Schrodinger Equation Questions and Answers

Q: A particle is subjected to a potential\[ V(x) = \begin{cases} \infty, & x \le 0 \\ V_{0}, & a \le x \le b \\ 0, & \text{elsewhere} \end{cases} \]Here, \(a \gt 0\) and \(b \gt a\). If the energy of the particle \(E \lt V_{0}\), which one of the following schematics is a valid quantum mechanical wavefunction \((\Psi)\) for the system?

The valid wavefunction is (B).The requirements are:Ψ = 0 at x = 0, because the potential is infinite for x ≤ 0;between 0 and a, where E > V = 0, Ψ oscillates (here it rises from zero to a maximum);inside the barrier, where E < V0, Ψ is a combination of rising and falling exponentials, so it cannot oscillate or develop a local peak;beyond b, Ψ oscillates again; Ψ and its slope are continuous at a and b.Sketch (B) meets all four. Sketch (A) has a bump inside the barrier, which an exponential cannot produce. Sketch (C) oscillates inside the barrier, as if E were above V0. Sketch (D) looks like (B) but is not zero at x = 0, violating the condition at the infinite wall.

Q: The wavefunction for a particle is given by the form \(e^{-(i\alpha x + \beta)}\), where \(\alpha\) and \(\beta\) are real constants. In which one of the following potentials \(V(x)\), the particle is moving?

The particle moves in the potential V(x) = 0.The wavefunction \(e^{-(i\alpha x + \beta)} = e^{-\beta}e^{-i\alpha x}\) is a plane wave with constant amplitude. Then \(\psi'' = -\alpha^{2}\psi\), and the time-independent Schrödinger equation gives\[ \frac{\hbar^{2}\alpha^{2}}{2m}\psi + V(x)\psi = E\psi \]For this to hold at every x, V must be the constant \(E - \hbar^{2}\alpha^{2}/2m\), which with the usual choice of zero is V = 0: a free particle with momentum \(\hbar\alpha\) moving in the -x direction.A harmonic, exponential or sinusoidal potential would vary with x and force the amplitude or wavelength of ψ to change, which a pure plane wave never does.

Q: The Hamiltonian \(H = \frac{-\hbar^{2}}{2m}\frac{d^{2}}{dx^{2}} - \alpha\delta(x)\) is given for a particle. Using the trial wave function \(\psi_{trial}(x) = Ae^{-bx^{2}}\) (with b as the variational parameter), the bound on the ground state energy is :(Given : \(\int_{-\infty}^{\infty} x^{2n}e^{-\alpha x^{2}}\,dx\) = \(\Gamma(n+1/2)/\alpha^{n+1/2}\))

The bound is \(-m\alpha^{2}/\pi\hbar^{2}\).Normalising the Gaussian gives \(|A|^{2} = \sqrt{2b/\pi}\). The kinetic and potential terms are:\[ \begin{aligned} \langle T \rangle &= \frac{\hbar^{2}b}{2m} \\ \langle V \rangle &= -\alpha|\psi(0)|^{2} = -\alpha\sqrt{\frac{2b}{\pi}} \end{aligned} \]Setting \(dE/db = 0\) gives \(\sqrt{b} = \frac{m\alpha}{\hbar^{2}}\sqrt{\frac{2}{\pi}}\), and then:\[ \begin{aligned} E_{min} &= \frac{m\alpha^{2}}{\pi\hbar^{2}} - \frac{2m\alpha^{2}}{\pi\hbar^{2}} \\ &= -\frac{m\alpha^{2}}{\pi\hbar^{2}} \end{aligned} \]The option \(-m\alpha^{2}/2\hbar^{2}\) is the exact ground-state energy of the delta well. The variational bound lies above it, as it must, since \(1/\pi \lt 1/2\). The positive option cannot describe a bound state.

Q: Which one of the following given below represents the bound state eigen function of the operator \(-\frac{d^{2}}{dx^{2}}\) in the region \(0 \le x \le \infty\) with eigen value -4 ?

The bound state eigenfunction is \(A_{0}e^{-2x}\).The eigenvalue equation is \(-\frac{d^{2}\psi}{dx^{2}} = -4\psi\), that is \(\psi'' = 4\psi\), whose solutions are combinations of \(e^{2x}\) and \(e^{-2x}\), or equivalently \(\cosh 2x\) and \(\sinh 2x\). All four options satisfy the equation.A bound state must be normalisable on \(0 \le x \lt \infty\), so \(\psi\) must vanish as \(x \to \infty\). Only \(e^{-2x}\) does; \(e^{2x}\), \(\cosh 2x\) and \(\sinh 2x\) all grow without limit, so they cannot describe a bound state.

A particle is subjected to a potential

\[ V(x) = \begin{cases} \infty, & x \le 0 \\ V_{0}, & a \le x \le b \\ 0, & \text{elsewhere} \end{cases} \]

Here, \(a \gt 0\) and \(b \gt a\). If the energy of the particle \(E \lt V_{0}\), which one of the following schematics is a valid quantum mechanical wavefunction \((\Psi)\) for the system?

Exam Year(s)

GATE Physics - 2024,

Answer: B

The valid wavefunction is (B).

The requirements are:

Ψ = 0 at x = 0, because the potential is infinite for x ≤ 0;
between 0 and a, where E > V = 0, Ψ oscillates (here it rises from zero to a maximum);
inside the barrier, where E < V₀, Ψ is a combination of rising and falling exponentials, so it cannot oscillate or develop a local peak;
beyond b, Ψ oscillates again; Ψ and its slope are continuous at a and b.

Sketch (B) meets all four. Sketch (A) has a bump inside the barrier, which an exponential cannot produce. Sketch (C) oscillates inside the barrier, as if E were above V₀. Sketch (D) looks like (B) but is not zero at x = 0, violating the condition at the infinite wall.

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The wavefunction for a particle is given by the form \(e^{-(i\alpha x + \beta)}\), where \(\alpha\) and \(\beta\) are real constants. In which one of the following potentials \(V(x)\), the particle is moving?

A.\(V(x) \propto \alpha^{2}x^{2}\)
B.\(V(x) \propto e^{-\alpha x}\)
C.\(V(x) = 0\)
D.\(V(x) \propto \sin(\alpha x)\)

Exam Year(s)

GATE Physics - 2024,

Answer: C

The particle moves in the potential V(x) = 0.

The wavefunction \(e^{-(i\alpha x + \beta)} = e^{-\beta}e^{-i\alpha x}\) is a plane wave with constant amplitude. Then \(\psi'' = -\alpha^{2}\psi\), and the time-independent Schrödinger equation gives

\[ \frac{\hbar^{2}\alpha^{2}}{2m}\psi + V(x)\psi = E\psi \]

For this to hold at every x, V must be the constant \(E - \hbar^{2}\alpha^{2}/2m\), which with the usual choice of zero is V = 0: a free particle with momentum \(\hbar\alpha\) moving in the -x direction.

A harmonic, exponential or sinusoidal potential would vary with x and force the amplitude or wavelength of ψ to change, which a pure plane wave never does.

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The Hamiltonian \(H = \frac{-\hbar^{2}}{2m}\frac{d^{2}}{dx^{2}} - \alpha\delta(x)\) is given for a particle. Using the trial wave function \(\psi_{trial}(x) = Ae^{-bx^{2}}\) (with b as the variational parameter), the bound on the ground state energy is :

(Given : \(\int_{-\infty}^{\infty} x^{2n}e^{-\alpha x^{2}}\,dx\) = \(\Gamma(n+1/2)/\alpha^{n+1/2}\))

A.\(-m\alpha^{2}/2\hbar^{2}\)
B.\(-2m\alpha^{2}/\pi\hbar^{2}\)
C.\(-m\alpha^{2}/\pi\hbar^{2}\)
D.\(m\alpha^{2}/\pi\hbar^{2}\)

Exam Year(s)

MH SET - 2019, SET Exam - 2019,

Answer: C

The bound is \(-m\alpha^{2}/\pi\hbar^{2}\).

Normalising the Gaussian gives \(|A|^{2} = \sqrt{2b/\pi}\). The kinetic and potential terms are:

\[ \begin{aligned} \langle T \rangle &= \frac{\hbar^{2}b}{2m} \\ \langle V \rangle &= -\alpha|\psi(0)|^{2} = -\alpha\sqrt{\frac{2b}{\pi}} \end{aligned} \]

Setting \(dE/db = 0\) gives \(\sqrt{b} = \frac{m\alpha}{\hbar^{2}}\sqrt{\frac{2}{\pi}}\), and then:

\[ \begin{aligned} E_{min} &= \frac{m\alpha^{2}}{\pi\hbar^{2}} - \frac{2m\alpha^{2}}{\pi\hbar^{2}} \\ &= -\frac{m\alpha^{2}}{\pi\hbar^{2}} \end{aligned} \]

The option \(-m\alpha^{2}/2\hbar^{2}\) is the exact ground-state energy of the delta well. The variational bound lies above it, as it must, since \(1/\pi \lt 1/2\). The positive option cannot describe a bound state.

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Which one of the following given below represents the bound state eigen function of the operator \(-\frac{d^{2}}{dx^{2}}\) in the region \(0 \le x \le \infty\) with eigen value -4 ?

A.\(A_{0}e^{2x}\)
B.\(A_{0}e^{-2x}\)
C.\(A_{0}\text{cosh}\, 2x\)
D.\(A_{0}\text{sinh}\, 2x\)

Exam Year(s)

MH SET - 2020, SET Exam - 2020,

Answer: B

The bound state eigenfunction is \(A_{0}e^{-2x}\).

The eigenvalue equation is \(-\frac{d^{2}\psi}{dx^{2}} = -4\psi\), that is \(\psi'' = 4\psi\), whose solutions are combinations of \(e^{2x}\) and \(e^{-2x}\), or equivalently \(\cosh 2x\) and \(\sinh 2x\). All four options satisfy the equation.

A bound state must be normalisable on \(0 \le x \lt \infty\), so \(\psi\) must vanish as \(x \to \infty\). Only \(e^{-2x}\) does; \(e^{2x}\), \(\cosh 2x\) and \(\sinh 2x\) all grow without limit, so they cannot describe a bound state.

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