QM A The operator ˆR is defined by R ˆ ψ( x) = Re[ ψ( x)] (it returns the real part of ψ ( x) ). Is ˆR a linear operator? Explain. SOLUTION ˆR is not linear. It s easy to find a counterexample against the hypothesis of linearity: Investigate the function f, for which f( x) = i (it always returns i, no matter what x is). Now R ˆ f ( x ) = Re[ f ( x )] = Re[ i ] = but Rˆ i f( x) = Re[ i f( x)] = Re[ ] = irf ˆ ( x) =
QM A 3 Find the energy levels of a spin s = particle whose Hamiltonian is given by ˆ α H Sˆ Sˆ Sˆ β Sˆ = + x y z z, where α and β are constants. SOLUTION We rewrite the Hamiltonian: ({ } ) { } ˆ α ˆ ˆ ˆ β ˆ α ˆ ˆ ˆ β H S S S S S S S Sˆ α Sˆ 3Sˆ β = + = = Sˆ x y z z z z z z z We see that the Hamiltonian is diagonal in the sm, basis: s ( ˆ ˆ ) ˆ { α( ss 3 m ) βm} sm, s s s ˆ α β α β H sm, = S 3 S S sm, ss 3 m m sm, s z z = s + = s s s = + So ( 3 ) 5 3 4 E= α s s+ m βm = α m βm and thus 5 5 s s s s Em ( = ) = α 3 + β = α + β = 3α + β s Em ( =+ ) = α 3 β = α β = 3α β s 3 5 9 3 3 3 4 4 4 3 5 9 3 3 3 4 4 4 Em ( = ) = α 3 + β = α + β = 3α + β s Em ( =+ ) = α 3 β = α β = 3α β s 4 4 4 4 4 4
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QM B Consider a system which is initially in the normalized state ψθφ (, ) = Y ( θφ, ) + ay ( θφ, ) + Y ( θφ, ),,, 5 5 in which a is a positive real constant. a. Find a. b. If L were measured, what values could one obtain, and with what probabilities? z We now measure L and find the value. z c. Calculate L and L. x y d. Calculate the uncertainties L and L and their product L L. You may use the x y x y equality L = L without proving it first. x y SOLUTION Part a. Writing ψθφ (, ) = cy ( θφ, ) we require m= m, m m= c m =, in other words 3 3 + a + = + a = a = a= 5 5 5 5 5 Part b. The probability to measure Lz = m equals c m, so 5 Pm ( = ) = c = = % 3 5 Pm ( = ) = c = = 6% + 5 Pm ( =+ ) = c = = %
Part c. The wave function has collapsed to Y ( θφ, ) = θφ,,. We have, L = ( L + L) and L = il ( L), so now x + y + L = ( L + L) and L = il ( L) x + y + ψ L ψ =, L, +, L, = C,, +, = x + + ψ L ψ = i, L, + i, L, = ic,, + i, = y where C is some constant and is the zero-length ket (for which = ). Now for = and m =, we have L = ( + ) = = L + L + L = L + L + L = L + L = L +( ) x y z x y z x z x L = L = so that x y ( L ) = ( L L ) = L L = L =, and, similarly, L = x x x x x x y We find L L = ( )( ) = x y
QM B The Hamiltonian for a one-dimensional harmonic oscillator is ˆ pˆ H m x m = + ω ˆ. We write its energy eigenkets as n ( n =,,,... ) for energy E = ( n+ ) ω. a. Suppose the system is in the normalized state ϕ given by ϕ = c + c, and that the expectation value of the energy is known to be ω. What are c and c? i b. Now choose c to be real and positive, but let c have any phase: c = c e θ. Suppose further that not only is the expectation value of the energy known to be ω, but the expectation value of x is also known: ϕ x ϕ =. Calculate the phase angle θ. mω c. Now suppose the system is in the state ϕ at time t =, i.e., ψ( t = ) = ϕ. Calculate ψ () t at some later time t. Use the values of c and c you found in parts a. and b. d. Also calculate the expectation value of x as a function of time. With what angular frequency does it oscillate? Again, use the values of c and c you found in parts a. and b. n SOLUTION Part a. 3 c c 3 3 ω = ϕ H ϕ = c E + c E = c ω+ c ω = ω c + c + = Normalization: c + c =, so we have + = c + c = 3 c c c = c =
Part b. Recall xˆ = ( aˆ+ aˆ ) mω, so we have xˆ = aˆ+ aˆ mω = ϕ x ϕ = ϕ aˆ ϕ + ϕ aˆ ϕ = mω mω mω * * * * ( c c ) aˆ ( c c ) ( c c ) aˆ ( c c ) = + + + + + mω mω Now ( c ) caˆ c aˆ ϕ = aˆ c + = = ( ) ( ) ( (...) ) aˆ ϕ = aˆ c + c = caˆ + caˆ = c + = ϕ x ϕ = ϕ aˆ ϕ + ϕ aˆ ϕ = mω mω mω * = c + c mω * * * ( c ) c + c mω * * * + c + (...) = because c c + c c = mω mω c + c c = We get = * * = c + c = c + c = c e + c e = mω mω iθ iθ cosθ = θ = π( = 45 ) 4 (cos θ ) Part c. 3 iet / iet / i ω t iπ /4 i ω t ψ () t = ce + ce = e + e e Part d. We had seen ϕ ϕ x * * * Re c c c c = + = mω mω c c, so now
iωt 3 3 iπ iωt iωt iπ iωt 4 4 ψ( t) x ψ( t) = Re( e e e ) = Re( e e e ) = mω mω = cos( ωt+ π) = cos( ωt π) 4 4 mω mω The angular frequency of oscillation is ω. (from http://dfcd.net/articles/firstyear/solutions/s6-.pdf)
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E&M A An isolated sphere of perfectly conducting material is surrounded by air. Though normally a good insulator, air breaks down (it becomes conductive) for electric fields beyond 3. kv/mm (the so-called dielectric strength of air). The sphere s radius is 5. cm. What is the maximum amount of electrostatic energy the sphere can store before breakdown occurs? Assume the electrostatic potential is zero at infinite distance from the sphere. SOLUTION For a charge q on the sphere we have q ER E= k q= R k q V = k = ER R. ER 3 6 3 So now U = qv = ER = πε E R = π (8.85 )(3 ) (5 ) =.63 = 63 mj k
E&M A The diagram shows part of an electronic circuit. Calculate the potential at point P. SOLUTION Left two inductors: 3 = + = + = = L = 4. mh eff L L L 6 4 eff di V 8 ( ) From the mh inductor we find = = = 3 A/s 3 dt L di 3 So V = V = L = 4 3 = V V = V P eff P dt
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E&M B A flat square loop of wire of length a on each side carries a stationary current I. Calculate the magnitude of the magnetic field at the center of the square. SOLUTION This field is 4 times the field of one side. The field of one side is the field at a point a distance a away from wire. µ Idl rˆ µ µ a µ Ia dy db(for one side) = = Idysinφ = Idy = 4π r 4π r 4π y + a y + a 4 π ( y + a ) 3/ a a a Ia dy Ia dy Ia y Ia a 3/ 3/ / / π a ( y a ) π ( y a ) π a ( y a ) π + + + a ( a + a ) µ µ µ µ B(of one side) = = = = = 4 4 µ I µ I µ I = = = / π ( a ) π a π a B tot µ I µ I = 4 = π a π a The integral must be provided in the cheat sheet: EM Hard
E&M B All space is filled with a material with uniform, fixed magnetization M, except for the region < z< a, in which there is vacuum. The magnetization is M= Mu, ˆ where û is a unit vector in the yz plane that makes an angle θ with the z-axis: uˆ = (sin θ) yˆ + (cos θ) z ˆ. Calculate the magnetic field B and the auxiliary field H everywhere. SOLUTION There are no free current densities, so if there s a B field it must be due to bound currents. The bound volume current density J = M= b because M is uniform. The bound surface ± sinθ current density is K = M n ˆ = M sinθ M b = for the surface at z = (plus sign) cosθ ± and the surface at z= a (minus sign). It follows from symmetry and Ampère s law that these bound currents give rise to a uniform magnetic field B= ( µ M sin θ) y ˆ inside the gap; outside the gap, B=. B The auxiliary field follows from B= µ ( H+ M) H= M µ B ( µ M sin θ) () In the gap: H= M= y ˆ = ( M sin θ ) y ˆ µ µ B () Outside the gap: H= M= M= ( Msin θ) yˆ ( Mcos θ) z ˆ µ To verify this result we can use the magnetostatic charge density. For instance, for the bottom surface we have σ = M n ˆ = M = Mcosθ M z. From Gauss s law for magnetostatics, we have H d a = q. Applying this to a pillbox of area A enclosing part of the surface, calling the M,encl S auxiliary field inside/outside gap, H i / H o, we find H zˆa+ H ( z ˆ) A= Aσ H H = σ = Mcosθ. i o M i, z o, z M This agrees with our result H = ( M sin θ ) y ˆ i and H = ( Msin θ) yˆ ( Mcos θ) zˆ = H ( Mcos θ) z ˆ o i H H = ( M cos θ ) z i o ˆ
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