Quantum Mechanics - Probability

The wave function for a particle constrained to move in one dimension is shown in the graph (Ψ = 0 for x ≤  0 and x ≥ 5). What is the probability that the particle would be found between x = 2 and x = 4?

A. 17/64
B. 25/64
C. 5/8
D. v(5/8)
E. 13/16

(GR9677 #17)

Solution:

Probability, P ∼ ψ²

Probability to find the particle between x = 2 and x = 4 (unnormalized probability):
ψ² = 2²  + 3² = 13

Total probability (normalized probability):
ψ² = 1²  +  1²  + 2²  + 3²  +  1² = 16

P = unnormalized probability / normalized probability = 13/16

Answer: E

Quantum Mechanics - Spherical Harmonics

A diatomic molecules is initially in the state Ψ(Θ, Φ) = (5Y₁¹ + 3Y₅¹ + 2Y₅⁻¹) / (38)^½ where Y_l^m is a spherical harmonics. If measurements are made of the total angular momentum quantum number l and of azimuthal angular momentum quantum number m, what is the probability of obtaining the results l = 5?

A. 36/1444
B. 9/38
C. 13/38
D. 5/(38)^½
E. 34/38

(GR9677 #33)

Solution:

l = 5 → Y₅¹ and Y₅⁻¹

Probability, P = ∑_i |c_i|² 

P = |3/√38|² + |2/√38|² =  9/38 + 4/38 =  13/38

Answer: C

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 - Probability

A system is know to be in the normalized state described by the wave function



Where  are the spherical harmonics. The probability of finding the system in a state with azimuthal orbital quantum number m = 3 is

A. 0
B. 1/15
C. 1/6
D. 1/3
E. 13/15

(GR9277 #28)

Solution:

For the state  ,  probability is  

States with  m = 3 → and  with  c₁ = 5/√30 and c₂ = 1/√30

Therefore,



Answer: E

Thermal Physics - Probability

Which of the following is true if the arrangement of an isolated thermodynamic system is of maximal probability?

A. Spontaneous change to a lower probability occurs
B. The entropy is minimum
C. Botzmann’s constant approaches zero
D. No spontaneous change occurs
E. The entropy is zero

(GR9277 #63)

Solution:

Maximal Probability → highest/maximum entropy → most stable state /equilibrium → No spontaneous change occurs. Spontaneous change occurs when the system is far from equilibrium.

Answer: D

Thermal Physics - Probability

A sample of N atoms of helium gas is confined in a 1.0 cubic meter volume. The probability that none of the helium atoms is in a 10⁻⁶ cubic meter volume of the container is

A. 0
B. (10⁻⁶)^N
C. (1 − 10⁻⁶)^N
D. 1 − (10⁻⁶)^N
E. 1

(GR8677 #15)

Solution:

Total Probability: P = P₁ + P₂ = 1
P₁ = The probability that one atom is in a 10⁻⁶ cubic meter volume of the container
P₂ =  The probability that none of the helium atoms is in a 10⁻⁶ cubic meter volume of the container
→ P₂ = 1 − P₁

P₁ = 1/n
n = number of 10⁻⁶ m³ cubes in the 1 m³ volume
10⁻⁶ × n = 1 → n = 1/10⁻⁶ = 10⁶ cubes
→ P₁ = 1/n = 1/10⁶ = 10⁻⁶
→ P₂ = 1 − P₁ = 1 − 10⁻⁶
For N atoms: P₂ = (1 − 10⁻⁶)^N

Answer: C

Quantum Mechanics - Bohr Radius

The solution to the Schrodinger equation for the ground state of hydrogen is



where a₀ is the Bohr radius and r is the distance from the origin. Which of the following is the most probable value for r?

A. 0
B. a₀ / 2
C. a₀
D. 2a₀
E. ∞

(GR0177 #93)

Solution:

Probability



The most probable value of r corresponds to the peak of the plot of P(r) versus r.
The slope of the curve at this point is zero.





For  

r = a₀

Answer: C