Wave Function and Probability Questions and Answers

Q: A quantum particle is confined in a one-dimensional space of \(0 \le x \le 2\). Consider a normalized wavefunction of the particle as\[ \psi(x) = \sqrt{p/5}\,[1 + \cos(\pi x/2)] \sin(\pi x/2). \]Which of the following options gives the correct value of \(p\)?

Normalisation gives \(p = 4\).The wavefunction must satisfy \(\int_0^2 |\psi|^2 dx = 1\). Substitute \(u = \pi x/2\), so \(dx = \frac{2}{\pi}du\) and \(u\) runs from 0 to \(\pi\):\[ \begin{aligned} 1 &= \frac{p}{5} \cdot \frac{2}{\pi} \int_0^{\pi} I(u)\, du \\ I(u) &= (1 + \cos u)^{2} \sin^{2} u \end{aligned} \]Expanding, \((1 + \cos u)^2 \sin^2 u = \sin^2 u + 2\cos u \sin^2 u + \cos^2 u \sin^2 u\). Over 0 to \(\pi\) these integrate to \(\frac{\pi}{2}\), 0 and \(\frac{\pi}{8}\):\[ \begin{aligned} 1 &= \frac{p}{5} \cdot \frac{2}{\pi} \cdot \frac{5\pi}{8} \\ &= \frac{p}{4}, \quad p = 4 \end{aligned} \]The middle term vanishes because \(\sin^3 u\) is the same at both limits. The values 1, 2 and 3 would leave the total probability at a quarter, a half or three quarters.

Q: A particle is represented by a plane wave in position space. Its wave function in momentum space is :

Its momentum-space wave function is a delta function.A plane wave \( e^{ip_{0}x/\hbar} \) has one definite momentum \( p_{0} \). Its Fourier transform, the momentum-space wave function, is \( \phi(p) \propto \delta(p - p_{0}) \): all the probability sits at \( p_{0} \). This is the uncertainty principle at its extreme, with \( \Delta x \) infinite and \( \Delta p = 0 \).A Gaussian in position transforms into a Gaussian in momentum (a wave packet), and a Lorentzian is the transform of an exponentially decaying function. A plane wave transforms into a plane wave only for a delta function in position, the reverse case.

Q: Which of the following functions can be an acceptable solution for Schrödinger equation for all values of x ?

The acceptable function is \(\psi(x) = Ae^{-\alpha x^{2}}\) (with \(\alpha \gt 0\)).An acceptable wave function must be finite, single-valued and continuous everywhere, and square integrable. The Gaussian is all of these: it is smooth and falls to zero at large |x|, so \(\int|\psi|^{2}dx\) is finite. It is the ground state of the harmonic oscillator.\(Ae^{\alpha x^{2}}\) blows up as \(x \to \pm\infty\) and cannot be normalised. A tan x and A sec x become infinite at \(x = \pi/2\) and its repeats, so they fail for some values of x.

Q: The normalized wave function of a particle can be written as\[ \begin{aligned} \psi(x) = N\sum_{n=0}^{\infty}&\left(\frac{1}{\sqrt{7}}\right)^{n} \\ &\phi_{n}(x) \end{aligned} \]where \(\phi_{n}(x)\) are the normalized energy eigenfunctions of a given Hamiltonian. The value of N is :

The normalisation constant is \(N = \sqrt{6/7}\).The eigenfunctions are orthonormal, so the norm of \(\psi\) is the sum of the squared moduli of its coefficients:\[ \begin{aligned} 1 &= N^{2}\sum_{n=0}^{\infty}\left(\frac{1}{7}\right)^{n} \\ &= N^{2}\cdot\frac{1}{1 - 1/7} = N^{2}\cdot\frac{7}{6} \end{aligned} \]so \(N^{2} = 6/7\) and \(N = \sqrt{6/7}\).The coefficients \((1/\sqrt{7})^{n}\) must be squared before summing, giving a geometric series of ratio 1/7. Forgetting to square them, or stopping after the first term (which gives \(N = 1\)), leads to the wrong values; \(\sqrt{1/7}\), \(\sqrt{3/7}\) and \(\sqrt{13/7}\) do not make the sum equal to 1.

A quantum particle is confined in a one-dimensional space of \(0 \le x \le 2\). Consider a normalized wavefunction of the particle as

\[ \psi(x) = \sqrt{p/5}\,[1 + \cos(\pi x/2)] \sin(\pi x/2). \]

Which of the following options gives the correct value of \(p\)?

Exam Year(s)

IIT JAM - 2026,

Answer: C

Normalisation gives \(p = 4\).

The wavefunction must satisfy \(\int_0^2 |\psi|^2 dx = 1\). Substitute \(u = \pi x/2\), so \(dx = \frac{2}{\pi}du\) and \(u\) runs from 0 to \(\pi\):

\[ \begin{aligned} 1 &= \frac{p}{5} \cdot \frac{2}{\pi} \int_0^{\pi} I(u)\, du \\ I(u) &= (1 + \cos u)^{2} \sin^{2} u \end{aligned} \]

Expanding, \((1 + \cos u)^2 \sin^2 u = \sin^2 u + 2\cos u \sin^2 u + \cos^2 u \sin^2 u\). Over 0 to \(\pi\) these integrate to \(\frac{\pi}{2}\), 0 and \(\frac{\pi}{8}\):

\[ \begin{aligned} 1 &= \frac{p}{5} \cdot \frac{2}{\pi} \cdot \frac{5\pi}{8} \\ &= \frac{p}{4}, \quad p = 4 \end{aligned} \]

The middle term vanishes because \(\sin^3 u\) is the same at both limits. The values 1, 2 and 3 would leave the total probability at a quarter, a half or three quarters.

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The condition for two distinct quantum mechanical states, represented by wave functions \(\psi_{1}(x)\) and \(\psi_{2}(x)\) respectively, to be orthogonal is :

A.\(\int_{-\infty}^{\infty} \psi_{1}^{*}(x)\psi_{2}(x)\, dx = 1\)
B.\(\int_{-\infty}^{\infty} \psi_{1}^{*}(x)\psi_{2}(x)\, dx = 0\)
C.\(\int_{-\infty}^{\infty} \psi_{1}^{*}(x)\psi_{2}(x)\, dx = \infty\)
D.\(\int_{-\infty}^{\infty} [\psi_{1}^{*}(x)\psi_{2}(x)\) \(+ \psi_{2}^{*}(x)\psi_{1}(x)]\, dx = 1\)

Exam Year(s)

MH SET - 2018, SET Exam - 2018,

Answer: B

The states are orthogonal when \(\int_{-\infty}^{\infty}\psi_{1}^{*}(x)\psi_{2}(x)\,dx = 0\).

Orthogonality means the inner product of the two wave functions vanishes. For functions of x the inner product is the overlap integral of the complex conjugate of one with the other, so the condition is that this integral is zero. Eigenfunctions of a Hermitian operator belonging to different eigenvalues always satisfy it.

An integral equal to 1 is the normalization condition, which applies when the two functions are the same normalized state. An infinite overlap signals non-normalizable functions, and the symmetric sum set equal to 1 is not a condition of orthogonality at all.

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A particle is represented by a plane wave in position space. Its wave function in momentum space is :

A.a delta function
B.a plane wave
C.a Gaussian function
D.a Lorentzian function

Exam Year(s)

MH SET - 2018, SET Exam - 2018,

Answer: A

Its momentum-space wave function is a delta function.

A plane wave \( e^{ip_{0}x/\hbar} \) has one definite momentum \( p_{0} \). Its Fourier transform, the momentum-space wave function, is \( \phi(p) \propto \delta(p - p_{0}) \): all the probability sits at \( p_{0} \). This is the uncertainty principle at its extreme, with \( \Delta x \) infinite and \( \Delta p = 0 \).

A Gaussian in position transforms into a Gaussian in momentum (a wave packet), and a Lorentzian is the transform of an exponentially decaying function. A plane wave transforms into a plane wave only for a delta function in position, the reverse case.

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Which of the following functions can be an acceptable solution for Schrödinger equation for all values of x ?

A.\(\psi(x) = Ae^{-\alpha x^{2}}\)
B.\(\psi(x) = Ae^{\alpha x^{2}}\)
C.\(\psi(x)\) = A tan x
D.\(\psi(x)\) = A sec x

Exam Year(s)

MH SET - 2019, SET Exam - 2019,

Answer: A

The acceptable function is \(\psi(x) = Ae^{-\alpha x^{2}}\) (with \(\alpha \gt 0\)).

An acceptable wave function must be finite, single-valued and continuous everywhere, and square integrable. The Gaussian is all of these: it is smooth and falls to zero at large |x|, so \(\int|\psi|^{2}dx\) is finite. It is the ground state of the harmonic oscillator.

\(Ae^{\alpha x^{2}}\) blows up as \(x \to \pm\infty\) and cannot be normalised. A tan x and A sec x become infinite at \(x = \pi/2\) and its repeats, so they fail for some values of x.

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For Klein-Gordon equation, the probability current density is given by :

A.\(\frac{i\hbar c}{2m}(\psi\nabla\psi^{*} - \psi^{*}\nabla\psi)\)
B.\(\frac{i\hbar}{2m}(\psi\nabla\psi^{*} - \psi^{*}\nabla\psi)\)
C.\(\frac{i\hbar}{2m}(\psi\nabla\psi^{*} + \psi^{*}\nabla\psi)\)
D.\(\frac{i\hbar}{2mc}(\psi\nabla\psi^{*} - \psi^{*}\nabla\psi)\)

Exam Year(s)

MH SET - 2023, SET Exam - 2023,

Answer: B

The Klein-Gordon probability current is \(\frac{i\hbar}{2m}(\psi\nabla\psi^{*} - \psi^{*}\nabla\psi)\).

Multiply the Klein-Gordon equation by \(\psi^{*}\), its conjugate by \(\psi\), and subtract. The result is a continuity equation \(\partial\rho/\partial t + \nabla\cdot\vec{j} = 0\), with

\[ \begin{aligned} \rho &= \frac{i\hbar}{2mc^{2}}\left(\psi^{*}\frac{\partial\psi}{\partial t} - \psi\frac{\partial\psi^{*}}{\partial t}\right) \\ \vec{j} &= \frac{i\hbar}{2m}(\psi\nabla\psi^{*} - \psi^{*}\nabla\psi) \end{aligned} \]

The current has the same form as the Schroedinger current; it is the density that differs and can become negative. The factors c and 1/c in the first and last options make the dimensions wrong, and a plus sign between the terms gives a real quantity, \(\nabla|\psi|^{2}\), which does not describe flow.

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The normalized wave function of a particle can be written as

\[ \begin{aligned} \psi(x) = N\sum_{n=0}^{\infty}&\left(\frac{1}{\sqrt{7}}\right)^{n} \\ &\phi_{n}(x) \end{aligned} \]

where \(\phi_{n}(x)\) are the normalized energy eigenfunctions of a given Hamiltonian. The value of N is :

A.\(\sqrt{\frac{6}{7}}\)
B.\(\sqrt{\frac{1}{7}}\)
C.\(\sqrt{\frac{3}{7}}\)
D.\(\sqrt{\frac{13}{7}}\)

Exam Year(s)

MH SET - 2025, SET Exam - 2025,

Answer: A

The normalisation constant is \(N = \sqrt{6/7}\).

The eigenfunctions are orthonormal, so the norm of \(\psi\) is the sum of the squared moduli of its coefficients:

\[ \begin{aligned} 1 &= N^{2}\sum_{n=0}^{\infty}\left(\frac{1}{7}\right)^{n} \\ &= N^{2}\cdot\frac{1}{1 - 1/7} = N^{2}\cdot\frac{7}{6} \end{aligned} \]

so \(N^{2} = 6/7\) and \(N = \sqrt{6/7}\).

The coefficients \((1/\sqrt{7})^{n}\) must be squared before summing, giving a geometric series of ratio 1/7. Forgetting to square them, or stopping after the first term (which gives \(N = 1\)), leads to the wrong values; \(\sqrt{1/7}\), \(\sqrt{3/7}\) and \(\sqrt{13/7}\) do not make the sum equal to 1.

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