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L.J. Sham April 27, 2009
Physics 130A Sample Mid-Term Examination
Name:
Problem 1(a) 1(b) 1(c) 1(d) 1(e) 2 or 3 Total
Grade
Out of 10 10 10 10 10 50 100
Useful information:
- 1 eV = 1. 60 × 10 −^19 J.
- 1 meV is equivalent to 0.24 THz or 8 cm−^1 or 12 K.
- ℏ = 1. 05 × 10 −^34 J · s = 6. 58 × 10 −^16 eV · s.
- 1 nm = 10−^9 m
Do problem 1 and choose either problem 2 or problem 3.
1 Samples of Problem 1
- Short questions – Answer with brief reasoning:
(a) There was a recent discovery of a quantum theory from the Lost World which used a property p as that world’s primary measure, where p is a real number. The only thing we have in common is time from looking at a common sun. Their quantum theory had a wave function Φ(p, t) at time t, with a conjugate property, xˆ = i ⊼ (^) ∂p∂ where ⊼ was their famous Kcnalp constant, as the distance operator, and a Regnideorhcs equation, [ V (ˆx) + p
2 2 m
]
Φ(p, t) = i ⊼ (^) ∂t∂ Φ(p, t). (1)
Decipher their theory in our language.
(b) Let the orthonormal set of energy eigenstates of a Hamiltonian Hˆ be denoted by ψn(x) with associated energy eigenvalue En. It is given that the state at time t is,
Ψ(x, t) =
n
ψn(x)e−iEnt/ℏcn. (2)
i. Give an expression for the expectation value of the energy of the state at time t. ii. Give an expression for the corresponding energy uncertainty. iii. What is the probability of finding the state having the energy eigenvalue En? (c) Consider the lowest 100 energy eigen-states of a symmetric one-dimensional potential well, i.e., V (−x) = V (x). If an impenetrable wall blocks off half of the well, how many bound states are there in the new half-well within the same energy range, i. if the potential is an infinite square? ii. if the potential is otherwise unknown? (d) For a simple harmonic oscillator with orthonormal energy eigenstates denoted by ψn with n = 0, 1 , 2 ,.. ., a state at time t = 0 is given by,
Ψ(x, 0) =
1 2 (ψ^0 (x) +^ ψ^3 (x))^.^ (3)
i. Find the time dependence of the expectation value in momentum pˆ. ii. Find the time dependence of the uncertainty in momentum pˆ if the corresponding uncertainty of the momentum in the energy eigenstate ψn is given as ∆n for all n’s. (e) In Figure 1 shown below, V 1 (x) is a quadratic potential and V 2 (x) is a series of steps. Which has a lower energy ground state? Why?
Figure 1: Two potentials.
2 Samples of Problems 2 and 3
See homework problem sets.
Solution – One of the two possible explanations will suffice: i. p corresponds to our momentum if ⊼ = ℏ. The two theories are then completely equivalent. The assertions can be seen from the Fourier transform relation between their Φ(p, t) and our wave function Ψ(x, t),
Ψ(x) =
dp 2 πℏ
eipx/ℏΦ(p, t). (2)
ii. Their p corresponds our position x˜, where we distinguish our dynamical variables with a tilde. Their wave function is our wave function related to their position wave function by,
Ψ(˜x, t) ≡ Φ(p, t) =
dx e−ikx^ Φ(˜ x), (3)
which leads to our momentum being given by p˜ = iℏ (^) ∂∂ ˜x which leads to the defini- tion of momentum being negative our own convention.
(b) Let the orthonormal set of energy eigenstates of a Hamiltonian Hˆ be denoted by ψn(x) with associated energy eigenvalue En. It is given that the state at time t is,
Ψ(x, t) =
n
ψn(x)e−iEnt/ℏcn. (4)
i. Give an expression for the expectation value of the energy of the state at time t. ii. Give an expression for the corresponding energy uncertainty. iii. What is the probability of finding the state having the energy eigenvalue En?
Solution – It is easier to do part (iii) first, i. → (iii). The probability of finding the state with En is |cn|^2. ii. → (i). The energy expectation value is
〈H〉 =
n
|cn|^2 En, (5)
by the usual probability weighting argument for the average. iii. → (ii). Uncertainty in energy ∆E is given by
(∆E)^2 = 〈H^2 〉 =
n
|cn|^2 (En)^2 −
[
n
|cn|^2 En
] 2
(c) Consider the lowest 100 energy eigen-states of a symmetric one-dimensional potential well, i.e., V (−x) = V (x). If an impenetrable wall blocks off half of the well, how many bound states are there in the new half-well within the same energy range, i. if the potential is an infinite square well? ii. if the potential is otherwise unknown?
Solution – i. If the old potential is an infinite square well, the parity of the eigenstates alternate in increasing energy from even to odd and odd to even, etc. In the half-blocked potential, the odd parity states which vanish at x = 0 automatically satisfy the boundary condition to the unblocked side in the blocked well. Then there will 50 states in the new well. ii. If the potential is otherwise unknown, the argument and result are unchanged be- cause of the general symmetry argument.
(d) For a simple harmonic oscillator with orthonormal energy eigenstates denoted by ψn with n = 0, 1 , 2 ,.. ., a state at time t = 0 is given by,
Ψ(x, 0) =
1 2 [ψ^0 (x) +^ ψ^3 (x)]^.^ (7)
i. Find the time dependence of the expectation value in momentum pˆ. ii. Find the time dependence of the uncertainty in momentum pˆ if the corresponding uncertainty of the momentum in the energy eigenstate ψn is given as ∆n for all n’s.
Solution – i. 〈Ψ(t)|pˆ|Ψ(t)〉 will contain diagonal elements 〈ψn|pˆ|ψn〉, n = 0, 3 and the off- diagonal element 〈ψ 3 |pˆ|ψ 0 〉 and its complex conjugate. The diagonal elements are zero because pˆ is odd under mirror operation x → −x and the product of the wave function and its conjugate, be them both even or odd, is even in x. Since pˆ is proportional to c − c†^ of the annihilation and creation operators, cψ 0 = 0 will contribution nothing to the off-diagonal element 〈ψ 3 |pˆ|ψ 0 〉 and c†ψ 0 = ψ 1 will also contribute zero since ψ 1 is orthgonal to ψ 3. Thus, its complex conjugate is also zero. The expectation value of the momentum is zero at all times. ii. pˆ^2 is proportional to c^2 − 2 c†c − 1 + c†^2. With the similar composition of diagonal terms and off-diagonal terms as the last part, the off-diagonal terms are zero for the same reasons that these operator terms cannot connect ψ 0 and ψ 3. Since the diagonal terms contain no time dependent factors, they give at all times,
(∆p)^2 = 〈Ψ(t)|pˆ^2 |Ψ(t)〉 = ∆^20 + ∆^23 , or ∆p =
∆^20 + ∆^23. (8)
