Physics Problems & Solutions: Schrodinger Equation

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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

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, $\psi_n=\sqrt{\frac{2}{a}} \sin \frac{n\pi x}{a}$
Time Independent Schrodinger Equation: $-\frac{\hbar^{2}}{2m}\psi''+ V\psi=E\psi$

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

$\\\psi=A\sin kx \\\psi'=Ak\cos kx \\\psi''=-Ak^2\sin kx$

$\\-\frac{\hbar^{2}}{2m}(-Ak^2 \sin kx)=EA\sin kx\rightarrow E=\frac{\hbar^{2}k^2}{2m}$

$\\k=\frac{n\pi}{a}\rightarrow E=\frac{n^2\pi^2 \hbar^{2}}{2ma^2} \\\\n=1,2,3, ...\rightarrow E\geqslant \frac{\pi^2 \hbar^{2}}{2ma^2}$

Answer: B

Quantum Mechanics - Schrodinger Equation

The wave function ψ(x) = A exp (−b²x²/2), where A and b are real constants, is a normalized eigenfunction of the Schrodinger equation for a particle of mass M and energy E in a one dimensional potential V(x) such that V(x) = 0 at x = 0. Which of the following is correct?

A. V = ħ²b⁴/2M
B. V = ħ²b⁴x²/2M
C. V = ħ²b⁶x⁴/2M
D. E = ħ²b²(1 − b²x²)
E. E = ħ²b⁴/2M

(GR8677 #18)

Solution:

Schrodinger Equation: $-\frac{\hbar^{2}}{2m}\frac{\partial^{2}\psi(x)}{\partial x^{2}}+V(x)\psi(x)=E\psi(x)$

$\begin{aligned} \frac{\partial\psi}{\partial x}&=\frac{\partial}{\partial x}e^{-b^2x^2/2}=-b^2x\psi \\\frac{\partial^{2}\psi}{\partial x^{2}}&=-b^2\frac{\partial}{\partial x}x\psi=-b^2\left( x\frac{\partial\psi}{\partial x}+\psi\frac{\partial x}{\partial x}\right ) \\\frac{\partial^{2}\psi}{\partial x^{2}}&=-b^2(-b^2x^2\psi+\psi)=(b^4x^2-b^2)\psi\end{aligned}$

Schrodinger Equation: $-\frac{\hbar^{2}}{2m}(b^4x^2-b^2)\psi(x)+V(x)\psi(x)=E\psi(x)$

To find E → V(x = 0) = 0:

$\\-\frac{\hbar^{2}}{2m}(0-b^2)+0=E \\\rightarrow E=\frac{\hbar^{2}b^2}{2m}$

To find V(x):
$\\-\frac{\hbar^{2}}{2m}(b^4x^2-b^2)+V(x)=E \\\\-\frac{\hbar^{2}b^4x^2}{2m}+\frac{\hbar^{2}b^2}{2m}+V(x)=E \\\\-\frac{\hbar^{2}b^4x^2}{2m}+E+V(x)=E \\\\\rightarrow V(x)=\frac{\hbar^{2}b^4x^2}{2m}$

Answer: B

Quantum Mechanics - Schrodinger Equation

The Hamiltonian operator in the Schrodinger equation can be formed from the classical Hamiltonian by substituting

A. Wavelength and frequency for momentum and energy
B. A differential operator for momentum
C. Transition probability for potential energy
D. Sums over discrete eigenvalues for integrals over continuous variables
E. Gaussian distributions of observables for exact values

(GR8677 #49)

Solution:

Schrodinger Equation: Hψ(x) = Eψ(x)

Hamiltonian: $H=\frac{\hat p^2}{2m}+V(x)$

Momentum operator : $\hat p=-i\hbar \frac{\partial }{\partial x}$

Answer: B

Quantum Mechanics - Expectation Value

If ψ is a normalized solution of the Schrodinger equation and Q is the operator corresponding to a physical observable x and the quantity ψ*Qψ may be integrated in order to obtain the

A. Normalization constant for ψ
B. Spatial overlap of Q with ψ
C. Mean value of x
D. Uncertainty in x
E. Time derivative of x

(GR8677 #56)

Solution:

Expectation value: $\langle Q\rangle=\int \psi(x)^*Q\psi(x)dx$

The predicted mean value of the result of an experiment.
The average value of the measurable quantity Q (observable x) on the state ψ

Answer: C

Quantum Mechanics - Bohr Radius

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

$\psi_0=\frac{1}{\sqrt{\pi a^3_0}}e^{-r/a_0}$

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

$\begin{aligned} P(r)dr &=|\psi_0|^2dV\\&=|\psi_0|^2d \left(\frac{4}{3}\pi r^3 \right)\\&=|\psi_0|^2 4\pi r^2 dr\\&=\frac{4 r^2}{a^3_0} e^{-2r/a_0}dr \end{aligned}$

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.

$\frac{dP(r)}{dr}=0$

$\begin{aligned} \frac{dP(r)}{dr} &=\frac{d}{dr}\frac{4 r^2}{a^3_0} e^{-2r/a_0} \\&=\frac{4}{a^3_0}\frac{d}{dr}r^2e^{-2r/a_0}\\&=\frac{4}{a^3_0}\left(2re^{-2r/a_0}+ \frac{-2}{a_0}e^{-2r/a_0}r^2\right)\\&=\frac{8 r}{a^3_0}e^{-2r/a_0}\left(1-\frac{r}{a_0} \right ) \end{aligned}$

For $\frac{dP(r)}{dr}=0$

$1-\frac{r}{a_0}=0\rightarrow r = a_0$

r = a₀

Answer: C