Assignment 7: Solution

Quantum Mechanics I / Quantum Theory I. 14 December, 2015

1. In class, we derived that the ground state of the harmonic oscillator |0 is given in the position space as

1/4 mω 2 −mωx /2~ ψ0(x) = x|0 = e . π~ (a) This question deals with finding the state in the momentum basis. Let’s start off by finding the position operatorx ˆ in the momentum basis. Show that ∂ p|xˆ|ψ = i p|ψ . ~∂p This is analogous to the expression of the momentum in the position basis, ∂ x|pˆ|ψ = −i~ ∂x x|ψ . You will use [Beck Eq. 10.56] expressing the momentum eigenstate in the position basis as well as the definition of the Fourier transform.

∂ (b) Sincea ˆ|0 = 0 and the position operator p|xˆ = i~ ∂p , construct a diﬀerential

equation to ﬁnd the wavefunction in momentum space p|0 = ψf0(p). Solve this

equation to ﬁnd ψf0(p). Plot ψf0(p) in the momentum space.

ψ0(x).

Answer

(a) Lets ﬁnd the position operator in momentum space

p|xˆ|ψ = p|xˆ1ˆ|ψ Z = dx p|xˆ|x x|ψ Z

= dx p|x x x|ψ ∵ xˆ|x = x|x .

Now the normalized wavefunction of the momentum eigenstate in the position representation is 1 x|p = √ eipx/~ Eq 10.56 in Beck 2π~ Z 1 −ipx/ ∴ p|xˆ|ψ = √ dxe ~x x|ψ . (1) 2π~

Due Date: December. 14, 2015, 05:00 pm 1 Quantum Mechanics I / Quantum Theory I. 14 December, 2015

(b) We need to solve the lowering operatora ˆ in momentum space as rmω i p|aˆ|0 = p| xˆ + pˆ |0 2~ mω rmω i = p|xˆ|0 + p|pˆ|0 . 2~ mω ∂ Sincea ˆ|0 and the position operator p|xˆ = i~ ∂p , r mω d i p|aˆ|0 = i~ p|0 + p p|0 2~ dp mω r mω d i 0 = i~ p|0 + p p|0 2~ dp mω

Let ψf0(p) = p|0 be the wavefunction in momentum space, rmω d i 0 = i~ ψf0(p) + pψf0(p) 2~ dp mω d i i ψf0(p) = − pψf0(p) ∴ ~dp mω

dψf0(p) 1 = − p dp ψf0(p) ~mω

Due Date: December. 14, 2015, 05:00 pm 2 Quantum Mechanics I / Quantum Theory I. 14 December, 2015

Z Z dψf0(p) 1 = − p dp ψf0(p) ~mω p2 ln ψf0(p) = − + ln A 2~mω 2 − p ψf0(p) = Ae 2~mω .

Now

Z Z 2 √ 2 2 − p 2 ψf0(p) dp = |A| e ~mω dp = |A| π~mω = 1 1 1/4 A = π~mω

1/4 2 1 − p ∴ ψf0(p) = e 2~mω . π~mω

(c) We need to ﬁnd the Fourier transform of ψ0(x) and compare it with ψf0(p) derived in (b). 1 Z ∞ −ipx/~ ψf0(p) = √ dxe ψ0(x) 2π~ −∞ 1/4 ∞ 1 mω Z 2 = √ dxe−ipx/~e−mωx /2~ by using Eq (1) 2π~ π~ −∞ 1/4 ∞ 1 mω Z 2 = √ dxe−(ipx/~+mωx /2~). 2π~ π~ −∞ Consider mωx2 ipx rmω 2 rmω ip ip 2 ip 2 + = x +2 x √ + √ − √ 2~ ~ 2~ 2~ 2mω~ 2mω~ 2mω~ rmω ip 2 ip 2 = x + √ − √ 2~ 2mω~ 2mω~ rmω ip 2 p2 = x + √ + · 2~ 2mω~ 2mω~

Due Date: December. 14, 2015, 05:00 pm 3 Quantum Mechanics I / Quantum Theory I. 14 December, 2015

Let rmω ip y = x + √ 2~ 2mω~ r rmω 2 ⇒ dy = dx or dx = ~ dy 2~ mω So the integral becomes

∞ r ∞ Z 2 2 Z 2 2 dxe−ipx/~emωx /2~ = ~ dye−y e−p /2mω~ −∞ mω −∞ r ∞ 2 2 Z 2 = ~ e−p /2mω~ dye−y mω −∞ r r 2 2 √ 2π 2 = ~ e−p /2mω~ π = ~e−p /2mω~. mω mω Hence 1/4r 1 mω 2π 2 ~ −p /2mω~ ψf0(p) = √ e 2π~ π~ mω 1/4 1 2 = e−p /2mω~, mωπ~ which is the required result.

2. A particle of mass m in a one-dimensional harmonic oscillator is in a state for which a measurement of the energy yields the values ~ω/2 or 3~ω/2, each with a probability 1/2 of one-half. The average value of momentum px at time t = 0 is (mω~/2) . This information speciﬁes the state of the particle completely. What is this state and what

is px at time t? Answer The initial state is a superposition of two adjacent states |0 and |1 , with eigenvalues

E0 = ~ω/2 and E1 = 3/2~ω, 1 |ψ(0) = √ |0 +eiφ|1 , 2 where eiφ is the phase factor and its magnitude can be ﬁnd as

Due Date: December. 14, 2015, 05:00 pm 4 Quantum Mechanics I / Quantum Theory I. 14 December, 2015

r mω 1 1 = −i ~ √ 0| + e−iφ 1| (ˆa − aˆ†) √ |0 +eiφ|1 . 2 2 2 Herea ˆ† anda ˆ are raising and lowering operators: √ aˆ|n = n|n − 1 √ aˆ†|n = n + 1|n + 1 .

Therefore, r i mω pˆ = − ~ 0| + e−iφ 1| eiφaˆ|1 −aˆ†|0 x 2 2 r i mω = − ~ 0| + e−iφ 1| eiφ|0 −|1 2 2 r i mω = − ~(eiφ − e−iφ) 2 2 r r mω (eiφ − e−iφ) mω = ~ = ~ sin φ. 2 2i 2 q mω~ Given that pˆx at t = 0 = 2 . So we have r r mω mω ~ = ~ sin φ 2 2 ⇒ sin φ = 1 or φ = π/2 1 ∴ |ψ(0) = √ |0 +i|1 , 2 is the required state.

If |ψ(0) = √1 (|0 +i|1 ), then after time t, the state evolves to 2

|ψ(t) = e−iHt/~|ψ(0) .

For the harmonic oscillator, the Hamiltonian is expressed as 1 1 H = ω(ˆa†aˆ + ) = ω(Nˆ + ) (as derived in class) ~ 2 ~ 2

with eigenvalues En = (n + 1/2)~ω for n = 0, 1. Therefore,

−i ω(Nˆ+ 1 )t/ |ψ(t) = e ~ 2 ~|ψ(0) = e−iωt/2e−iωtNˆ |ψ(0) 1 = e−iωt/2e−iωtNˆ √ |0 +i|1 2 −iωt/2 e −iωt ˆ = √ |0 +ie |1 ∵ N|n = n|n . 2

Due Date: December. 14, 2015, 05:00 pm 5 Quantum Mechanics I / Quantum Theory I. 14 December, 2015

So the expectation value of momentum operatorp ˆx at time t is

3. Calculate the probability that a particle in the ground state of the harmonic oscillator is located in a classically disallowed region, namely, where V (x) > E. Obtain a numerical value for the probability. Suggestion: Express your integral in terms of a dimensionless variable and compare with the tabulated values of the error function. Answer The ground state of the harmonic oscillator (or the minimum uncertainty state)

1/4 mω 2 −mωx /2~ ψ0(x) = e π~ is a Gaussian function as drawn in ﬁgure.

Here ±xc are classical turning points (or critical points), at these points the energy of

Due Date: December. 14, 2015, 05:00 pm 6 Quantum Mechanics I / Quantum Theory I. 14 December, 2015

the particle equals to its potential energy, 1 1 ω = mω2x2 2~ 2 c r ⇒ x = ~ · c mω

In one of the classically forbidden regions (x > xc), where V (x) > E, the probability of ﬁnding a particle in ground state of the harmonic oscillator is ∞ r ∞ Z mω Z 2 2 −mωx /~ dx ψ0(x) = e dx. xc π~ xc Let rmω 1 a = , ⇒ xc = · ~ a By these substitutions, integral becomes ∞ √ a Z 2 2 a π 1 1 √ e−a x dx = √ − Erf π 1/a π 2 a a 1 = (1 − Erf(1)) ≈ 7.9%. 2 But this is only half of the desired result since we have two classically forbidden regions,

xc > a and xc < a. Hence the required probability is 16%.

4. Calculate E for the coherent state |α . Answer The energy eigenvalue corresponds to hamiltonian for harmonic oscillator is p2 1 E = + mω2x2 2m 2 1 1 E = pˆ2 + mω2 xˆ2 . (2) ∴ 2m 2 The expectation values ofp ˆ2 andx ˆ2 for the coherent state |α as derived in class mω pˆ2 = ~ 4|α|2 sin2(ωt − δ) + 1 2 xˆ2 = ~ 4|α|2 cos2(ωt − δ) + 1 . 2mω Plugging these values in Eq(2), 1 mω 1 E = ~ 4|α|2 sin2(ωt − δ) + 1 + mω2 ~ 4|α|2 cos2(ωt − δ) + 1 2m 2 2 2mω ω ω = ~ 4|α|2 sin2(ωt − δ) + 1 + ~ 4|α|2 cos2(ωt − δ) + 1 4 4 ω 1 = ~ 4|α|2 + 2 = ω |α|2 + , 4 ~ 2

Due Date: December. 14, 2015, 05:00 pm 7 Quantum Mechanics I / Quantum Theory I. 14 December, 2015

is the required result. Therefore, |α|2 is a measure of the average number of photons.

∞ n 2 X (α) |n |α = e−|α| /2 √ n=0 n! ∞ n 2 X (β) |n |β = e−|β| /2 √ n=0 n! ∞ m ∞ n |α|2+|β|2 − X (β) m| X (α) |n Now β|α = e 2 √ √ m=0 m! n=0 n! ∞ |α|2+|β|2 n − X (αβ) = e 2 (n!) n=0 The coherent states are the eigenstates of annihilation operatora ˆ:

aˆ|α = α|α

α|aˆ† = α∗ α| ⇒ β|aˆ† = β∗ β|.

Let

β|aˆ†aˆ|α = β∗ β|α|α = β∗α β|α . (3)

From L.H.S,

∞ n 2 X (α) |n β|aˆ†aˆ|α = β|Nˆ|α = β|Nˆ e−|α| /2 √ n=0 n! ∞ n 2 X (α) = e−|α| /2 √ β|Nˆ|n n=0 n! ∞ n 2 X (α) = e−|α| /2 √ n β|n . n=0 n! Now

∞ ∗ m 2 X (β ) m| β|n = e−|β| /2 √ |n m=0 m!

Due Date: December. 14, 2015, 05:00 pm 8 Quantum Mechanics I / Quantum Theory I. 14 December, 2015

∞ 2 X (β∗)ne−|β| /2 = √ n=0 n! ∞ n ∞ ∗ n 2 2 X (α) X (β ) ⇒ β|aˆ†aˆ|α = e−|α| /2e−|β| /2 √ n √ n=0 n! n=0 n! ∞ ∗ n 2 2 X (αβ ) = e−|α| /2e−|β| /2 n . n! n=0 Using Eq(3)

∞ ∗ n 2 2 X (αβ ) β∗α β|α = e−|α| /2e−|β| /2 n n! n=0 ∞ ∗ n 1 2 2 X (αβ ) β|α = e−|α| /2e−|β| /2 n (αβ∗) n! n=0 ∞ ∗ n−1 2 2 X n(αβ ) = e−|α| /2e−|β| /2 n! n=0 ∞ ∗ n−1 2 2 X n(αβ ) = e−|α| /2e−|β| /2 n(n − 1)! n=0

Since for n = 0 the above term vanishes, so we have

∞ ∗ n−1 2 2 X (αβ ) β|α = e−|α| /2e−|β| /2 (n − 1)! n=1 ∞ ∗ j 2 2 X (αβ ) = e−|α| /2e−|β| /2 let (n − 1) = j j! j=0 ∞ n 2 2 ∗ X x = e−|α| /2e−|β| /2e(αβ ) = ex ∵ n! n=0 2 −|α|2 −|β|2 (αβ∗+α∗β) ⇒ α|β β|α = α|β = e e e = e−(|α|2+|β|2−(αβ∗+α∗β)

= e−(|α−β|2) 2 ∴ α|β → 0 if |α − β| 1.

6. (a) Find the mean number of photons in the coherent state |α .

(b) Find the probability of ﬁnding n photons in the coherent state |α .

(c) Show that the probability distribution obtained above in (b) is Poissonian. Look up and state the discrete Poissonian probability distribution function.

Due Date: December. 14, 2015, 05:00 pm 9 Quantum Mechanics I / Quantum Theory I. 14 December, 2015

Answer

(a) The mean number of photons in the coherent state |α are

Nˆ = α|Nˆ|α = α|aˆ†aˆ|α = α|α∗α|α

= α∗α α|α = |α|2.

(b) The probability of ﬁnding the n photons in the coherent state |α is given by 2 P n |α = n|α

n α 2 n|α = √ e−|α| /2 n! 2 n 2 (|α| ) −|α|2 n|α = e . ∴ n!

λke−λ Pr(k) = . k!

If λ = |α|2 and k = n, then

(|α|2)ne−|α|2 Pr(n) = n!

is the probability distribution function derived in (b).

7. A one-dimensional harmonic oscillator is in the n = 1 state, for which

2 1/2 −1/2 −0.5/(x/x0) ψn=1(x) = 2(2π x0) (x/x0)e

1/2 with x0 = (~/mω) . Calculate the probability of ﬁnding the particle in the interval x to x + dx. Show that, according to classical mechanics, the probability is

1 P (x)dx = (A2 − x2)−1/2dx classical π 0

for −A0 < x < A0 and zero elsewhere (A0 is the classical amplitude). With the aid of sketches compare this probability distribution with the quantum mechanical one. Locate the maxima for the probability distribution for the n = 1 quantum state

Due Date: December. 14, 2015, 05:00 pm 10

Quantum Mechanics I / Quantum Theory I. 14 December, 2015

8. Field emission is a common technique to eject electrons from a metal surface. In a

metal, electrons are ﬁlled upto the Fermi level EF .

The minimum energy required to eject the electron from the metal is W , called the

work function. If a nearby metal is placed at a positive voltage V0 w.r.t. the bulk metal tip, an electric field is setup. As a result the potential energy profile changes as shown in figure (c). The Fermi electron can now easily tunnel out of the metal tip.

(a) If |E| is the constant electric ﬁeld between the tip and the nearby metal, express

V (x) − EF in terms of the electric ﬁeld. Use the notation V (x)= potential energy, v=electric voltage,

EF =Fermi energy = energy of tunneling electron.

(b) Show that the tunneling property of the Fermi electron is √ −4 2mW 3/2 T ≈ exp . 3e|E|~ Use the approximate approximation √ −2 2m Z b T ≈ exp dxpV (x) − E , ~ a choosing the approximate integration limits. If you encounter an integral of the √ form R dz 1 − z2, make the substitution z = cos θ.

Answer

Due Date: December. 14, 2015, 05:00 pm 12 Quantum Mechanics I / Quantum Theory I. 14 December, 2015

(a) From the ﬁgure (c), it is clear that the potential energy varies in the region x > 0 as

V (x) = EF + W − e|E|x

V (x) − EF = W − e|E|x.

(b) The approximate expression for tunneling coeﬃcient is given by √ −2 2m Z b T ≈ exp dxpV (x) − E . ~ a First we have to calculate the integral,

Z b I = dxpV (x) − E a Z b = dxpW − e|E|x a r √ Z b e|E|x = W dx 1 − · a W Here [a, b] corresponds to classically forbidden region (or tunneling allowed re-

gion). So it is clear that a = 0, b is the value of x when V (x) = EF , i.e., W W − e|E|b = 0, ⇒ b = · e|E|

So we have

W r √ Z e|E| e|E|x I = W dx 1 − · 0 W Let

e|E|x W = z2, ⇒ x = z2 W e|E| 2W z ⇒ dx = dz. e|E|

W When x = 0, z = 0 and when x = e|E| , z = 1. By these substitutions, integral becomes

√ Z 1 2W √ I = W z 1 − z2dz. 0 e|E|

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Put z = cos θ, ⇒ dz = − sin θdθ. When z = 0, θ = π/2 and when z = 1, θ = 0. Therefore,

2(W )3/2 Z 0 I = cos θ sin θ(− sin θdθ) e|E| π/2 2(W )3/2 Z π/2 = cos θ sin2 θdθ e|E| 0 3/2 3 π/2 3/2 2(W ) sin θ 2(W ) = = · e|E| 3 0 3e|E| By inserting it in tunneling expression, √ −2 2m 2(W )3/2 T ≈ exp 3e|E| √~ −4 2m(W )3/2 = exp , 3e~|E| which is the desired result.

Due Date: December. 14, 2015, 05:00 pm 14