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Today’s Outline - December 02, 2015

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Review session by grader & ARC tutor, Johny EcheversThursday, December 3, 2015, 17:00 – 19:00Room 220 Engineering 1 Building

Final Exam:Monday, December 7, 14:00 – 16:00Room 204 Stuart Building

C. Segre (IIT) PHYS 405 - Fall 2015 December 02, 2015 1 / 18

Problem 3.11

Find the momentum space wave function, Φ(p, t), for a particle in theground state of the harmonic oscillator. What is the probability that ameasurement of p on a particle in this state would yield a value outsidethe classical range?

The ground state of the harmonic oscillator is

Ψ0(x , t) =(mωπ~

)14e−mωx

2/2~e−iωt/2

Φ(p, t) =1√2π~

(mωπ~

)14e−iωt/2

∫ ∞−∞

e−ipx/~e−mωx2/2~ dx

=1√2π~

(mωπ~

)14e−iωt/2

√2π~mω

e−p2/2mω~

πmω~

e−p2/2mω~e−iωt/2

Problem 3.11

Ψ0(x , t) =(mωπ~

)14e−mωx

2/2~e−iωt/2

Φ(p, t) =1√2π~

(mωπ~

)14e−iωt/2

∫ ∞−∞

=1√2π~

(mωπ~

)14e−iωt/2

√2π~mω

e−p2/2mω~

πmω~

Problem 3.11

Ψ0(x , t) =(mωπ~

)14e−mωx

2/2~e−iωt/2

Φ(p, t) =1√2π~

(mωπ~

)14e−iωt/2

∫ ∞−∞

=1√2π~

(mωπ~

)14e−iωt/2

√2π~mω

e−p2/2mω~

πmω~

Problem 3.11

Ψ0(x , t) =(mωπ~

)14e−mωx

2/2~e−iωt/2

Φ(p, t) =1√2π~

(mωπ~

)14e−iωt/2

∫ ∞−∞

=1√2π~

(mωπ~

)14e−iωt/2

√2π~mω

e−p2/2mω~

πmω~

Problem 3.11

Ψ0(x , t) =(mωπ~

)14e−mωx

2/2~e−iωt/2

Φ(p, t) =1√2π~

(mωπ~

)14e−iωt/2

∫ ∞−∞

=1√2π~

(mωπ~

)14e−iωt/2

√2π~mω

e−p2/2mω~

πmω~

Problem 3.11

Ψ0(x , t) =(mωπ~

)14e−mωx

2/2~e−iωt/2

Φ(p, t) =1√2π~

(mωπ~

)14e−iωt/2

∫ ∞−∞

=1√2π~

(mωπ~

)14e−iωt/2

√2π~mω

e−p2/2mω~

πmω~

Problem 3.11 (cont.)

The maximum classical momentum

2m= E = 1

2~ω → p =√mω~

the probablity of finding the system outside the maximum classical limits

∫ −√mω~−∞

|Φ|2 dp +

∫ ∞√mω~|Φ|2 dp = 2

∫ ∞√mω~|Φ|2 dp

= 1− 2

∫ √mω~0

|Φ|2 dp = 1− 2√πmω~

∫ √mω~0

e−p2/mω~ dp

substituting z ≡ p/√mω~ and dp =

√mω~ dz

P = 1− 2√π

0e−z

2dx = 1− erf(1) = 1− 0.843 = 0.16

The maximum classical momentum p2

2m= E = 1

→ p =√mω~

∫ −√mω~−∞

|Φ|2 dp +

∫ ∞√mω~|Φ|2 dp = 2

∫ ∞√mω~|Φ|2 dp

= 1− 2

∫ √mω~0

|Φ|2 dp = 1− 2√πmω~

∫ √mω~0

e−p2/mω~ dp

√mω~ dz

P = 1− 2√π

0e−z

2dx = 1− erf(1) = 1− 0.843 = 0.16

2m= E = 1

2~ω → p =√mω~

∫ −√mω~−∞

|Φ|2 dp +

∫ ∞√mω~|Φ|2 dp = 2

∫ ∞√mω~|Φ|2 dp

= 1− 2

∫ √mω~0

|Φ|2 dp = 1− 2√πmω~

∫ √mω~0

e−p2/mω~ dp

√mω~ dz

P = 1− 2√π

0e−z

2dx = 1− erf(1) = 1− 0.843 = 0.16

2m= E = 1

2~ω → p =√mω~

∫ −√mω~−∞

|Φ|2 dp +

∫ ∞√mω~|Φ|2 dp = 2

∫ ∞√mω~|Φ|2 dp

= 1− 2

∫ √mω~0

|Φ|2 dp = 1− 2√πmω~

∫ √mω~0

e−p2/mω~ dp

√mω~ dz

P = 1− 2√π

0e−z

2dx = 1− erf(1) = 1− 0.843 = 0.16

2m= E = 1

2~ω → p =√mω~

∫ −√mω~−∞

|Φ|2 dp +

∫ ∞√mω~|Φ|2 dp

= 1− 2

∫ √mω~0

|Φ|2 dp = 1− 2√πmω~

∫ √mω~0

e−p2/mω~ dp

√mω~ dz

P = 1− 2√π

0e−z

2dx = 1− erf(1) = 1− 0.843 = 0.16

2m= E = 1

2~ω → p =√mω~

∫ −√mω~−∞

|Φ|2 dp +

∫ ∞√mω~|Φ|2 dp = 2

∫ ∞√mω~|Φ|2 dp

= 1− 2

∫ √mω~0

|Φ|2 dp = 1− 2√πmω~

∫ √mω~0

e−p2/mω~ dp

√mω~ dz

P = 1− 2√π

0e−z

2dx = 1− erf(1) = 1− 0.843 = 0.16

2m= E = 1

2~ω → p =√mω~

∫ −√mω~−∞

|Φ|2 dp +

∫ ∞√mω~|Φ|2 dp = 2

∫ ∞√mω~|Φ|2 dp

= 1− 2

∫ √mω~0

|Φ|2 dp

= 1− 2√πmω~

∫ √mω~0

e−p2/mω~ dp

√mω~ dz

P = 1− 2√π

0e−z

2dx = 1− erf(1) = 1− 0.843 = 0.16

2m= E = 1

2~ω → p =√mω~

∫ −√mω~−∞

|Φ|2 dp +

∫ ∞√mω~|Φ|2 dp = 2

∫ ∞√mω~|Φ|2 dp

= 1− 2

∫ √mω~0

|Φ|2 dp = 1− 2√πmω~

∫ √mω~0

e−p2/mω~ dp

√mω~ dz

P = 1− 2√π

0e−z

2dx = 1− erf(1) = 1− 0.843 = 0.16

2m= E = 1

2~ω → p =√mω~

∫ −√mω~−∞

|Φ|2 dp +

∫ ∞√mω~|Φ|2 dp = 2

∫ ∞√mω~|Φ|2 dp

= 1− 2

∫ √mω~0

|Φ|2 dp = 1− 2√πmω~

∫ √mω~0

e−p2/mω~ dp

√mω~ dz

P = 1− 2√π

0e−z

2dx = 1− erf(1) = 1− 0.843 = 0.16

2m= E = 1

2~ω → p =√mω~

∫ −√mω~−∞

|Φ|2 dp +

∫ ∞√mω~|Φ|2 dp = 2

∫ ∞√mω~|Φ|2 dp

= 1− 2

∫ √mω~0

|Φ|2 dp = 1− 2√πmω~

∫ √mω~0

e−p2/mω~ dp

√mω~ dz

P = 1− 2√π

0e−z

= 1− erf(1) = 1− 0.843 = 0.16

2m= E = 1

2~ω → p =√mω~

∫ −√mω~−∞

|Φ|2 dp +

∫ ∞√mω~|Φ|2 dp = 2

∫ ∞√mω~|Φ|2 dp

= 1− 2

∫ √mω~0

|Φ|2 dp = 1− 2√πmω~

∫ √mω~0

e−p2/mω~ dp

√mω~ dz

P = 1− 2√π

0e−z

2dx = 1− erf(1)

= 1− 0.843 = 0.16

2m= E = 1

2~ω → p =√mω~

∫ −√mω~−∞

|Φ|2 dp +

∫ ∞√mω~|Φ|2 dp = 2

∫ ∞√mω~|Φ|2 dp

= 1− 2

∫ √mω~0

|Φ|2 dp = 1− 2√πmω~

∫ √mω~0

e−p2/mω~ dp

√mω~ dz

P = 1− 2√π

0e−z

2dx = 1− erf(1) = 1− 0.843 = 0.16

Problem 3.27

An operator A, representing observable A has two normalized eigenstatesψ1 and ψ2, with eigenvalues a1 and a2, respectively. Operator B,representing observable B has two normalized eigenstates φ1 and φ2, witheigenvalues b1 and b2. The eigenstates are related by

ψ1 = 15(3φ1 + 4φ2) ψ2 = 1

5(4φ1 − 3φ2)

(a) Observable A is measured, and the value a1 is obtained. What is thestate of the system immediately after this measurement?

(b) If B is now measured, what are the possible results, and what are theirprobabilities?

(c) Right after the measurement of B, A is measured again. What is theprobability of getting a1?

Problem 3.27 – solution

ψ1 = 15(3φ1 + 4φ2)

ψ2 = 15(4φ1 − 3φ2)

φ1 = 15(3ψ1 + 4ψ2)

φ2 = 15(4ψ1 − 3ψ2)

(a) Since the expectation value measured is a1, the system must be in thecorresponding eigenstate ψ1

(b) The probabilities can be obtained by taking the expectation value of Bwith the state ψ1

〈ψ1|Bψ1〉 = 925b1〈φ1|φ1〉+ 16

25b2〈φ2|φ2〉 = 925b1 + 16

(c) We have to consider two cases just after B is measured

the system is in state φ1 with prob-ablity 9

25 and then we measure

Pa1(φ1) = 925

Pa1(φ2) = 1625

Pa1(total) = 925 ·

925 + 16

25 ·1625

= 337625 = 0.5392

ψ1 = 15(3φ1 + 4φ2)

ψ2 = 15(4φ1 − 3φ2)

φ1 = 15(3ψ1 + 4ψ2)

φ2 = 15(4ψ1 − 3ψ2)

〈ψ1|Bψ1〉 = 925b1〈φ1|φ1〉+ 16

25b2〈φ2|φ2〉 = 925b1 + 16

Pa1(φ1) = 925

Pa1(φ2) = 1625

Pa1(total) = 925 ·

925 + 16

25 ·1625

= 337625 = 0.5392

ψ1 = 15(3φ1 + 4φ2)

ψ2 = 15(4φ1 − 3φ2)

φ1 = 15(3ψ1 + 4ψ2)

φ2 = 15(4ψ1 − 3ψ2)

〈ψ1|Bψ1〉 = 925b1〈φ1|φ1〉+ 16

25b2〈φ2|φ2〉 = 925b1 + 16

Pa1(φ1) = 925

Pa1(φ2) = 1625

Pa1(total) = 925 ·

925 + 16

25 ·1625

= 337625 = 0.5392

ψ1 = 15(3φ1 + 4φ2)

ψ2 = 15(4φ1 − 3φ2)

φ1 = 15(3ψ1 + 4ψ2)

φ2 = 15(4ψ1 − 3ψ2)

〈ψ1|Bψ1〉 = 925b1〈φ1|φ1〉+ 16

25b2〈φ2|φ2〉 = 925b1 + 16

Pa1(φ1) = 925

Pa1(φ2) = 1625

Pa1(total) = 925 ·

925 + 16

25 ·1625

= 337625 = 0.5392

ψ1 = 15(3φ1 + 4φ2)

ψ2 = 15(4φ1 − 3φ2)

φ1 = 15(3ψ1 + 4ψ2)

φ2 = 15(4ψ1 − 3ψ2)

〈ψ1|Bψ1〉 = 925b1〈φ1|φ1〉+ 16

25b2〈φ2|φ2〉 = 925b1 + 16

Pa1(φ1) = 925

Pa1(φ2) = 1625

Pa1(total) = 925 ·

925 + 16

25 ·1625

= 337625 = 0.5392

ψ1 = 15(3φ1 + 4φ2)

ψ2 = 15(4φ1 − 3φ2)

φ1 = 15(3ψ1 + 4ψ2)

φ2 = 15(4ψ1 − 3ψ2)

〈ψ1|Bψ1〉

= 925b1〈φ1|φ1〉+ 16

25b2〈φ2|φ2〉 = 925b1 + 16

Pa1(φ1) = 925

Pa1(φ2) = 1625

Pa1(total) = 925 ·

925 + 16

25 ·1625

= 337625 = 0.5392

ψ1 = 15(3φ1 + 4φ2)

ψ2 = 15(4φ1 − 3φ2)

φ1 = 15(3ψ1 + 4ψ2)

φ2 = 15(4ψ1 − 3ψ2)

〈ψ1|Bψ1〉 = 925b1〈φ1|φ1〉

+ 1625b2〈φ2|φ2〉 = 9

25b1 + 1625b2

Pa1(φ1) = 925

Pa1(φ2) = 1625

Pa1(total) = 925 ·

925 + 16

25 ·1625

= 337625 = 0.5392

ψ1 = 15(3φ1 + 4φ2)

ψ2 = 15(4φ1 − 3φ2)

φ1 = 15(3ψ1 + 4ψ2)

φ2 = 15(4ψ1 − 3ψ2)

〈ψ1|Bψ1〉 = 925b1〈φ1|φ1〉+ 16

25b2〈φ2|φ2〉

= 925b1 + 16

Pa1(φ1) = 925

Pa1(φ2) = 1625

Pa1(total) = 925 ·

925 + 16

25 ·1625

= 337625 = 0.5392

ψ1 = 15(3φ1 + 4φ2)

ψ2 = 15(4φ1 − 3φ2)

φ1 = 15(3ψ1 + 4ψ2)

φ2 = 15(4ψ1 − 3ψ2)

〈ψ1|Bψ1〉 = 925b1〈φ1|φ1〉+ 16

25b2〈φ2|φ2〉 = 925b1 + 16

Pa1(φ1) = 925

Pa1(φ2) = 1625

Pa1(total) = 925 ·

925 + 16

25 ·1625

= 337625 = 0.5392

ψ1 = 15(3φ1 + 4φ2)

ψ2 = 15(4φ1 − 3φ2)

φ1 = 15(3ψ1 + 4ψ2)

φ2 = 15(4ψ1 − 3ψ2)

〈ψ1|Bψ1〉 = 925b1〈φ1|φ1〉+ 16

25b2〈φ2|φ2〉 = 925b1 + 16

Pa1(φ1) = 925

Pa1(φ2) = 1625

Pa1(total) = 925 ·

925 + 16

25 ·1625

= 337625 = 0.5392

ψ1 = 15(3φ1 + 4φ2)

ψ2 = 15(4φ1 − 3φ2)

φ1 = 15(3ψ1 + 4ψ2)

φ2 = 15(4ψ1 − 3ψ2)

〈ψ1|Bψ1〉 = 925b1〈φ1|φ1〉+ 16

25b2〈φ2|φ2〉 = 925b1 + 16

Pa1(φ1) = 925

Pa1(φ2) = 1625

Pa1(total) = 925 ·

925 + 16

25 ·1625

= 337625 = 0.5392

ψ1 = 15(3φ1 + 4φ2)

ψ2 = 15(4φ1 − 3φ2)

φ1 = 15(3ψ1 + 4ψ2)

φ2 = 15(4ψ1 − 3ψ2)

〈ψ1|Bψ1〉 = 925b1〈φ1|φ1〉+ 16

25b2〈φ2|φ2〉 = 925b1 + 16

Pa1(φ1) = 925

Pa1(φ2) = 1625

Pa1(total) = 925 ·

925 + 16

25 ·1625

= 337625 = 0.5392

ψ1 = 15(3φ1 + 4φ2)

ψ2 = 15(4φ1 − 3φ2)

φ1 = 15(3ψ1 + 4ψ2)

φ2 = 15(4ψ1 − 3ψ2)

〈ψ1|Bψ1〉 = 925b1〈φ1|φ1〉+ 16

25b2〈φ2|φ2〉 = 925b1 + 16

Pa1(φ1) = 925

Pa1(φ2) = 1625

Pa1(total) = 925 ·

925 + 16

25 ·1625

= 337625 = 0.5392

ψ1 = 15(3φ1 + 4φ2)

ψ2 = 15(4φ1 − 3φ2)

φ1 = 15(3ψ1 + 4ψ2)

φ2 = 15(4ψ1 − 3ψ2)

〈ψ1|Bψ1〉 = 925b1〈φ1|φ1〉+ 16

25b2〈φ2|φ2〉 = 925b1 + 16

Pa1(φ1) = 925

Pa1(φ2) = 1625

Pa1(total) = 925 ·

925 + 16

25 ·1625

= 337625 = 0.5392

ψ1 = 15(3φ1 + 4φ2)

ψ2 = 15(4φ1 − 3φ2)

φ1 = 15(3ψ1 + 4ψ2)

φ2 = 15(4ψ1 − 3ψ2)

〈ψ1|Bψ1〉 = 925b1〈φ1|φ1〉+ 16

25b2〈φ2|φ2〉 = 925b1 + 16

Pa1(φ1) = 925

Pa1(φ2) = 1625

Pa1(total) = 925 ·

925 + 16

25 ·1625

= 337625 = 0.5392

Problem 3.32

In an interesting version of the energy-time uncertainty principle∆t = τ/π, where τ is the time it takes Ψ(x , t) to evolve into a stateorthogonal to Ψ(x , 0). Test this out, using a wave function that is an equaladmixture of two orthnormal stationary states of some arbitrary potential:

Ψ(x , 0) =√

12 [ψ1(x) + ψ2(x)]

Assume that the two states have energies E1 and E2, respectively. Thetime dependent wave function is thus

Ψ(x , t) =√

12(ψ1e

−iE1t/~ + ψ2e−iE2t/~)

if we assume that at a time τ this wave function is orthogonal to the wavefunction at t = 0, we can solve for τ by taking the inner product andsetting it to zero

Problem 3.32

Ψ(x , 0) =√

12 [ψ1(x) + ψ2(x)]

Ψ(x , t) =√

12(ψ1e

−iE1t/~ + ψ2e−iE2t/~)

Problem 3.32

Ψ(x , 0) =√

12 [ψ1(x) + ψ2(x)]

Ψ(x , t) =√

12(ψ1e

−iE1t/~ + ψ2e−iE2t/~)

Problem 3.32

Ψ(x , 0) =√

12 [ψ1(x) + ψ2(x)]

Ψ(x , t) =√

12(ψ1e

−iE1t/~ + ψ2e−iE2t/~)

0 = 〈Ψ(x , τ)|Ψ(x , 0)〉

(e iE1τ/~〈ψ1|ψ1〉+ e iE1τ/~

��〈ψ1|ψ2〉+ e iE2τ/~��〈ψ2|ψ1〉+ e iE2τ/~〈ψ2|ψ2〉

(e iE1τ/~ + e iE2τ/~

)−→ e iE2τ/~ = −e iE1τ/~

− 1 = e i(E2−E1)τ/~ = e iπ −→ (E2 − E1)τ

∆t =τ

π~(E2 − E1)

but from problem 3.18, we have that

∆E = σH =(E2 − E1)

2−→ ∆t∆E =

0 = 〈Ψ(x , τ)|Ψ(x , 0)〉

(e iE1τ/~〈ψ1|ψ1〉+ e iE1τ/~〈ψ1|ψ2〉+ e iE2τ/~〈ψ2|ψ1〉+ e iE2τ/~〈ψ2|ψ2〉

)−→ e iE2τ/~ = −e iE1τ/~

− 1 = e i(E2−E1)τ/~ = e iπ −→ (E2 − E1)τ

∆t =τ

π~(E2 − E1)

∆E = σH =(E2 − E1)

2−→ ∆t∆E =

0 = 〈Ψ(x , τ)|Ψ(x , 0)〉

(e iE1τ/~〈ψ1|ψ1〉+ e iE1τ/~

��〈ψ1|ψ2〉+ e iE2τ/~��〈ψ2|ψ1〉+ e iE2τ/~〈ψ2|ψ2〉

)−→ e iE2τ/~ = −e iE1τ/~

− 1 = e i(E2−E1)τ/~ = e iπ −→ (E2 − E1)τ

∆t =τ

π~(E2 − E1)

∆E = σH =(E2 − E1)

2−→ ∆t∆E =

0 = 〈Ψ(x , τ)|Ψ(x , 0)〉

(e iE1τ/~〈ψ1|ψ1〉+ e iE1τ/~

��〈ψ1|ψ2〉+ e iE2τ/~��〈ψ2|ψ1〉+ e iE2τ/~〈ψ2|ψ2〉

−→ e iE2τ/~ = −e iE1τ/~

− 1 = e i(E2−E1)τ/~ = e iπ −→ (E2 − E1)τ

∆t =τ

π~(E2 − E1)

∆E = σH =(E2 − E1)

2−→ ∆t∆E =

0 = 〈Ψ(x , τ)|Ψ(x , 0)〉

(e iE1τ/~〈ψ1|ψ1〉+ e iE1τ/~

��〈ψ1|ψ2〉+ e iE2τ/~��〈ψ2|ψ1〉+ e iE2τ/~〈ψ2|ψ2〉

)−→ e iE2τ/~ = −e iE1τ/~

− 1 = e i(E2−E1)τ/~ = e iπ −→ (E2 − E1)τ

∆t =τ

π~(E2 − E1)

∆E = σH =(E2 − E1)

2−→ ∆t∆E =

0 = 〈Ψ(x , τ)|Ψ(x , 0)〉

(e iE1τ/~〈ψ1|ψ1〉+ e iE1τ/~

��〈ψ1|ψ2〉+ e iE2τ/~��〈ψ2|ψ1〉+ e iE2τ/~〈ψ2|ψ2〉

)−→ e iE2τ/~ = −e iE1τ/~

− 1 = e i(E2−E1)τ/~

= e iπ −→ (E2 − E1)τ

∆t =τ

π~(E2 − E1)

∆E = σH =(E2 − E1)

2−→ ∆t∆E =

0 = 〈Ψ(x , τ)|Ψ(x , 0)〉

(e iE1τ/~〈ψ1|ψ1〉+ e iE1τ/~

��〈ψ1|ψ2〉+ e iE2τ/~��〈ψ2|ψ1〉+ e iE2τ/~〈ψ2|ψ2〉

)−→ e iE2τ/~ = −e iE1τ/~

− 1 = e i(E2−E1)τ/~ = e iπ

−→ (E2 − E1)τ

∆t =τ

π~(E2 − E1)

∆E = σH =(E2 − E1)

2−→ ∆t∆E =

0 = 〈Ψ(x , τ)|Ψ(x , 0)〉

(e iE1τ/~〈ψ1|ψ1〉+ e iE1τ/~

��〈ψ1|ψ2〉+ e iE2τ/~��〈ψ2|ψ1〉+ e iE2τ/~〈ψ2|ψ2〉

)−→ e iE2τ/~ = −e iE1τ/~

− 1 = e i(E2−E1)τ/~ = e iπ −→ (E2 − E1)τ

∆t =τ

π~(E2 − E1)

∆E = σH =(E2 − E1)

2−→ ∆t∆E =

0 = 〈Ψ(x , τ)|Ψ(x , 0)〉

(e iE1τ/~〈ψ1|ψ1〉+ e iE1τ/~

��〈ψ1|ψ2〉+ e iE2τ/~��〈ψ2|ψ1〉+ e iE2τ/~〈ψ2|ψ2〉

)−→ e iE2τ/~ = −e iE1τ/~

− 1 = e i(E2−E1)τ/~ = e iπ −→ (E2 − E1)τ

∆t =τ

π~(E2 − E1)

∆E = σH =(E2 − E1)

2−→ ∆t∆E =

0 = 〈Ψ(x , τ)|Ψ(x , 0)〉

(e iE1τ/~〈ψ1|ψ1〉+ e iE1τ/~

��〈ψ1|ψ2〉+ e iE2τ/~��〈ψ2|ψ1〉+ e iE2τ/~〈ψ2|ψ2〉

)−→ e iE2τ/~ = −e iE1τ/~

− 1 = e i(E2−E1)τ/~ = e iπ −→ (E2 − E1)τ

∆t =τ

π~(E2 − E1)

∆E = σH =(E2 − E1)

2−→ ∆t∆E =

0 = 〈Ψ(x , τ)|Ψ(x , 0)〉

(e iE1τ/~〈ψ1|ψ1〉+ e iE1τ/~

��〈ψ1|ψ2〉+ e iE2τ/~��〈ψ2|ψ1〉+ e iE2τ/~〈ψ2|ψ2〉

)−→ e iE2τ/~ = −e iE1τ/~

− 1 = e i(E2−E1)τ/~ = e iπ −→ (E2 − E1)τ

∆t =τ

π~(E2 − E1)

∆E = σH

=(E2 − E1)

2−→ ∆t∆E =

0 = 〈Ψ(x , τ)|Ψ(x , 0)〉

(e iE1τ/~〈ψ1|ψ1〉+ e iE1τ/~

��〈ψ1|ψ2〉+ e iE2τ/~��〈ψ2|ψ1〉+ e iE2τ/~〈ψ2|ψ2〉

)−→ e iE2τ/~ = −e iE1τ/~

− 1 = e i(E2−E1)τ/~ = e iπ −→ (E2 − E1)τ

∆t =τ

π~(E2 − E1)

∆E = σH =(E2 − E1)

−→ ∆t∆E =~2

0 = 〈Ψ(x , τ)|Ψ(x , 0)〉

(e iE1τ/~〈ψ1|ψ1〉+ e iE1τ/~

��〈ψ1|ψ2〉+ e iE2τ/~��〈ψ2|ψ1〉+ e iE2τ/~〈ψ2|ψ2〉

)−→ e iE2τ/~ = −e iE1τ/~

− 1 = e i(E2−E1)τ/~ = e iπ −→ (E2 − E1)τ

∆t =τ

π~(E2 − E1)

∆E = σH =(E2 − E1)

2−→ ∆t∆E =

Problem 4.38

Consider the three-dimensional harmonic oscillator, with potential

V (r) =1

2mω2r2

(a) Show that separation of variables in cartesian coordinates turns thisinto three one-dimensional oscillators, and exploit your knowledge ofthe latter to determine the allowed energies.

(b) Determine the degeneracy d(n) of En.

(a) The Schrodinger equation for a 3D harmonic oscillator is

− ~2

(∂2ψ

∂x2+∂2ψ

∂y2+∂2ψ

(x2 + y2 + z2

)ψ = Eψ

this can be separated by choosing ψ = X (x)Y (y)Z (z)

E XYZ =− ~2

YZ∂2X

∂x2+ XZ

∂y2+ XY

(x2 + y2 + z2

(− ~2

∂x2+

2mω2x2

(− ~2

∂y2+

2mω2y2

(− ~2

∂z2+

2mω2z2

the equation is thus separated and can be solved in the usual way

− ~2

(∂2ψ

∂x2+∂2ψ

∂y2+∂2ψ

(x2 + y2 + z2

)ψ = Eψ

E XYZ =− ~2

YZ∂2X

∂x2+ XZ

∂y2+ XY

(x2 + y2 + z2

(− ~2

∂x2+

2mω2x2

(− ~2

∂y2+

2mω2y2

(− ~2

∂z2+

2mω2z2

− ~2

(∂2ψ

∂x2+∂2ψ

∂y2+∂2ψ

(x2 + y2 + z2

)ψ = Eψ

E XYZ =− ~2

YZ∂2X

∂x2+ XZ

∂y2+ XY

(x2 + y2 + z2

(− ~2

∂x2+

2mω2x2

(− ~2

∂y2+

2mω2y2

(− ~2

∂z2+

2mω2z2

− ~2

(∂2ψ

∂x2+∂2ψ

∂y2+∂2ψ

(x2 + y2 + z2

)ψ = Eψ

E XYZ =− ~2

YZ∂2X

∂x2+ XZ

∂y2+ XY

(x2 + y2 + z2

(− ~2

∂x2+

2mω2x2

(− ~2

∂y2+

2mω2y2

(− ~2

∂z2+

2mω2z2

− ~2

(∂2ψ

∂x2+∂2ψ

∂y2+∂2ψ

(x2 + y2 + z2

)ψ = Eψ

E XYZ =− ~2

+ XZ∂2Y

∂y2+ XY

(x2 + y2 + z2

(− ~2

∂x2+

2mω2x2

(− ~2

∂y2+

2mω2y2

(− ~2

∂z2+

2mω2z2

− ~2

(∂2ψ

∂x2+∂2ψ

∂y2+∂2ψ

(x2 + y2 + z2

)ψ = Eψ

E XYZ =− ~2

∂x2+ XZ

+ XY∂2Z

(x2 + y2 + z2

(− ~2

∂x2+

2mω2x2

(− ~2

∂y2+

2mω2y2

(− ~2

∂z2+

2mω2z2

− ~2

(∂2ψ

∂x2+∂2ψ

∂y2+∂2ψ

(x2 + y2 + z2

)ψ = Eψ

E XYZ =− ~2

∂x2+ XZ

∂y2+ XY

(x2 + y2 + z2

(− ~2

∂x2+

2mω2x2

(− ~2

∂y2+

2mω2y2

(− ~2

∂z2+

2mω2z2

− ~2

(∂2ψ

∂x2+∂2ψ

∂y2+∂2ψ

(x2 + y2 + z2

)ψ = Eψ

E XYZ =− ~2

∂x2+ XZ

∂y2+ XY

(x2 + y2 + z2

(− ~2

∂x2+

2mω2x2

(− ~2

∂y2+

2mω2y2

(− ~2

∂z2+

2mω2z2

− ~2

(∂2ψ

∂x2+∂2ψ

∂y2+∂2ψ

(x2 + y2 + z2

)ψ = Eψ

E XYZ =− ~2

∂x2+ XZ

∂y2+ XY

(x2 + y2 + z2

(− ~2

∂x2+

2mω2x2

(− ~2

∂y2+

2mω2y2

(− ~2

∂z2+

2mω2z2

)the equation is thus separated and can be solved in the usual way

− ~2

(∂2ψ

∂x2+∂2ψ

∂y2+∂2ψ

(x2 + y2 + z2

)ψ = Eψ

E XYZ =− ~2

∂x2+ XZ

∂y2+ XY

(x2 + y2 + z2

(− ~2

∂x2+

2mω2x2

(− ~2

∂y2+

2mω2y2

(− ~2

∂z2+

2mω2z2

− ~2

(∂2ψ

∂x2+∂2ψ

∂y2+∂2ψ

(x2 + y2 + z2

)ψ = Eψ

E XYZ =− ~2

∂x2+ XZ

∂y2+ XY

(x2 + y2 + z2

(− ~2

∂x2+

2mω2x2

(− ~2

∂y2+

2mω2y2

(− ~2

∂z2+

2mω2z2

− ~2

(∂2ψ

∂x2+∂2ψ

∂y2+∂2ψ

(x2 + y2 + z2

)ψ = Eψ

E XYZ =− ~2

∂x2+ XZ

∂y2+ XY

(x2 + y2 + z2

(− ~2

∂x2+

2mω2x2

(− ~2

∂y2+

2mω2y2

(− ~2

∂z2+

2mω2z2

− ~2

(∂2ψ

∂x2+∂2ψ

∂y2+∂2ψ

(x2 + y2 + z2

)ψ = Eψ

E XYZ =− ~2

∂x2+ XZ

∂y2+ XY

(x2 + y2 + z2

(− ~2

∂x2+

2mω2x2

(− ~2

∂y2+

2mω2y2

(− ~2

∂z2+

2mω2z2

The three separation constants arecalled Ex ,Ey ,Ez and yield three dif-ferent equations to solve.

Each of these three equations is a1D harmonic oscillator with eigen-values

)~ω, nx = 0, 1, 2, . . .

)~ω, ny = 0, 1, 2, . . .

)~ω, nz = 0, 1, 2, . . .

ExX = − ~2

∂x2+

2mω2x2X

EyY = − ~2

∂y2+

2mω2y2Y

EzZ = − ~2

∂z2+

2mω2z2Z

The total energy is thus

(nx + ny + nz +

)~ω, n≡nx +ny +nz

)~ω, nx = 0, 1, 2, . . .

)~ω, ny = 0, 1, 2, . . .

)~ω, nz = 0, 1, 2, . . .

ExX = − ~2

∂x2+

2mω2x2X

EyY = − ~2

∂y2+

2mω2y2Y

EzZ = − ~2

∂z2+

2mω2z2Z

(nx + ny + nz +

)~ω, nx = 0, 1, 2, . . .

)~ω, ny = 0, 1, 2, . . .

)~ω, nz = 0, 1, 2, . . .

ExX = − ~2

∂x2+

2mω2x2X

EyY = − ~2

∂y2+

2mω2y2Y

EzZ = − ~2

∂z2+

2mω2z2Z

(nx + ny + nz +

)~ω, nx = 0, 1, 2, . . .

)~ω, ny = 0, 1, 2, . . .

)~ω, nz = 0, 1, 2, . . .

ExX = − ~2

∂x2+

2mω2x2X

EyY = − ~2

∂y2+

2mω2y2Y

EzZ = − ~2

∂z2+

2mω2z2Z

(nx + ny + nz +

)~ω, nx = 0, 1, 2, . . .

)~ω, ny = 0, 1, 2, . . .

)~ω, nz = 0, 1, 2, . . .

ExX = − ~2

∂x2+

2mω2x2X

EyY = − ~2

∂y2+

2mω2y2Y

EzZ = − ~2

∂z2+

2mω2z2Z

(nx + ny + nz +

)~ω, nx = 0, 1, 2, . . .

)~ω, ny = 0, 1, 2, . . .

)~ω, nz = 0, 1, 2, . . .

ExX = − ~2

∂x2+

2mω2x2X

EyY = − ~2

∂y2+

2mω2y2Y

EzZ = − ~2

∂z2+

2mω2z2Z

(nx + ny + nz +

)~ω, nx = 0, 1, 2, . . .

)~ω, ny = 0, 1, 2, . . .

)~ω, nz = 0, 1, 2, . . .

ExX = − ~2

∂x2+

2mω2x2X

EyY = − ~2

∂y2+

2mω2y2Y

EzZ = − ~2

∂z2+

2mω2z2Z

(nx + ny + nz +

(b) The degeneracy is governed by how many ways one can make thesame value of n using the three individual quantum numbers nx , ny , andnz given the constraint n≡nx +ny +nz

approach this by looking atthe first few cases and thenextrapolate

thus the sum of all theseways to make n is

d(n) = 1 + 2 + · · ·+ (n + 1)

=(n + 1)(n + 2)

nx ny nzn 0 0 1 way

n-11 0

2 ways0 1

n-22 0

3 ways1 10 2

n+1 waysn − 1 1

· · ·1 n − 10 n

d(n) = 1 + 2 + · · ·+ (n + 1)

=(n + 1)(n + 2)

nx ny nz

n 0 0 1 way

n-11 0

2 ways0 1

n-22 0

3 ways1 10 2

n+1 waysn − 1 1

· · ·1 n − 10 n

d(n) = 1 + 2 + · · ·+ (n + 1)

=(n + 1)(n + 2)

nx ny nzn 0 0 1 way

n-11 0

2 ways0 1

n-22 0

3 ways1 10 2

n+1 waysn − 1 1

· · ·1 n − 10 n

d(n) = 1 + 2 + · · ·+ (n + 1)

=(n + 1)(n + 2)

nx ny nzn 0 0 1 way

n-11 0

2 ways0 1

n-22 0

3 ways1 10 2

n+1 waysn − 1 1

· · ·1 n − 10 n

d(n) = 1 + 2 + · · ·+ (n + 1)

=(n + 1)(n + 2)

nx ny nzn 0 0 1 way

n-11 0

2 ways0 1

n-22 0

3 ways1 10 2

n+1 waysn − 1 1

· · ·1 n − 10 n

d(n) = 1 + 2 + · · ·+ (n + 1)

=(n + 1)(n + 2)

nx ny nzn 0 0 1 way

n-11 0

2 ways0 1

n-22 0

3 ways1 10 2

n+1 waysn − 1 1

· · ·1 n − 10 n

d(n) = 1 + 2 + · · ·+ (n + 1)

=(n + 1)(n + 2)

nx ny nzn 0 0 1 way

n-11 0

2 ways0 1

n-22 0

3 ways1 10 2

n+1 waysn − 1 1

· · ·1 n − 10 n

d(n) = 1 + 2 + · · ·+ (n + 1)

=(n + 1)(n + 2)

nx ny nzn 0 0 1 way

n-11 0

2 ways0 1

n-22 0

3 ways1 10 2

n+1 waysn − 1 1

· · ·1 n − 10 n

d(n) = 1 + 2 + · · ·+ (n + 1)

=(n + 1)(n + 2)

nx ny nzn 0 0 1 way

n-11 0

2 ways0 1

n-22 0

3 ways1 10 2

n+1 waysn − 1 1

· · ·1 n − 10 n

Problem 4.2

Use separation of variables in cartesian coordinates to solve the infinitecubical well.

V (x , y , z) =

{0, 0 < x , y , z < a

∞, otherwise

(a) Find the stationary states and the corresponding energies

(b) Call the distince energies E1,E2,E3, . . . , in order of increasing energyand determine the degeneracies of the first 6 states

(c) What is the degeneracy of E14 and why is this case interesting?

(a) The solutions are given by

ψ(x , y , z) =

)3/2sin(nxπ

sin(nyπ

sin(nzπ

Problem 4.2

V (x , y , z) =

{0, 0 < x , y , z < a

∞, otherwise

ψ(x , y , z) =

)3/2sin(nxπ

sin(nyπ

sin(nzπ

Problem 4.2

V (x , y , z) =

{0, 0 < x , y , z < a

∞, otherwise

ψ(x , y , z) =

)3/2sin(nxπ

sin(nyπ

sin(nzπ

Problem 4.2 (cont.)

The energies are thus

a2(n2x + n2y + n2z)

, nx , ny , nz = 1, 2, 3, . . .

(b) We can now enumerate the different states and their degeneracies

nx ny nz N d

E1 1 1 1 3 1

2 1 16 31 2 1

2 2 19 32 1 2

3 1 111 31 3 1

nx ny nz N d

E5 2 2 2 12 1

1 3 22 3 13 1 22 3 11 2 3

Problem 4.2 (cont.)

a2(n2x + n2y + n2z)

, nx , ny , nz = 1, 2, 3, . . .

nx ny nz N d

E1 1 1 1 3 1

2 1 16 31 2 1

2 2 19 32 1 2

3 1 111 31 3 1

nx ny nz N d

E5 2 2 2 12 1

1 3 22 3 13 1 22 3 11 2 3

Problem 4.2 (cont.)

a2(n2x + n2y + n2z) =

a2N, nx , ny , nz = 1, 2, 3, . . .

nx ny nz N d

E1 1 1 1 3 1

2 1 16 31 2 1

2 2 19 32 1 2

3 1 111 31 3 1

nx ny nz N d

E5 2 2 2 12 1

1 3 22 3 13 1 22 3 11 2 3

Problem 4.2 (cont.)

a2(n2x + n2y + n2z) =

a2N, nx , ny , nz = 1, 2, 3, . . .

nx ny nz N d

E1 1 1 1 3 1

2 1 16 31 2 1

2 2 19 32 1 2

3 1 111 31 3 1

nx ny nz N d

E5 2 2 2 12 1

1 3 22 3 13 1 22 3 11 2 3

Problem 4.2 (cont.)

a2(n2x + n2y + n2z) =

a2N, nx , ny , nz = 1, 2, 3, . . .

nx ny nz N d

E1 1 1 1 3 1

2 1 16 31 2 1

2 2 19 32 1 2

3 1 111 31 3 1

nx ny nz N d

E5 2 2 2 12 1

1 3 22 3 13 1 22 3 11 2 3

Problem 4.2 (cont.)

a2(n2x + n2y + n2z) =

a2N, nx , ny , nz = 1, 2, 3, . . .

nx ny nz N d

E1 1 1 1 3 1

2 1 16 31 2 1

2 2 19 32 1 2

3 1 111 31 3 1

nx ny nz N d

E5 2 2 2 12 1

1 3 22 3 13 1 22 3 11 2 3

Problem 4.2 (cont.)

a2(n2x + n2y + n2z) =

a2N, nx , ny , nz = 1, 2, 3, . . .

nx ny nz N d

E1 1 1 1 3 1

2 1 16 31 2 1

2 2 19 32 1 2

3 1 111 31 3 1

nx ny nz N d

E5 2 2 2 12 1

1 3 22 3 13 1 22 3 11 2 3

Problem 4.2 (cont.)

a2(n2x + n2y + n2z) =

a2N, nx , ny , nz = 1, 2, 3, . . .

nx ny nz N d

E1 1 1 1 3 1

2 1 16 31 2 1

2 2 19 32 1 2

3 1 111 31 3 1

nx ny nz N d

E5 2 2 2 12 1

1 3 22 3 13 1 22 3 11 2 3

Problem 4.2 (cont.)

a2(n2x + n2y + n2z) =

a2N, nx , ny , nz = 1, 2, 3, . . .

nx ny nz N d

E1 1 1 1 3 1

2 1 16 31 2 1

2 2 19 32 1 2

3 1 111 31 3 1

nx ny nz N d

E5 2 2 2 12 1

1 3 22 3 13 1 22 3 11 2 3

Problem 4.2 (cont.)

a2(n2x + n2y + n2z) =

a2N, nx , ny , nz = 1, 2, 3, . . .

nx ny nz N d

E1 1 1 1 3 1

2 1 16 31 2 1

2 2 19 32 1 2

3 1 111 31 3 1

nx ny nz N d

E5 2 2 2 12 1

1 3 22 3 13 1 22 3 11 2 3

Problem 4.2 (cont.)

a2(n2x + n2y + n2z) =

a2N, nx , ny , nz = 1, 2, 3, . . .

nx ny nz N d

E1 1 1 1 3 1

2 1 16 31 2 1

2 2 19 32 1 2

3 1 111 31 3 1

nx ny nz N d

E5 2 2 2 12 1

1 3 22 3 13 1 22 3 11 2 3

Problem 4.2 (cont.)

(c) The next 7 states and their associatedN’s and degeneracies are

for E14, however, there are two integercombinations which give the same valueof N = 27, namely

(3 3 3) with d = 1

(5 1 1) with d = 3

for a total degeneracy of 4

E1 1 1 1 3 1

E2 2 1 1 6 3

E3 2 2 1 9 3

E4 3 1 1 11 3

E5 2 2 2 12 1

E6 3 2 1 14 6

E7 3 2 2 17 3

E8 4 1 1 18 3

E9 3 3 1 19 3

E10 4 2 1 21 6

E11 3 3 2 22 3

E12 4 2 2 24 3

E13 4 3 1 26 6

E143 3 3

27 45 1 1

Problem 4.2 (cont.)

(3 3 3) with d = 1

(5 1 1) with d = 3

E1 1 1 1 3 1

E2 2 1 1 6 3

E3 2 2 1 9 3

E4 3 1 1 11 3

E5 2 2 2 12 1

E6 3 2 1 14 6

E7 3 2 2 17 3

E8 4 1 1 18 3

E9 3 3 1 19 3

E10 4 2 1 21 6

E11 3 3 2 22 3

E12 4 2 2 24 3

E13 4 3 1 26 6

E143 3 3

27 45 1 1

Problem 4.2 (cont.)

(3 3 3) with d = 1

(5 1 1) with d = 3

E1 1 1 1 3 1

E2 2 1 1 6 3

E3 2 2 1 9 3

E4 3 1 1 11 3

E5 2 2 2 12 1

E6 3 2 1 14 6

E7 3 2 2 17 3

E8 4 1 1 18 3

E9 3 3 1 19 3

E10 4 2 1 21 6

E11 3 3 2 22 3

E12 4 2 2 24 3

E13 4 3 1 26 6

E143 3 3

27 45 1 1

Problem 4.2 (cont.)

(3 3 3) with d = 1

(5 1 1) with d = 3

E1 1 1 1 3 1

E2 2 1 1 6 3

E3 2 2 1 9 3

E4 3 1 1 11 3

E5 2 2 2 12 1

E6 3 2 1 14 6

E7 3 2 2 17 3

E8 4 1 1 18 3

E9 3 3 1 19 3

E10 4 2 1 21 6

E11 3 3 2 22 3

E12 4 2 2 24 3

E13 4 3 1 26 6

E143 3 3

27 45 1 1

Problem 4.2 (cont.)

(3 3 3) with d = 1

(5 1 1) with d = 3

E1 1 1 1 3 1

E2 2 1 1 6 3

E3 2 2 1 9 3

E4 3 1 1 11 3

E5 2 2 2 12 1

E6 3 2 1 14 6

E7 3 2 2 17 3

E8 4 1 1 18 3

E9 3 3 1 19 3

E10 4 2 1 21 6

E11 3 3 2 22 3

E12 4 2 2 24 3

E13 4 3 1 26 6

E143 3 3

27 45 1 1

Problem 4.2 (cont.)

(3 3 3) with d = 1

(5 1 1) with d = 3

E1 1 1 1 3 1

E2 2 1 1 6 3

E3 2 2 1 9 3

E4 3 1 1 11 3

E5 2 2 2 12 1

E6 3 2 1 14 6

E7 3 2 2 17 3

E8 4 1 1 18 3

E9 3 3 1 19 3

E10 4 2 1 21 6

E11 3 3 2 22 3

E12 4 2 2 24 3

E13 4 3 1 26 6

E143 3 3

27 45 1 1

Problem 4.2 (cont.)

(3 3 3) with d = 1

(5 1 1) with d = 3

E1 1 1 1 3 1

E2 2 1 1 6 3

E3 2 2 1 9 3

E4 3 1 1 11 3

E5 2 2 2 12 1

E6 3 2 1 14 6

E7 3 2 2 17 3

E8 4 1 1 18 3

E9 3 3 1 19 3

E10 4 2 1 21 6

E11 3 3 2 22 3

E12 4 2 2 24 3

E13 4 3 1 26 6

E143 3 3

27 45 1 1

Problem 4.2 (cont.)

(3 3 3) with d = 1

(5 1 1) with d = 3

E1 1 1 1 3 1

E2 2 1 1 6 3

E3 2 2 1 9 3

E4 3 1 1 11 3

E5 2 2 2 12 1

E6 3 2 1 14 6

E7 3 2 2 17 3

E8 4 1 1 18 3

E9 3 3 1 19 3

E10 4 2 1 21 6

E11 3 3 2 22 3

E12 4 2 2 24 3

E13 4 3 1 26 6

E143 3 3

27 45 1 1

Problem 4.2 (cont.)

(3 3 3) with d = 1

(5 1 1) with d = 3

E1 1 1 1 3 1

E2 2 1 1 6 3

E3 2 2 1 9 3

E4 3 1 1 11 3

E5 2 2 2 12 1

E6 3 2 1 14 6

E7 3 2 2 17 3

E8 4 1 1 18 3

E9 3 3 1 19 3

E10 4 2 1 21 6

E11 3 3 2 22 3

E12 4 2 2 24 3

E13 4 3 1 26 6

E143 3 3

27 45 1 1

Problem 4.2 (cont.)

(3 3 3) with d = 1

(5 1 1) with d = 3

E1 1 1 1 3 1

E2 2 1 1 6 3

E3 2 2 1 9 3

E4 3 1 1 11 3

E5 2 2 2 12 1

E6 3 2 1 14 6

E7 3 2 2 17 3

E8 4 1 1 18 3

E9 3 3 1 19 3

E10 4 2 1 21 6

E11 3 3 2 22 3

E12 4 2 2 24 3

E13 4 3 1 26 6

E143 3 3

27 45 1 1

Problem 4.2 (cont.)

(3 3 3) with d = 1

(5 1 1) with d = 3

E1 1 1 1 3 1

E2 2 1 1 6 3

E3 2 2 1 9 3

E4 3 1 1 11 3

E5 2 2 2 12 1

E6 3 2 1 14 6

E7 3 2 2 17 3

E8 4 1 1 18 3

E9 3 3 1 19 3

E10 4 2 1 21 6

E11 3 3 2 22 3

E12 4 2 2 24 3

E13 4 3 1 26 6

E143 3 3

27 45 1 1

Problem 4.2 (cont.)

(3 3 3) with d = 1

(5 1 1) with d = 3

E1 1 1 1 3 1

E2 2 1 1 6 3

E3 2 2 1 9 3

E4 3 1 1 11 3

E5 2 2 2 12 1

E6 3 2 1 14 6

E7 3 2 2 17 3

E8 4 1 1 18 3

E9 3 3 1 19 3

E10 4 2 1 21 6

E11 3 3 2 22 3

E12 4 2 2 24 3

E13 4 3 1 26 6

E143 3 3

27 45 1 1

Problem 5.10

Qualitatively discuss the energy level scheme for helium if (a) electronswere spin 1

2 bosons, and (b) spin 12 distinguishable particles

(a) Bosons

the ground state is spatially symmetricand so must have a symmetric (triplet)spin state and be orthohelium

the excited states come both in ortho(triplet) and para (singlet) form and theformer have symmetric spatial functionswhile the latter have antisymmetric spa-tial function

because all orthohelium states have spa-tially symmetric functions, they will havehigher energies than the parahelium states

(b) Distinguishable particles

every state has both orthoand para configurations andare therefore quadruply de-generate since the spatiallysymmetric and antisymmet-ric functions can be associ-ated with any spin state.

Problem 5.10

(a) Bosons

Problem 5.10

(a) Bosons

Problem 5.10

(a) Bosons

Problem 5.10

(a) Bosons

Problem 5.10

(a) Bosons

Problem 5.10

(a) Bosons

Problem 5.15

Find the average energy per free electron (Etot/Nq), as a fraction of theFermi energy

For the free electron gas, the Fermienergy is defined by the filled oc-tant of the sphere in reciprocalspace

the energy in one octant of a shellof radius k is given by

so the total energy can be com-puted

3πk3F

)kF = (3ρπ2)1/3, ρ ≡ Nq

EF =~2k2F2m

2m(3ρπ2)2/3

dE =~2k2

π2k2 dk

Etot =~2V

∫ kF

0k4 dk =

~2k5FV10π2m

Etot/Nq

~2k5FV10π2m

~2k2F=

5π2Nq=

3ρπ2

5π2ρ=

Problem 5.15

3πk3F

)kF = (3ρπ2)1/3, ρ ≡ Nq

EF =~2k2F2m

2m(3ρπ2)2/3

dE =~2k2

π2k2 dk

Etot =~2V

∫ kF

0k4 dk =

~2k5FV10π2m

Etot/Nq

~2k5FV10π2m

~2k2F=

5π2Nq=

3ρπ2

5π2ρ=

Problem 5.15

3πk3F

kF = (3ρπ2)1/3, ρ ≡ Nq

EF =~2k2F2m

2m(3ρπ2)2/3

dE =~2k2

π2k2 dk

Etot =~2V

∫ kF

0k4 dk =

~2k5FV10π2m

Etot/Nq

~2k5FV10π2m

~2k2F=

5π2Nq=

3ρπ2

5π2ρ=

Problem 5.15

3πk3F

)kF = (3ρπ2)1/3, ρ ≡ Nq

EF =~2k2F2m

2m(3ρπ2)2/3

dE =~2k2

π2k2 dk

Etot =~2V

∫ kF

0k4 dk =

~2k5FV10π2m

Etot/Nq

~2k5FV10π2m

~2k2F=

5π2Nq=

3ρπ2

5π2ρ=

Problem 5.15

3πk3F

)kF = (3ρπ2)1/3, ρ ≡ Nq

EF =~2k2F2m

2m(3ρπ2)2/3

dE =~2k2

π2k2 dk

Etot =~2V

∫ kF

0k4 dk =

~2k5FV10π2m

Etot/Nq

~2k5FV10π2m

~2k2F=

5π2Nq=

3ρπ2

5π2ρ=

Problem 5.15

3πk3F

)kF = (3ρπ2)1/3, ρ ≡ Nq

EF =~2k2F2m

2m(3ρπ2)2/3

dE =~2k2

π2k2 dk

Etot =~2V

∫ kF

0k4 dk =

~2k5FV10π2m

Etot/Nq

~2k5FV10π2m

~2k2F=

5π2Nq=

3ρπ2

5π2ρ=

Problem 5.15

3πk3F

)kF = (3ρπ2)1/3, ρ ≡ Nq

EF =~2k2F2m

2m(3ρπ2)2/3

dE =~2k2

π2k2 dk

Etot =~2V

∫ kF

0k4 dk =

~2k5FV10π2m

Etot/Nq

~2k5FV10π2m

~2k2F=

5π2Nq=

3ρπ2

5π2ρ=

Problem 5.15

3πk3F

)kF = (3ρπ2)1/3, ρ ≡ Nq

EF =~2k2F2m

2m(3ρπ2)2/3

dE =~2k2

π2k2 dk

Etot =~2V

∫ kF

0k4 dk =

~2k5FV10π2m

Etot/Nq

~2k5FV10π2m

~2k2F=

5π2Nq=

3ρπ2

5π2ρ=

Problem 5.15

3πk3F

)kF = (3ρπ2)1/3, ρ ≡ Nq

EF =~2k2F2m

2m(3ρπ2)2/3

dE =~2k2

π2k2 dk

Etot =~2V

∫ kF

0k4 dk =

~2k5FV10π2m

Etot/Nq

~2k5FV10π2m

~2k2F=

5π2Nq=

3ρπ2

5π2ρ=

Problem 5.15

3πk3F

)kF = (3ρπ2)1/3, ρ ≡ Nq

EF =~2k2F2m

2m(3ρπ2)2/3

dE =~2k2

π2k2 dk

Etot =~2V

∫ kF

0k4 dk

=~2k5FV10π2m

Etot/Nq

~2k5FV10π2m

~2k2F=

5π2Nq=

3ρπ2

5π2ρ=

Problem 5.15

3πk3F

)kF = (3ρπ2)1/3, ρ ≡ Nq

EF =~2k2F2m

2m(3ρπ2)2/3

dE =~2k2

π2k2 dk

Etot =~2V

∫ kF

0k4 dk =

~2k5FV10π2m

Etot/Nq

~2k5FV10π2m

~2k2F=

5π2Nq=

3ρπ2

5π2ρ=

Problem 5.15

3πk3F

)kF = (3ρπ2)1/3, ρ ≡ Nq

EF =~2k2F2m

2m(3ρπ2)2/3

dE =~2k2

π2k2 dk

Etot =~2V

∫ kF

0k4 dk =

~2k5FV10π2m

Etot/Nq

~2k5FV10π2m

5π2Nq=

3ρπ2

5π2ρ=

Problem 5.15

3πk3F

)kF = (3ρπ2)1/3, ρ ≡ Nq

EF =~2k2F2m

2m(3ρπ2)2/3

dE =~2k2

π2k2 dk

Etot =~2V

∫ kF

0k4 dk =

~2k5FV10π2m

Etot/Nq

~2k5FV10π2m

~2k2F=

5π2Nq

=3ρπ2

5π2ρ=

Problem 5.15

3πk3F

)kF = (3ρπ2)1/3, ρ ≡ Nq

EF =~2k2F2m

2m(3ρπ2)2/3

dE =~2k2

π2k2 dk

Etot =~2V

∫ kF

0k4 dk =

~2k5FV10π2m

Etot/Nq

~2k5FV10π2m

~2k2F=

5π2Nq=

3ρπ2

5π2ρ

Problem 5.15

3πk3F

)kF = (3ρπ2)1/3, ρ ≡ Nq

EF =~2k2F2m

2m(3ρπ2)2/3

dE =~2k2

π2k2 dk

Etot =~2V

∫ kF

0k4 dk =

~2k5FV10π2m

Etot/Nq

~2k5FV10π2m

~2k2F=

5π2Nq=

3ρπ2

5π2ρ=

Problem 4.31

Construct the spin matrices (Sx , Sy , and Sz) for a particle of spin 1.

With spin 1, there are three possible values of m: -1, 0, 1 which meansthat we have spin eigenstates as follows

, χ0 =

, χ− =

Since these spinors are eigenstates of Sz , we can construct the matrix

representation of the operator as follows:

Szχ+ = ~χ+

Szχ0 = 0

Szχ− = −~χ−

~ 0 00 0 00 0 − ~

1 0 00 0 00 0 −1

Problem 4.31

, χ0 =

, χ− =

Szχ+ = ~χ+

Szχ0 = 0

Szχ− = −~χ−

~ 0 00 0 00 0 − ~

1 0 00 0 00 0 −1

Problem 4.31

, χ0 =

, χ− =

Since these spinors are eigenstates of Sz , we can construct the matrixrepresentation of the operator as follows:

Szχ+ = ~χ+

Szχ0 = 0

Szχ− = −~χ−

~ 0 00 0 00 0 − ~

1 0 00 0 00 0 −1

Problem 4.31

, χ0 =

, χ− =

Szχ+ = ~χ+

Szχ0 = 0

Szχ− = −~χ−

~ 0 00 0 00 0 − ~

1 0 00 0 00 0 −1

Problem 4.31

, χ0 =

, χ− =

Szχ+ = ~χ+

Szχ0 = 0

Szχ− = −~χ−

0 − ~

1 0 00 0 00 0 −1

Problem 4.31

, χ0 =

, χ− =

Szχ+ = ~χ+

Szχ0 = 0

Szχ− = −~χ−

1 0 00 0 00 0 −1

Problem 4.31

, χ0 =

, χ− =

Szχ+ = ~χ+

Szχ0 = 0

Szχ− = −~χ−

~ 0 00 0 00 0 − ~

1 0 00 0 00 0 −1

Problem 4.31

, χ0 =

, χ− =

Szχ+ = ~χ+

Szχ0 = 0

Szχ− = −~χ−

~ 0 00 0 00 0 − ~

1 0 00 0 00 0 −1

Now build the raising and lowering operators

S+χ+ = 0

S+χ0 = ~√

S+χ− = ~√

0 1 00 0 10 0 0

S−χ+ = ~√

S−χ0 = ~√

2χ−

S−χ− = 0

S− =

0 0 01 0 00 1 0

finally, use Sx = 1

2(S+ + S−) and Sy = 12i (S+ − S−)

Sx =~√2

0 1 01 0 10 1 0

Sy =i~√

0 −1 01 0 −10 1 0

S+χ+ = 0

S+χ0 = ~√

S+χ− = ~√

S−χ+ = ~√

S−χ0 = ~√

2χ−

S−χ− = 0

S− =

0 0 01 0 00 1 0

finally, use Sx = 1

2(S+ + S−) and Sy = 12i (S+ − S−)

Sx =~√2

0 1 01 0 10 1 0

Sy =i~√

0 −1 01 0 −10 1 0

S+χ+ = 0

S+χ0 = ~√

S+χ− = ~√

S+ =√

S−χ+ = ~√

S−χ0 = ~√

2χ−

S−χ− = 0

S− =

0 0 01 0 00 1 0

finally, use Sx = 1

2(S+ + S−) and Sy = 12i (S+ − S−)

Sx =~√2

0 1 01 0 10 1 0

Sy =i~√

0 −1 01 0 −10 1 0

S+χ+ = 0

S+χ0 = ~√

S+χ− = ~√

S+ =√

0 1 00 0 10 0 0

S−χ+ = ~√

S−χ0 = ~√

2χ−

S−χ− = 0

S− =

0 0 01 0 00 1 0

finally, use Sx = 1

2(S+ + S−) and Sy = 12i (S+ − S−)

Sx =~√2

0 1 01 0 10 1 0

Sy =i~√

0 −1 01 0 −10 1 0

S+χ+ = 0

S+χ0 = ~√

S+χ− = ~√

S+ =√

0 1 00 0 10 0 0

S−χ+ = ~√

S−χ0 = ~√

2χ−

S−χ− = 0

S− =

0 0 01 0 00 1 0

finally, use Sx = 12(S+ + S−) and Sy = 1

2i (S+ − S−)

Sx =~√2

0 1 01 0 10 1 0

Sy =i~√

0 −1 01 0 −10 1 0

S+χ+ = 0

S+χ0 = ~√

S+χ− = ~√

S+ =√

0 1 00 0 10 0 0

S−χ+ = ~√

S−χ0 = ~√

2χ−

S−χ− = 0

S− =√

2i (S+ − S−)

Sx =~√2

0 1 01 0 10 1 0

Sy =i~√

0 −1 01 0 −10 1 0

S+χ+ = 0

S+χ0 = ~√

S+χ− = ~√

S+ =√

0 1 00 0 10 0 0

S−χ+ = ~√

S−χ0 = ~√

2χ−

S−χ− = 0

S− =√

2i (S+ − S−)

Sx =~√2

0 1 01 0 10 1 0

Sy =i~√

0 −1 01 0 −10 1 0

S+χ+ = 0

S+χ0 = ~√

S+χ− = ~√

S+ =√

0 1 00 0 10 0 0

S−χ+ = ~√

S−χ0 = ~√

2χ−

S−χ− = 0

S− =√

0 0 01 0 00 1 0

2i (S+ − S−)

Sx =~√2

0 1 01 0 10 1 0

Sy =i~√

0 −1 01 0 −10 1 0

S+χ+ = 0

S+χ0 = ~√

S+χ− = ~√

S+ =√

0 1 00 0 10 0 0

S−χ+ = ~√

S−χ0 = ~√

2χ−

S−χ− = 0

S− =√

0 0 01 0 00 1 0

finally, use Sx = 1

2(S+ + S−) and Sy = 12i (S+ − S−)

Sx =~√2

0 1 01 0 10 1 0

Sy =i~√

0 −1 01 0 −10 1 0

S+χ+ = 0

S+χ0 = ~√

S+χ− = ~√

S+ =√

0 1 00 0 10 0 0

S−χ+ = ~√

S−χ0 = ~√

2χ−

S−χ− = 0

S− =√

0 0 01 0 00 1 0

finally, use Sx = 1

2(S+ + S−) and Sy = 12i (S+ − S−)

Sx =~√2

0 1 01 0 10 1 0

Sy =i~√

0 −1 01 0 −10 1 0

S+χ+ = 0

S+χ0 = ~√

S+χ− = ~√

S+ =√

0 1 00 0 10 0 0

S−χ+ = ~√

S−χ0 = ~√

2χ−

S−χ− = 0

S− =√

0 0 01 0 00 1 0

finally, use Sx = 1

2(S+ + S−) and Sy = 12i (S+ − S−)

Sx =~√2

0 1 01 0 10 1 0

Sy =i~√

0 −1 01 0 −10 1 0

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