n ϕ. So n n n n ˆ n P = Pη = ( 1) n n P = P P + P = + = = n Part c. n n The time evolution may be described as follows: time T 3 elapses 3 ( ) ( ) ( ) ( ) ( ) ( ) ψ( x,) = A ϕ + ϕ + ϕ ψ = ϕ + ϕ + ϕ 1 ( xt, ) B i 1 1 in which the numbers in boxes are the phase factors by which the individual expansion coefficients change as a result of this time evolution (relative to the -th expansion coefficient).

8 A phase factor potentially accumulated in the -th expansion coefficient may have been absorbed in the constant B. From the general time-evolution of the expansion coefficients, c ( T) = c () e = c () e = c () e iet n / in ( + 1/) ωt/ in ( + 1/) ωt n n n n we now deduce: c ( T) c () e c () 1 (5/) iωt iω T iωt iωt = = e = e e = i(1/) ωt ( ) () c () e c c T 1 and c ( T) c () e c () 1 = = e = e e = i c T (5/) iωt iω T iωt iωt (3/) ωt ( ) () 1 c () e c i 3 The ω angular frequency is slower than the ω frequency, so we find the earliest time for the former frequency: 3 iω π+ π T i( N ) 3 e = i= e T = + N ω π π T = 3 ω π

9 QM B3 Parts a. and b. Excluding the origin, the TISE is me ± κ x e ψ = ψ= κψ ψ The wavefunction must have zero asymptotes at x = ±, so we set ψ ( x) = κ x ψ x = Ae L x κ x ψ x = Be R x ( ) ( ) and start by requiring ψ ( x= ) = ψ ( x= ) A= B, so we now have L R ψ ( x) = κ x ψ x = Ae L x κ x ψ x = Ae R x ( ) ( ) The derivatives at the origin are ψ ( x= ) = κa L ψ ( x= ) = κa R ma With the given expression for the kink, lim ψ + ε ψ ε = ψ ε ( ) ( ) (), this leads to ma ma ( κ A ) ( κa) = A κ = kg.j.m kg.m m kg.m 1 1 (unit check: a is in J.m, so κ is in = = = m, correct) J s J.s m J.s m We can now find the energy of the ground state: E κ ma ma = = = m m Next, we normalize the wavefunction:

10 ψ ( x) κ x ψ x = Ae x L κ x ψ x = Ae x R = ( ) ( ) ψ ( x) = A e dx + A e dx = A e dx = A = κ κ x κx κx κx e x= x= κ x e 1 = A = A = 1 A= κ κ x= κ x= κ x ψ ( x) = κ e x So ground state is ψ ( x) = L κ ψ ( ) = κ x x e x R ma and its energy is E= E = (Check units: a is in J.m, so E is in kg J J s m kg m = = J, which is correct.) s Part c. x = ψ( xψ) dx = (odd integrand) Part d. = κ x x = ψ( x ψ) dx = x ψ dx = κ x e dx = κ 3 = = 4κ κ ma (the definite integral is in cheat sheet) 4 m a (Check units: a is in J.m, so x = ma Js Js Jsm is in = = = 4 m, which is correct.) kg J m kg m kg m Part e. ( x) = ( x x ) = x xx x x + x = x x = m a Part f. p = ψψ ( ) dx i, and the integrand is odd, so p =. 4

11 Or use integration by parts: p = ψψ ( ) dx = ψ ψ dx ψ = = = ( ) dx p p i i x i Part g. p = ψ( ψ ) dx = m ψ + δ ψ = + ψ E ( x) a ( x) ( x) dx m E a () = ma ma ma ma ma m E+ aκ = m + a = + = ma Check units: p = is in Part h. ma ( p) = p p = = = s s kg J m kg m kg m Js, correct. Part i. 4 ma 1 ( x ) ( p) = = ( x)( p) = m a

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13 E&M A1 Two forces act on the small sphere: gravitational F = mg and electrostatic g F = qe. e 9 σ.5 1 E = = = N/C, so F = = 1 e ε Fg = mg = = N N And so tanθ F e = = = θ = 5 Fg F qe qσ / ε tanθ = = = F mg mg g e E&M A

14 E&M A3

15 E&M A4

16 E&M B1 This solution is not in SI, most candidates will use SI.

17 E&M B TG (5/19/15): In my solution for E&M B, the answer I gave should have had a multiplicative factor of the height a (.5 m)included.

18 E&M B3

19 E&M B4 Part a. V = IR, V = IX = Iω L R L L d Part b. Z = R + ( X X ) = R + X = R + ω L L C L d Part c. The power dissipated in the lightbulb is PL ( ) = = rms rms Z R + ωdl The maximum power of 1 W is obtained for the minimum value of Z, so when L.. Part d. We require P I R Z R R Z P Z 1 P This implies rms rms R R 3 avg = rms = = = avg,max = avg,max. R Z = 1 Z = R + ( ωdl) = 3 R L= 3 R/ ωd = 3 =.58 H. π 6 Note that R may be calculated from rms rms (1) P = 1 W R 144 R = = P = 1 = Ω.

20 E&M PREVIOUS A1 Step 1 - Apply Gauss Law to a cylinder that is inside the conductor everywhere. Flux = so net charge is. The +7 uc/m on the central line thus attracts 7 uc/m on the inner surface. Total charge of conductor is 5 uc/m, so on outer surface there is 5 ( 7) = + uc/m. The area of the outer surface for one running meter of the cylinder is A = pi x (outer radius) x 1 =.157 m. The charge on this 1 meter is uc, so sigma = E-6 /.157 = 1.59E-5 C. Step - Again Gauss law with surface a cylinder of radius r and take its length L. Call charge per meter labda, this equals + uc/m Then pi r L E = L x (labda) / eps E(r) = (labda) / (eps pi r) = / r E(r =.5 m) = /.5 = 719 kn/c. Because net charge is positive, field points away from cylinder, perpendicular to its symmetry axis.