Quantum Mechanics, Particle in a box

An attractive, one-dimensional square well has depth V₀ as shown above. Which of the following best shows a possible wave function for a bound state?

(GR9277 #29)

Solution:

1-D square well with depth  V₀  → finite potential well.

Wave function of a particle in finite potential well should have these following properties:

• ψ →0 as x → ± ∞ or get further into regions where the classical particle cannot penetrate at all due to its inadequate energy
• ψ is always a decaying exponential function
• ψ must be continuous and differentiable

Therefore:

A. FALSE. ψ does not go to zero as x → ±∞

C. FALSE. ψ is not continuous

D and E are FALSE. ψ are not decaying exponential functions

Answer: B

Quantum Mechanics - Infinite Potential Well

Questions 51-53

A particle of mass m is confined to an infinitely deep square-well potential:

V(x) = ∞, x ≤ 0, x ≥ a
V(x) = 0, 0  x  a

The normalized eigenfunction, labeled by the quantum number n, are



For any state n, the expectation value of the momentum of the particle is

A. 0

B. 

C. 

D. 

E. 

(GR9277 #51)

Solution:

Infinitely deep square-well potential
→ there is zero probability for particle to be outside the well
→ 〈〉= 0

If 〈〉 ≠ 0 the particle would tend to go to the right or left and leave the well, which is impossible for infinitely deep square-well potential.

Answer: A


Alternative Answer #1:

Eigen function,  ψ_n = λψ_n
The eigenvalue, λ is associated with expectation value:〈Â〉= 〈ψ | Â | ψ〉

Since  is imaginer and ψ_n is real →  the eigenvalue, λ is imaginer = not real = not observable
→ 〈〉= 0


Alternative Answer #2:

since sine and cosine are orthogonal the whole period.

Quantum Mechanics - Infinite Potential Well

See Problem 51

The eigenfunctions satisfy the condition

₀∫^aψ_n*(x) ψ_l (x) dx = δ_nl

δ_nl = 1 if n = l, otherwise δ_nl = 0. This is a statement that the eigenfunctions are

A. Solutions to the Schrodinger equation 
B. Orthonormal
C. Bounded 
D. Linearly dependent
E. Symmetric

(GR9277 #52)

Solution:

Orthonormality = orthogonal and normal

₀∫^aψ_n*(x) ψ_l (x) dx = 〈ψ_n|ψ_l〉= δ_nl

Orthogonal,  δ_nl = 0
Normal, δ_nl = 1 if n = l

Answer: B

Quantum Mechanics - Particle in a Box

The energy eigenstates for a particle of mass m in a box of length L have wave functions



and energies



where =1,2,3,... . At time t = 0, the particle is in a state described as follows.



Which of the following is a possible result of a measurement of energy for the state

A. 2E₁
B. 5E₁
C. 7E₁
D. 9E₁
E. 14E₁

(GR0177 #44)

Solution:

The energy:



So, the possible result of a measurement of energy is a squared quantity:


... and so on.

The only choice is (D)

Answer: D

Quantum Mechanics – Perturbation Theory

A particle of mass M is in an infinitely deep square well potential   where

      for    ,         and
    for    .

A very small perturbing potential '  is superimposed on   such that

      for            , and

                         for       .

If      are the energy eigenfunctions for a particle in the infinitely deep square well potential, with      being the ground state, which of the following statements is correct about the eigenfunction     of a particle in the perturbed potential ?

A.      

B.       with        for all odd values of n.

C.       with        for all even values of n.

D.        with       for all values of n.

E.    None of the above

(GR8677 #96)

Solution:

The eigenfunctions of the unperturbed infinite deep well form a complete set (or complete basis).

This means that any function can be represented by the old unperturbed infinite deep well.

Thus, the solution to the perturbed eigenfunction should look like .



Also, since the perturbed potential is also symmetrical with respect to the origin (as the original unperturbed potential was, too), one knows that all the odd terms should go to 0.

Answer: B