Quantum Mechanics - Schrodinger Equation

The solution to the Schrödinger equation for a particle bound in a one-dimensional, infinitely deep potential well, indexed by quantum number n, indicates that in the middle of the well the probability density vanishes for

A. The ground state (n = 1) only
B. States of even n (n = 2, 4, ...)
C. States of odd n (n = 1, 3, ...)
D. All states (n = 1, 2, 3, ...)
E. All states except the ground state

(GR9677 #51)

Solution:

Wavefunctions for the first 5 states for a particle bound in a 1-D, infinitely deep potential well

Pic: ecee.colorado.edu

The even wave functions (n = 2, 4, ...) always have nodes in the middle.

→ probability density for states of even n vanishes.

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 - Infinite Potential Well

See Problem #51

A measurement of energy E will always satisfy which of the following relationships?

A. E ≤ π²ħ²/8ma²

B. E ≥ π²ħ²/2ma²
C. E = π²ħ²/8ma²
D. E = n²π²ħ²/8ma²
E. E = π²ħ²/2ma²

(GR9277 #53)

Solution:

Wave function, 
Time Independent Schrodinger Equation:  

Inside the well, V(x) = 0, and







Answer: B

Quantum Mechanics - Infinite Potential Well

The figure above shows one of the possible energy eigenfunctions ψ(x) for a particle bouncing freely back and forth along the x-axis between impenetrable walls located at x = −a and x = +a. The potential energy equals zero for |x| > a. If the energy of the particle is 2 electron volts when it is in the quantum state associated with this eigenfunction, what is its energy when it is in the quantum state of lowest possible energy? 

A. 0 eV
B. 1/√2 eV
C. 1/2 eV
D. 1 eV
E. 2 eV

(GR8677 #90)

Solution:

Impenetrable walls = infinite potential walls.
The initial wavefunctions for the first four states in the system:

So, 

represents n = 2.

The energies for infinite potential walls:  E_n = n² E₁

Since E₂ = 2 eV → 2² E₁ = 2 → E₁ = 2/4 = 1/2 eV

Answer: C