Answer: Particle in a one-dimensional box

Consider a particle of mass (m) confined in a one-dimensional box of length (L). The potential is

$$V(x)= \begin{cases} 0, & 0<x<L\\ \infty, & x\le 0 \text{ or } x\ge L \end{cases}$$

Inside the box, the time-independent Schrödinger equation is

$$-\frac{\hbar^2}{2m}\frac{d^2\psi}{dx^2}=E\psi$$

or

$$\frac{d^2\psi}{dx^2}+k^2\psi=0,\qquad k^2=\frac{2mE}{\hbar^2}$$

The general solution is

$$\psi(x)=A\sin kx+B\cos kx$$

Boundary conditions:

At (x=0), the wave function must vanish:

$$\psi(0)=0 \Rightarrow B=0$$

So

$$\psi(x)=A\sin kx$$

At (x=L),

$$\psi(L)=0 \Rightarrow A\sin kL=0$$

For a non-trivial solution, (A\neq 0), hence

$$\sin kL=0$$

Therefore,

$$kL=n\pi,\qquad n=1,2,3,\dots$$

or

$$k_n=\frac{n\pi}{L}$$

Energy eigenvalues:

Using (k_n^2=\frac{2mE_n}{\hbar^2}),

$$E_n=\frac{\hbar^2 k_n^2}{2m} =\frac{n^2\pi^2\hbar^2}{2mL^2}$$

Since (\hbar=\frac{h}{2\pi}),

$$E_n=\frac{n^2 h^2}{8mL^2},\qquad n=1,2,3,\dots$$

Thus the energy of the particle in a one-dimensional box is quantized. The lowest energy, called the zero-point energy, is

$$E_1=\frac{h^2}{8mL^2}$$

Normalized wave function:

The wave function is

$$\psi_n(x)=A\sin\left(\frac{n\pi x}{L}\right)$$

Normalization gives

$$\int_0^L |\psi_n(x)|^2 dx=1$$ $$|A|^2\int_0^L \sin^2\left(\frac{n\pi x}{L}\right)dx=1$$

Since

$$\int_0^L \sin^2\left(\frac{n\pi x}{L}\right)dx=\frac{L}{2}$$

we get

$$|A|^2\frac{L}{2}=1 \Rightarrow A=\sqrt{\frac{2}{L}}$$

Hence the normalized wave function is

$$\boxed{\psi_n(x)=\sqrt{\frac{2}{L}}\sin\left(\frac{n\pi x}{L}\right),\qquad 0<x<L}$$

and

$$\psi_n(x)=0,\qquad x\le 0 \text{ or } x\ge L$$

The functions satisfy the orthonormality condition

$$\int_0^L \psi_m^*(x)\psi_n(x)dx=\delta_{mn}$$

So the energy eigenvalues and eigenfunctions for a particle in a one-dimensional box are

$$\boxed{E_n=\frac{n^2 h^2}{8mL^2}}$$ $$\boxed{\psi_n(x)=\sqrt{\frac{2}{L}}\sin\left(\frac{n\pi x}{L}\right)}$$

with (n=1,2,3,\dots)

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OR

Creation and annihilation operators

For a linear harmonic oscillator of mass (m) and angular frequency (\omega), the annihilation operator (a) and creation operator (a^\dagger) are defined as

$$a=\sqrt{\frac{m\omega}{2\hbar}}\left(x+\frac{ip}{m\omega}\right)$$ $$a^\dagger=\sqrt{\frac{m\omega}{2\hbar}}\left(x-\frac{ip}{m\omega}\right)$$

They satisfy the commutation relation

$$[a,a^\dagger]=1$$

The Hamiltonian of the harmonic oscillator is

$$H=\frac{p^2}{2m}+\frac12 m\omega^2 x^2$$

In terms of (a) and (a^\dagger),

$$\boxed{H=\hbar\omega\left(a^\dagger a+\frac12\right)}$$

Let

$$N=a^\dagger a$$

be the number operator. If (|n\rangle) is an eigenstate of (N), then

$$N|n\rangle=n|n\rangle$$

where (n=0,1,2,3,\dots)

Therefore,

$$H|n\rangle=\hbar\omega\left(n+\frac12\right)|n\rangle$$

Hence the energy eigenvalues of the linear harmonic oscillator are

$$\boxed{E_n=\left(n+\frac12\right)\hbar\omega,\qquad n=0,1,2,3,\dots}$$

The ground-state energy is

$$E_0=\frac12\hbar\omega$$

This is the zero-point energy. The operator (a) lowers the energy by one quantum, while (a^\dagger) raises it by one quantum.

Physical Implications of the Schrödinger Equation

1. The Equation Itself

For a particle of mass $m$ moving in a potential $V(\mathbf{r})$, the time-dependent Schrödinger equation is

$i\hbar\frac{\partial \Psi(\mathbf{r},t)}{\partial t} = \left[-\frac{\hbar^2}{2m}\nabla^2 + V(\mathbf{r})\right]\Psi(\mathbf{r},t)$

and for stationary states (definite energy $E$) it reduces to the time-independent form

$\nabla^2\psi + \frac{2m}{\hbar^2}\big[E - V(\mathbf{r})\big]\psi = 0$

2. Physical Implications

(i) Wave–particle duality and the wavefunction. The equation treats a material particle as being described by a wavefunction $\Psi(\mathbf{r},t)$. It is the fundamental dynamical equation of quantum mechanics, playing the same role that Newton's second law plays in classical mechanics. Given $V$ and the initial state, the future state is completely determined.

(ii) Probabilistic interpretation of $\Psi$. $\Psi$ itself is not directly measurable; it is a complex probability amplitude. The measurable quantity is

$P(\mathbf{r},t)\,d^3r = |\Psi(\mathbf{r},t)|^2\,d^3r$

the probability of finding the particle in the volume element $d^3r$ about $\mathbf{r}$. This is Born's interpretation. It replaces the deterministic trajectory of classical mechanics with a statistical description.

(iii) Normalization and conservation of probability. Since the particle must be somewhere,

$\int_{-\infty}^{\infty}|\Psi|^2\,d^3r = 1$

The equation guarantees the continuity equation

$\frac{\partial\rho}{\partial t} + \nabla\cdot\mathbf{J} = 0,\qquad \rho = |\Psi|^2,\qquad \mathbf{J} = \frac{\hbar}{2mi}\big(\Psi^*\nabla\Psi - \Psi\nabla\Psi^*\big)$

so probability is locally conserved — probability never appears or disappears, it only flows.

(iv) Quantization of energy and other observables. Acceptable solutions exist only for certain values of $E$ (eigenvalues), determined by the boundary conditions on $\psi$. This is the origin of discrete energy levels in atoms, molecules and nuclei — a result classical physics cannot give. The corresponding $\psi_n$ are stationary states; in them $|\Psi|^2$ is time-independent, so the probability distribution does not change with time.

(v) Superposition and interference. Because the equation is linear, any linear combination $\Psi = \sum_n c_n\psi_n$ is also a solution. This gives quantum superposition, interference of amplitudes, and the possibility of transitions between states. The coefficients satisfy $\sum|c_n|^2 = 1$, so $|c_n|^2$ is the probability of measuring energy $E_n$.

(vi) Expectation values and measurement. For any observable $A$ with Hermitian operator $\hat{A}$,

$\langle A\rangle = \int \Psi^*\hat{A}\,\Psi\,d^3r$

Measurement outcomes are the eigenvalues of $\hat{A}$; the uncertainty principle $\Delta x\,\Delta p \ge \hbar/2$ follows directly from the operator algebra of the equation.

(vii) Correspondence with classical mechanics. For large quantum numbers and macroscopic systems, the wave packet follows Newton's laws (Ehrenfest theorem), and the equation reproduces classical mechanics as a limiting case. This is the correspondence principle.

(viii) Tunnelling and other non-classical effects. Because $\psi$ need not vanish in a region where $E < V$, the equation predicts barrier penetration (alpha decay, tunnel diodes, STM) — impossible classically.

(ix) Stationary-state energy condition. Writing $\Psi = \psi(\mathbf{r})e^{-iEt/\hbar}$, the equation shows that only states of sharply defined energy have a time-independent probability density; the phase rotates at angular frequency $\omega = E/\hbar$.

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OR

Radial Part of the Schrödinger Equation for the Hydrogen Atom

1. Separation of Variables

For the electron (charge $-e$) in the Coulomb field of the proton,

$V(r) = -\frac{e^2}{4\pi\varepsilon_0 r} = -\frac{k e^2}{r}$

Using spherical polar coordinates and writing

$\psi(r,\theta,\phi) = R(r)\,\Theta(\theta)\,\Phi(\phi) = R(r)\,Y_{lm}(\theta,\phi)$

the time-independent Schrödinger equation separates into an angular part (giving spherical harmonics, with quantum numbers $l$ and $m$) and a radial equation:

$-\frac{\hbar^2}{2m}\frac{1}{r^2}\frac{d}{dr}\!\left(r^2\frac{dR}{dr}\right) + \left[V(r) + \frac{\hbar^2 l(l+1)}{2mr^2}\right]R = E R$

The term $\dfrac{\hbar^2 l(l+1)}{2mr^2}$ is the centrifugal barrier potential.

2. Substitution and the Asymptotic Form

Put $u(r) = rR(r)$. The equation becomes one-dimensional:

$\frac{d^2u}{dr^2} + \frac{2m}{\hbar^2}\left[E + \frac{ke^2}{r} - \frac{\hbar^2 l(l+1)}{2mr^2}\right]u = 0$

For a bound state ($E<0$), define

$\kappa = \frac{\sqrt{-2mE}}{\hbar},\qquad \rho = 2\kappa r,\qquad \lambda = \frac{mke^2}{\hbar^2\kappa}$

giving

$\frac{d^2u}{d\rho^2} + \left[-\frac{1}{4} + \frac{\lambda}{\rho} - \frac{l(l+1)}{\rho^2}\right]u = 0$

As $\rho\to\infty$, $u \sim e^{-\rho/2}$; as $\rho\to 0$, $u \sim \rho^{\,l+1}$. So try

$u(\rho) = \rho^{\,l+1}e^{-\rho/2}\,v(\rho)$

3. The Associated Laguerre Equation

Substituting this into the radial equation gives

$\rho\frac{d^2v}{d\rho^2} + \big[2(l+1)-\rho\big]\frac{dv}{d\rho} + \big[\lambda - (l+1)\big]v = 0$

This is the associated Laguerre equation. Its power-series solution

$v(\rho) = \sum_{j=0}^{\infty} c_j \rho^{\,j}$

terminates (becomes a polynomial) only if

$\lambda - (l+1) = n_r,\qquad n_r = 0,1,2,\dots$

Otherwise the series behaves like $e^{\rho}$ and $u$ diverges — not an acceptable wavefunction.

4. Energy Eigenvalues

With $\lambda = n_r + l + 1 = n$ (the principal quantum number), and using $\lambda = mke^2/\hbar^2\kappa$,

$E_n = -\frac{m k^2 e^4}{2\hbar^2 n^2} = -\frac{m e^4}{32\pi^2\varepsilon_0^2\hbar^2}\cdot\frac{1}{n^2} = -\frac{13.6}{n^2}\ \text{eV}$

This exactly reproduces the Bohr energy levels — but here it emerges from a proper wave-mechanical treatment. Note that $E$ depends only on $n$, so states with the same $n$ but different $l$ are degenerate.

5. The Radial Wavefunctions

The normalized solutions are

$R_{nl}(r) = \left(\frac{2}{n a_0}\right)^{3/2}\sqrt{\frac{(n-l-1)!}{2n\,[(n+l)!]^3}}\;e^{-\rho/2}\rho^{\,l}\,L^{2l+1}_{n-l-1}(\rho)$

where $a_0 = \dfrac{4\pi\varepsilon_0\hbar^2}{me^2} = 0.529\ \text{Å}$ is the Bohr radius and $L^{2l+1}_{n-l-1}$ is the associated Laguerre polynomial. The first few are:

$R_{10} = 2a_0^{-3/2}e^{-r/a_0}$

$R_{20} = \frac{1}{2\sqrt{2}}\,a_0^{-3/2}\left(2-\frac{r}{a_0}\right)e^{-r/2a_0}$

$R_{21} = \frac{1}{2\sqrt{6}}\,a_0^{-3/2}\,\frac{r}{a_0}\,e^{-r/2a_0}$

6. Discussion of the Solution

• Allowed values: $n = 1,2,3,\dots$; $l = 0,1,\dots,n-1$; $m = -l,\dots,+l$. The radial equation fixes $n$ and $l$; the angular equation fixes $l$ and $m$.
• Nodes: $R_{nl}$ has $n-l-1$ radial nodes (zeros) and behaves as $\rho^{\,l}$ near the origin. Only $l=0$ states have nonzero probability at $r=0$.
• Radial probability density: $P(r) = r^2|R_{nl}|^2$ gives the probability of finding the electron between $r$ and $r+dr$. For the ground state this peaks at $r = a_0$.
• Degeneracy: the level $E_n$ has $n^2$-fold degeneracy (factor 2 for electron spin gives $2n^2$).
• Bound states only: the series must terminate; otherwise the wavefunction is non-normalizable. This is why the hydrogen atom has discrete, negative energies — the electron is bound.
• Continuum: for $E>0$ the series does not terminate and one obtains unbound (scattering) states with a continuous energy spectrum.

Using the Uncertainty Principle to obtain the Bohr radius and ground-state energy of hydrogen

1. Setting up the uncertainty relation

For an electron confined to a region of size $r$ (the orbit radius), the uncertainty in its position is of the order of $r$:

$\Delta x \approx r$

According to Heisenberg's uncertainty principle,

$\Delta x \,\Delta p \ge \frac{\hbar}{2}$

Taking the minimum uncertainty (the most probable, lowest-energy case),

$\Delta p \approx \frac{\hbar}{2r} \quad \text{or simply} \quad p \approx \frac{\hbar}{r}$

where $\hbar = h/2\pi$.

2. Total energy of the electron

The electron is bound to the proton by the Coulomb force. Its total energy is the sum of kinetic and potential energy:

$E = \frac{p^2}{2m} - \frac{e^2}{4\pi\varepsilon_0 r}$

Using $p \approx \hbar/r$,

$E(r) = \frac{\hbar^2}{2mr^2} - \frac{e^2}{4\pi\varepsilon_0 r}$

3. Minimising the energy to find the radius

The electron settles at the radius that minimises $E$. Setting $\dfrac{dE}{dr} = 0$:

$\frac{d}{dr}\left(\frac{\hbar^2}{2mr^2} - \frac{e^2}{4\pi\varepsilon_0 r}\right) = 0$

$-\frac{\hbar^2}{mr^3} + \frac{e^2}{4\pi\varepsilon_0 r^2} = 0$

$\frac{\hbar^2}{mr^3} = \frac{e^2}{4\pi\varepsilon_0 r^2}$

$\boxed{\,r = \frac{4\pi\varepsilon_0 \hbar^2}{m e^2}\,}$

This is exactly the Bohr radius $a_0$:

$a_0 = \frac{4\pi\varepsilon_0 \hbar^2}{m e^2} \approx 0.53\ \text{Å} = 5.3 \times 10^{-11}\ \text{m}$

4. Ground-state energy

Substituting $r = a_0$ back into $E(r)$:

$E = \frac{\hbar^2}{2m a_0^2} - \frac{e^2}{4\pi\varepsilon_0 a_0}$

From the minimisation condition, $\dfrac{e^2}{4\pi\varepsilon_0 a_0} = \dfrac{\hbar^2}{m a_0^2}$. Hence

$E = \frac{\hbar^2}{2m a_0^2} - \frac{\hbar^2}{m a_0^2} = -\frac{\hbar^2}{2m a_0^2}$

Using $a_0 = \dfrac{4\pi\varepsilon_0 \hbar^2}{m e^2}$,

$E = -\frac{m e^4}{2(4\pi\varepsilon_0)^2 \hbar^2}$

$\boxed{\,E = -\frac{m e^4}{32\pi^2 \varepsilon_0^2 \hbar^2}\,}$

5. Numerical value

Putting $m = 9.1\times10^{-31}$ kg, $e = 1.6\times10^{-19}$ C, $\varepsilon_0 = 8.85\times10^{-12}$ F/m, $\hbar = 1.055\times10^{-34}$ J·s:

$E = -\frac{(9.1\times10^{-31})(1.6\times10^{-19})^4}{32\pi^2(8.85\times10^{-12})^2(1.055\times10^{-34})^2}$

$E \approx -2.18\times10^{-18}\ \text{J} = -13.6\ \text{eV}$

Result: The uncertainty principle gives the Bohr radius $r = \dfrac{4\pi\varepsilon_0\hbar^2}{me^2} \approx 0.53\ \text{Å}$ and the ground-state energy $E = -\dfrac{me^4}{32\pi^2\varepsilon_0^2\hbar^2} \approx -13.6\ \text{eV}$, in agreement with the Bohr model. The negative sign shows the electron is bound.

Ehrenfest's Theorem

Statement. The expectation (average) values of position and momentum in a quantum state obey the same equations of motion as the corresponding classical quantities:

$\frac{d\langle x\rangle}{dt}=\frac{\langle p\rangle}{m},\qquad \frac{d\langle p\rangle}{dt}=\left\langle -\frac{\partial V}{\partial x}\right\rangle=\langle F\rangle$

where $V(x)$ is the potential energy and $F$ the force. Thus Newton's second law holds for averages.

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Proof

Step 1: Equation of motion for the expectation value of any operator.

Let $\hat A$ be an operator (possibly time dependent). Its expectation value in the state $\psi$ is

$\langle A\rangle=\int \psi^{*}\hat A\,\psi\,dx .$

Differentiating with respect to $t$ (the integration limits are fixed),

$\frac{d\langle A\rangle}{dt}=\int\!\left(\frac{\partial \psi^{*}}{\partial t}\hat A\psi+\psi^{*}\frac{\partial \hat A}{\partial t}\psi+\psi^{*}\hat A\frac{\partial \psi}{\partial t}\right)dx .$

Use the Schrödinger equation and its complex conjugate (with $\hat H$ Hermitian, so $V$ is real):

$i\hbar\frac{\partial \psi}{\partial t}=\hat H\psi,\qquad -i\hbar\frac{\partial \psi^{*}}{\partial t}=\hat H\psi^{*}.$

Hence

$\frac{\partial \psi}{\partial t}=\frac{1}{i\hbar}\hat H\psi,\qquad \frac{\partial \psi^{*}}{\partial t}=-\frac{1}{i\hbar}\hat H\psi^{*}.$

Substituting,

$\frac{d\langle A\rangle}{dt}=\frac{i}{\hbar}\int \psi^{*}\hat H\hat A\psi\,dx-\frac{i}{\hbar}\int \psi^{*}\hat A\hat H\psi\,dx+\left\langle\frac{\partial \hat A}{\partial t}\right\rangle .$

Since $\hat H$ is Hermitian, $\int \psi^{*}\hat H\hat A\psi\,dx=\int(\hat H\psi)^{*}\hat A\psi\,dx$, so the first two terms combine into the commutator:

$\boxed{\ \frac{d\langle A\rangle}{dt}=\frac{i}{\hbar}\big\langle[\hat H,\hat A]\big\rangle+\left\langle\frac{\partial \hat A}{\partial t}\right\rangle\ }$

This is the general Ehrenfest relation.

Step 2: Apply to the position operator $\hat x$.

Since $\hat x$ does not depend explicitly on $t$, $\partial \hat x/\partial t=0$, and

$\frac{d\langle x\rangle}{dt}=\frac{i}{\hbar}\big\langle[\hat H,\hat x]\big\rangle .$

With $\hat H=\dfrac{\hat p^{2}}{2m}+V(\hat x)$ and $[V(\hat x),\hat x]=0$,

$[\hat H,\hat x]=\frac{1}{2m}[\hat p^{2},\hat x]=\frac{1}{2m}\Big(\hat p[\hat p,\hat x]+[\hat p,\hat x]\hat p\Big).$

Using $[\hat x,\hat p]=i\hbar$, i.e. $[\hat p,\hat x]=-i\hbar$,

$[\hat p^{2},\hat x]=-2i\hbar\,\hat p .$

Therefore

$\frac{d\langle x\rangle}{dt}=\frac{i}{\hbar}\cdot\left(-\frac{i\hbar}{m}\right)\langle p\rangle=\frac{\langle p\rangle}{m}. \qquad \blacksquare$

Step 3: Apply to the momentum operator $\hat p$.

Again $\partial \hat p/\partial t=0$, so

$\frac{d\langle p\rangle}{dt}=\frac{i}{\hbar}\big\langle[\hat H,\hat p]\big\rangle .$

Here $[\hat p^{2},\hat p]=0$, so only the potential term survives:

$[V(\hat x),\hat p]\psi=V(-i\hbar\psi')-(-i\hbar)(V\psi)'=-i\hbar V\psi'+i\hbar V'\psi+i\hbar V\psi'=i\hbar V'\psi .$

Hence $[V(\hat x),\hat p]=i\hbar\dfrac{\partial V}{\partial x}$, and

$\frac{d\langle p\rangle}{dt}=\frac{i}{\hbar}\cdot i\hbar\left\langle\frac{\partial V}{\partial x}\right\rangle=-\left\langle\frac{\partial V}{\partial x}\right\rangle=\langle F\rangle . \qquad \blacksquare$

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Remarks (for full marks)

1. Classical limit. If the wave packet is narrow compared with the distance over which $V$ varies, then $\langle \partial V/\partial x\rangle\approx \partial V/\partial\langle x\rangle$, and the two results reduce exactly to Newton's equations for the mean position — this is the correspondence principle.

2. Ehrenfest's theorem does not say that $\langle x\rangle$ follows a classical trajectory in general; the equality is exact only for averages, and the potential term is generally nonlinear in $\hat x$.

3. Conservation laws follow directly. For a free particle ($V=$ constant), $[\hat H,\hat p]=0$, so $\langle p\rangle$ is constant and $\langle x\rangle$ moves uniformly. For a harmonic oscillator, $V=\tfrac12 m\omega^{2}x^{2}$, giving $\dfrac{d\langle p\rangle}{dt}=-m\omega^{2}\langle x\rangle$, so $\langle x\rangle$ executes simple harmonic motion of angular frequency $\omega$.

Free Particle: Wave Function and Energy

For a free particle, the potential energy is zero everywhere:

[ V(x)=0 ]

The time-independent Schrödinger wave equation in one dimension is

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

Putting (V(x)=0),

[ -\frac{\hbar^2}{2m}\frac{d^2\psi}{dx^2}=E\psi ]

or

[ \frac{d^2\psi}{dx^2}+\frac{2mE}{\hbar^2}\psi=0 ]

Let

[ k^2=\frac{2mE}{\hbar^2} ]

Then the equation becomes

[ \frac{d^2\psi}{dx^2}+k^2\psi=0 ]

This is the standard equation of simple harmonic motion. Its general solution is

[ \psi(x)=A e^{ikx}+B e^{-ikx} ]

where (A) and (B) are arbitrary constants. Equivalently,

[ \psi(x)=C\cos kx+D\sin kx ]

The time-dependent wave function is obtained by multiplying by the time factor (e^{-iEt/\hbar}):

[ \Psi(x,t)=\psi(x)e^{-iEt/\hbar} ]

Therefore,

[ \Psi(x,t)=A e^{i(kx-Et/\hbar)}+B e^{-i(kx+Et/\hbar)} ]

Using

[ E=\hbar\omega ]

we get

[ \Psi(x,t)=A e^{i(kx-\omega t)}+B e^{-i(kx+\omega t)} ]

The first term represents a plane wave travelling in the positive (x)-direction, and the second term represents a plane wave travelling in the negative (x)-direction.

For a free particle moving in the positive (x)-direction, we take (B=0). Then

[ \Psi(x,t)=A e^{i(kx-\omega t)} ]

The energy of the free particle is

[ E=\frac{\hbar^2 k^2}{2m} ]

Since the particle is free, there is no boundary condition restricting (k). Therefore, (k) can take any real positive value, and hence the energy (E) is continuous, not quantized:

[ E=\frac{\hbar^2 k^2}{2m},\qquad 0<E<\infty ]

The corresponding momentum is

[ p=\hbar k ]

so that

[ E=\frac{p^2}{2m} ]

which is the usual classical kinetic energy expression.

The probability density is

[ |\Psi(x,t)|^2=|A|^2 ]

which is constant in space and time. This means the free particle has equal probability of being found anywhere in space. Such a plane wave is not normalizable over infinite space, so for strict normalization one uses a finite box or delta-function normalization.

Thus, for a free particle the wave function is a plane wave,

[ \Psi(x,t)=A e^{i(kx-\omega t)} ]

and its energy is

[ \boxed{E=\frac{\hbar^2 k^2}{2m}} ]

with (k) continuous. Hence the energy spectrum of a free particle is continuous.

Probability Density and Probability Current Density

1. Probability Density

In quantum mechanics, the state of a particle is described by a wave function $\Psi(\mathbf{r}, t)$. Since $\Psi$ is generally complex, it cannot itself be a physical observable. The physically meaningful quantity is the probability density, defined as

$$\rho(\mathbf{r}, t) = |\Psi(\mathbf{r}, t)|^2 = \Psi^*(\mathbf{r}, t)\,\Psi(\mathbf{r}, t)$$

Meaning: $\rho(\mathbf{r},t)\,d^3r = |\Psi|^2\,d^3r$ gives the probability of finding the particle at time $t$ within the volume element $d^3r$ about the point $\mathbf{r}$.

Normalization: Since the particle must be found somewhere in space,

$$\int_{-\infty}^{\infty} |\Psi|^2\,d^3r = 1$$

This total probability is conserved (independent of time) for a normalizable wave function.

2. Probability Current Density

The probability current density $\mathbf{J}(\mathbf{r}, t)$ is a vector quantity that measures the rate of flow of probability per unit area per unit time. It is defined as

$$\mathbf{J} = \frac{\hbar}{2mi}\left(\Psi^* \nabla\Psi - \Psi \nabla\Psi^*\right)$$

or equivalently,

$$\mathbf{J} = \frac{1}{m}\,\mathrm{Re}\!\left(\Psi^* \frac{\hbar}{i}\nabla\Psi\right)$$

Meaning: $\mathbf{J}\cdot d\mathbf{S}$ gives the probability flowing per unit time through the surface element $d\mathbf{S}$. For a plane wave $\Psi = Ae^{i(\mathbf{k}\cdot\mathbf{r}-\omega t)}$, we get $\mathbf{J} = \frac{\hbar k}{m}|A|^2 = v|A|^2$, i.e. probability density moving with velocity $v$ — exactly as expected classically.

3. Equation of Continuity

Statement

The probability density $\rho$ and probability current density $\mathbf{J}$ satisfy

$$\boxed{\;\frac{\partial \rho}{\partial t} + \nabla \cdot \mathbf{J} = 0\;}$$

This is the equation of continuity — the quantum-mechanical statement of conservation of probability. It says that probability can neither be created nor destroyed; any decrease of probability in a volume element is exactly compensated by a net outflow of probability current through its surface.

Derivation

Step 1 — Write the Schrödinger equation and its complex conjugate.

The time-dependent Schrödinger equation is

$$i\hbar \frac{\partial \Psi}{\partial t} = -\frac{\hbar^2}{2m}\nabla^2\Psi + V\Psi \tag{1}$$

Taking the complex conjugate (with $V$ real):

$$-i\hbar \frac{\partial \Psi^*}{\partial t} = -\frac{\hbar^2}{2m}\nabla^2\Psi^* + V\Psi^* \tag{2}$$

Step 2 — Multiply (1) by $\Psi^*$ and (2) by $\Psi$.

$$i\hbar\,\Psi^*\frac{\partial \Psi}{\partial t} = -\frac{\hbar^2}{2m}\Psi^*\nabla^2\Psi + V\Psi^*\Psi \tag{3}$$ $$-i\hbar\,\Psi\frac{\partial \Psi^*}{\partial t} = -\frac{\hbar^2}{2m}\Psi\nabla^2\Psi^* + V\Psi\Psi^* \tag{4}$$

Step 3 — Subtract (4) from (3).

The potential terms cancel since $V\Psi^*\Psi = V\Psi\Psi^*$:

$$i\hbar\left(\Psi^*\frac{\partial \Psi}{\partial t} + \Psi\frac{\partial \Psi^*}{\partial t}\right) = -\frac{\hbar^2}{2m}\left(\Psi^*\nabla^2\Psi - \Psi\nabla^2\Psi^*\right)$$

The left side is $i\hbar\,\dfrac{\partial}{\partial t}(\Psi^*\Psi) = i\hbar\,\dfrac{\partial \rho}{\partial t}$. Hence

$$i\hbar \frac{\partial \rho}{\partial t} = -\frac{\hbar^2}{2m}\left(\Psi^*\nabla^2\Psi - \Psi\nabla^2\Psi^*\right)$$

Step 4 — Use the vector identity.

Recall the identity

$$\nabla\cdot\left(\Psi^*\nabla\Psi - \Psi\nabla\Psi^*\right) = \Psi^*\nabla^2\Psi - \Psi\nabla^2\Psi^*$$

because the terms involving $\nabla\Psi^*\cdot\nabla\Psi$ cancel. Therefore,

$$i\hbar \frac{\partial \rho}{\partial t} = -\frac{\hbar^2}{2m}\,\nabla\cdot\left(\Psi^*\nabla\Psi - \Psi\nabla\Psi^*\right)$$

Step 5 — Rearrange into the continuity equation.

Dividing throughout by $i\hbar$,

$$\frac{\partial \rho}{\partial t} = -\frac{\hbar}{2mi}\,\nabla\cdot\left(\Psi^*\nabla\Psi - \Psi\nabla\Psi^*\right)$$ $$\frac{\partial \rho}{\partial t} = -\,\nabla\cdot\left[\frac{\hbar}{2mi}\left(\Psi^*\nabla\Psi - \Psi\nabla\Psi^*\right)\right]$$

The quantity in the square bracket is precisely the probability current density $\mathbf{J}$. Hence

$$\frac{\partial \rho}{\partial t} + \nabla\cdot\mathbf{J} = 0$$

which is the required equation of continuity. $\blacksquare$

Integral Form (Conservation Statement)

Integrating over a fixed volume $V$ bounded by surface $S$ and using the divergence theorem,

$$\frac{d}{dt}\int_V \rho\,d^3r = -\oint_S \mathbf{J}\cdot d\mathbf{S}$$

This states that the rate of decrease of probability inside $V$ equals the net probability flux leaving through $S$. For the whole of space the surface term vanishes (since $\Psi \to 0$ at infinity), giving

$$\frac{d}{dt}\int_{-\infty}^{\infty} |\Psi|^2\,d^3r = 0$$

so the normalization of the wave function is preserved for all time — the physical content of the equation of continuity.

(a) Prove that the particle velocity is equal to the group velocity of a wave packet.

A wave packet is formed by the superposition of a group of plane waves having wave numbers close to a central value (k_0). The velocity with which the envelope of the wave packet moves is called the group velocity:

[ v_g=\frac{d\omega}{dk} ]

For a free particle, the de Broglie relations are

[ E=\hbar\omega,\qquad p=\hbar k ]

Non-relativistic case:

[ E=\frac{p^2}{2m}=\frac{\hbar^2 k^2}{2m} ]

So,

[ \omega=\frac{E}{\hbar}=\frac{\hbar k^2}{2m} ]

Therefore,

[ v_g=\frac{d\omega}{dk} =\frac{d}{dk}\left(\frac{\hbar k^2}{2m}\right) =\frac{\hbar k}{m} =\frac{p}{m} =v ]

where (v) is the particle velocity. Hence,

[ \boxed{v_g=v} ]

Thus, the group velocity of the wave packet is equal to the particle velocity.

OR

What is degeneracy? How will you calculate the total number of eigen functions for a given principal quantum number? Explain.

Degeneracy: If two or more independent eigen functions correspond to the same energy eigenvalue, the energy level is said to be degenerate. The number of independent eigen functions having the same energy is called the degree of degeneracy.

For the hydrogen atom, the energy depends only on the principal quantum number (n):

[ E_n=-\frac{13.6}{n^2}\ \text{eV} ]

For a given (n), the orbital quantum number (l) can take values

[ l=0,1,2,\dots,(n-1) ]

For each (l), the magnetic quantum number (m_l) can take

[ m_l=-l,-l+1,\dots,0,\dots,l-1,l ]

so there are ((2l+1)) states for each (l).

Therefore, the total number of eigen functions without spin is

[ \sum_{l=0}^{n-1}(2l+1) =1+3+5+\cdots+(2n-1) =n^2 ]

If electron spin is included, each state has two spin orientations (m_s=+\frac12,-\frac12). Hence the total degeneracy becomes

[ \boxed{2n^2} ]

So, for a given principal quantum number (n), the total number of eigen functions is (n^2) without spin and (2n^2) with spin.

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(b) Show that the momentum operator is Hermitian.

An operator (\hat{A}) is Hermitian if

[ \int \psi^* \hat{A}\phi,dx

\int (\hat{A}\psi)^*\phi,dx ]

The momentum operator is

[ \hat{p}_x=-i\hbar\frac{\partial}{\partial x} ]

Consider

[ \int \psi^* \hat{p}_x \phi,dx

-i\hbar\int \psi^* \frac{\partial \phi}{\partial x},dx ]

Integrating by parts,

[

-i\hbar\left[\psi^\phi\right]_{-\infty}^{\infty} + i\hbar\int \frac{\partial \psi^}{\partial x}\phi,dx ]

For square-integrable wave functions, the boundary term vanishes:

[ \left[\psi^*\phi\right]_{-\infty}^{\infty}=0 ]

Thus,

[ \int \psi^* \hat{p}_x \phi,dx

i\hbar\int \frac{\partial \psi^*}{\partial x}\phi,dx ]

But

[ (\hat{p}_x\psi)^*

\left(-i\hbar\frac{\partial \psi}{\partial x}\right)^*

i\hbar\frac{\partial \psi^*}{\partial x} ]

Therefore,

[ \int \psi^* \hat{p}_x \phi,dx

\int (\hat{p}_x\psi)^*\phi,dx ]

Hence,

[ \boxed{\hat{p}_x=\hat{p}_x^\dagger} ]

So the momentum operator is Hermitian. Similarly, (\hat{p}_y) and (\hat{p}_z) are also Hermitian.

OR

Can we observe the energy states for a ball of mass 10 gm moving in a box of length 10 cm? Justify your answer.

For a particle in a one-dimensional box, the energy levels are

[ E_n=\frac{n^2h^2}{8mL^2} ]

Given:

[ m=10\ \text{gm}=0.01\ \text{kg} ]

[ L=10\ \text{cm}=0.1\ \text{m} ]

[ h=6.626\times10^{-34}\ \text{J s} ]

For (n=1),

[ E_1=\frac{h^2}{8mL^2} ]

[ E_1=\frac{(6.626\times10^{-34})^2}{8\times0.01\times(0.1)^2} ]

[ E_1\approx 5.49\times10^{-64}\ \text{J} ]

In electron volt,

[ E_1\approx 3.43\times10^{-45}\ \text{eV} ]

The energy spacing between two successive levels is

[ \Delta E=E_{n+1}-E_n=(2n+1)E_1 ]

For (n=1),

[ \Delta E\approx 3E_1\approx 1.65\times10^{-63}\ \text{J} ]

This is extremely small. At room temperature, the thermal energy is

[ kT\approx 4.14\times10^{-21}\ \text{J} ]

The corresponding quantum number is

[ n\approx \sqrt{\frac{kT}{E_1}} \approx \sqrt{\frac{4.14\times10^{-21}}{5.49\times10^{-64}}} \approx 2.75\times10^{21} ]

So the ball occupies a very high quantum state, and the energy levels are practically continuous. The spacing is far below any measurable limit. Therefore,

[ \boxed{\text{The discrete energy states of the ball cannot be observed.}} ]

The motion of the ball is effectively classical.

(a) General results for two particles bound states

For two particles of masses $m_1$ and $m_2$ interacting through a potential that depends only on their relative separation, $V = V(\mathbf r_1-\mathbf r_2)$, the following general results hold:

1. Separation of the centre-of-mass motion. Introducing $\mathbf R=\frac{m_1\mathbf r_1+m_2\mathbf r_2}{M},\qquad \mathbf r=\mathbf r_1-\mathbf r_2,\qquad M=m_1+m_2,\qquad \mu=\frac{m_1m_2}{M},$ the Hamiltonian splits into a free CM part and a relative part: $\hat H=-\frac{\hbar^2}{2M}\nabla_R^2-\frac{\hbar^2}{2\mu}\nabla_r^2+V(\mathbf r).$ The wave function factorises as $\psi(\mathbf r_1,\mathbf r_2)=e^{i\mathbf K\cdot\mathbf R}\phi(\mathbf r)$.

2. Energy spectrum. $E=\dfrac{\hbar^2K^2}{2M}+E_n$, where $E_n$ are the discrete eigenvalues of the relative-motion equation for a particle of reduced mass $\mu$ in $V(\mathbf r)$. Bound states have $E_n<0$ (below the zero of the relative energy); the CM motion remains free and continuous. Thus the two-body bound-state problem is exactly equivalent to a one-body problem with $m\to\mu$.

3. Existence and number of bound states. For a short-range potential in 3D a minimum strength (depth $\times$ range$^2$) is required for even one bound state; in 1D and 2D an arbitrarily weak attraction always binds. The number of bound states grows with the strength parameter; for a square well of depth $V_0$ and width $a$, $N\approx \dfrac{a}{\pi\hbar}\sqrt{2\mu V_0}$.

4. Identical particles. The total wave function must be symmetric (bosons) or antisymmetric (fermions) under $\mathbf r\to-\mathbf r$, which ties the spatial symmetry to the spin.

The bound-state wave functions are normalisable and decay exponentially outside the range of the potential; the binding energy is $B=-E_n$.

(b) Virtual binding (virtual bound state)

A virtual state (virtual bound state) is a pole of the scattering amplitude lying at a negative energy but on the unphysical (second) Riemann sheet of the complex momentum plane — that is, at $k=-i\kappa$ ($\kappa>0$ real) on the negative imaginary axis, instead of at $k=+i\kappa$ as for a true bound state.

It is called "virtual binding" because the system is almost bound but not quite: the corresponding wave function behaves asymptotically as $e^{+\kappa r}/r$, which grows at large $r$ and therefore cannot be normalised. Hence no genuine, physically observable bound state exists; the system appears only as a sharp enhancement (a low-energy resonance) in the scattering cross-section. For s-waves, $\sigma\to 4\pi a^2$ as $k\to0$, so a large scattering length $a$ (large and negative) signals a nearby virtual state.

Example: the neutron–proton system in the $^1S_0$ (singlet) channel has a virtual state at about $-66\ \text{keV}$ with $a\approx-23.7\ \text{fm}$, whereas the triplet channel gives the truly bound deuteron at $-2.22\ \text{MeV}$.

Importance: it explains the very large low-energy cross-sections of nucleon–nucleon scattering, the Wigner threshold behaviour, and the smooth continuity of observables as a system passes from just-bound to just-unbound (as $|a|\to\infty$ the virtual state moves to zero energy and turns into a bound state).

(c) Conditions the wave function must satisfy

A physically acceptable solution $\psi(\mathbf r)$ of the Schrödinger equation must satisfy:

1. Finiteness and single-valuedness: $\psi$ must be finite and single-valued at every point; it must be a well-behaved, continuous function.
2. Continuity: $\psi$ and its first spatial derivatives must be continuous across any boundary at which the potential is finite. Where $V\to\infty$ (rigid wall) $\psi=0$; across a delta-function potential the derivative suffers a finite jump, $\psi'_{right}-\psi'_{left}=\dfrac{2m}{\hbar^2}\lambda\,\psi$.
3. Normalisability (square integrability): $\displaystyle\int|\psi|^2\,d^3r=1<\infty$, so that $\psi\to0$ as $r\to\infty$. This requirement is what makes the bound-state energies discrete (quantised).
4. Satisfaction of the Schrödinger equation: $\hat H\psi=E\psi$ must hold in every region, with the appropriate boundary conditions at the edges of each region.
5. Correct asymptotic form: for scattering (continuum) states, $\psi$ must reduce at large $r$ to an incident plane wave plus an outgoing spherical wave; for bound states it must decay exponentially.
6. Exchange symmetry: for identical particles, $\psi$ must be symmetric (bosons) or antisymmetric (fermions) under interchange of the particles.
7. Continuity of probability density and current: the probability current must be continuous, so that no particles are created or destroyed at a boundary.

(d) Meaning and importance of parity

Meaning. Parity is the property of a wave function under spatial inversion, $\mathbf r\to-\mathbf r$. The parity operator is defined by $\hat P\,\psi(\mathbf r)=\psi(-\mathbf r).$ Since $\hat P^2=1$ and $\hat P$ is Hermitian and unitary, its eigenvalues are only $\pm1$:

• even parity ($+1$): $\psi(-\mathbf r)=+\psi(\mathbf r)$,
• odd parity ($-1$): $\psi(-\mathbf r)=-\psi(\mathbf r)$.

Parity is a multiplicative quantum number. If the potential is symmetric, $V(-\mathbf r)=V(\mathbf r)$, then $[\hat P,\hat H]=0$, so parity is a conserved quantity and every stationary state has a definite parity. For a state of orbital angular momentum $l$, the parity is $(-1)^l$; for a two-particle system the total parity is $P=P_1P_2(-1)^l$.

Importance.

1. Selection rules. Electric dipole (E1) transitions are allowed only between states of opposite parity (Laporte rule); transitions between states of the same parity are forbidden. This determines which spectral lines are observed in atoms and nuclei.
2. Conservation law. Parity is strictly conserved in strong and electromagnetic interactions. Its non-conservation in weak interactions (Wu's experiment on $\beta$-decay of $^{60}$Co, 1957) was a landmark discovery in physics.
3. Classification of states. States are labelled by spin–parity $J^P$ in nuclear and particle physics; the parity of a level is found from angular distributions of decay products.
4. Simplification of calculations. Matrix elements of parity-odd operators vanish between states of equal parity and vice versa, so many integrals can be discarded at once.
5. Symmetry classification in spectroscopy. In molecular and atomic spectroscopy levels are classified as gerade/ungerade (even/odd) according to parity, which governs the allowed radiative transitions.

Given:

• Mass, (m = 1\ \text{gm} = 1 \times 10^{-3}\ \text{kg})
• Force, (F = 0.1\ \text{N})
• Extension, (x = 1\ \text{cm} = 1 \times 10^{-2}\ \text{m})

Spring constant: [ k=\frac{F}{x}=\frac{0.1}{1\times 10^{-2}}=10\ \text{N/m} ]

Angular frequency: [ \omega=\sqrt{\frac{k}{m}} =\sqrt{\frac{10}{1\times 10^{-3}}} =\sqrt{10^4} =100\ \text{rad/s} ]

Zero point energy of a harmonic oscillator: [ E_0=\frac{1}{2}\hbar\omega ]

Substituting (\hbar = 1.0546\times 10^{-34}\ \text{J s}): [ E_0=\frac{1}{2}(1.0546\times 10^{-34})(100) ]

[ E_0=5.273\times 10^{-33}\ \text{J} ]

In electron volt: [ E_0=\frac{5.273\times 10^{-33}}{1.602\times 10^{-19}} =3.29\times 10^{-14}\ \text{eV} ]

[ \boxed{E_0 \approx 5.27\times 10^{-33}\ \text{J} \approx 3.29\times 10^{-14}\ \text{eV}} ]

So, the zero point energy of the system is approximately
[ \boxed{5.27\times 10^{-33}\ \text{J}} ]

Given:

• Side of cubical box, (L = 1,\text{Å} = 1.0 \times 10^{-10},\text{m})
• Mass of electron, (m = 9.11 \times 10^{-31},\text{kg})
• Planck’s constant, (h = 6.626 \times 10^{-34},\text{J s})

For a particle in a 3D cubical box, the energy is

[ E = \frac{h^2}{8mL^2}(n_x^2+n_y^2+n_z^2) ]

The lowest energy occurs for the ground state:

[ n_x=n_y=n_z=1 ]

So,

[ E_{\min} = \frac{h^2}{8mL^2}(1^2+1^2+1^2) ]

[ E_{\min} = \frac{3h^2}{8mL^2} ]

Substituting values:

[ E_{\min} = \frac{3(6.626 \times 10^{-34})^2}{8(9.11 \times 10^{-31})(1.0 \times 10^{-10})^2} ]

[ E_{\min} = 1.81 \times 10^{-17},\text{J} ]

In electron-volt:

[ E_{\min} = \frac{1.81 \times 10^{-17}}{1.602 \times 10^{-19}} ]

[ E_{\min} \approx 113,\text{eV} ]

[ \boxed{E_{\min} \approx 1.81 \times 10^{-17},\text{J} \approx 113,\text{eV}} ]

Thus, the lowest energy of the electron confined in a cubical box of side (1,\text{Å}) is about (1.81 \times 10^{-17},\text{J}) or (113,\text{eV}).

Given: Two operators $A$ and $B$ with

$[A, B] = AB - BA = \alpha B$

where $\alpha$ is a number (a c-number, commuting with $A$ and $B$).

To show: $e^{A} B e^{-A} = e^{\alpha} B$.

---

Proof. Define the operator-valued function of the parameter $s$:

$F(s) = e^{sA} B e^{-sA}$

so that the required result is $F(1) = e^{A} B e^{-A}$.

Step 1 — Differentiate with respect to $s$:

$\frac{dF}{ds} = \frac{d}{ds}\left(e^{sA}\right) B e^{-sA} + e^{sA} B \frac{d}{ds}\left(e^{-sA}\right)$

Since $\dfrac{d}{ds}e^{sA} = A\,e^{sA} = e^{sA}A$ and $\dfrac{d}{ds}e^{-sA} = -A\,e^{-sA} = -e^{-sA}A$,

$\frac{dF}{ds} = A e^{sA} B e^{-sA} - e^{sA} B e^{-sA} A = A F(s) - F(s) A = [A, F(s)]$

Step 2 — Evaluate the commutator $[A, F(s)]$:

$[A, F(s)] = \left[A,\; e^{sA} B e^{-sA}\right] = e^{sA}[A, B]\,e^{-sA}$

since $A$ commutes with $e^{\pm sA}$. Using the given relation $[A, B] = \alpha B$,

$[A, F(s)] = e^{sA}(\alpha B)e^{-sA} = \alpha\, e^{sA} B e^{-sA} = \alpha F(s)$

Step 3 — Solve the differential equation:

$\frac{dF}{ds} = \alpha F(s) \quad\Longrightarrow\quad F(s) = F(0)\,e^{\alpha s}$

Step 4 — Apply the initial condition. At $s = 0$,

$F(0) = e^{0}B e^{0} = B$

Hence

$F(s) = B\,e^{\alpha s}$

Step 5 — Put $s = 1$:

$F(1) = e^{A} B e^{-A} = e^{\alpha} B$

---

$\boxed{\,e^{A} B e^{-A} = e^{\alpha} B\,}$

which is the required result. (This is the operator form of the Baker–Campbell–Hausdorff / Weyl identity for the case $[A,B] = \alpha B$.)

For a rigid diatomic molecule such as (H_2), the rotational energy levels are

[ E_l=\frac{\hbar^2}{2I},l(l+1) ]

where (I) is the moment of inertia. Let the rotational constant be

[ L=\frac{\hbar^2}{2I}. ]

Then

[ E_l=L,l(l+1). ]

For (l=1),

[ E_1=L(1)(2)=2L. ]

For (l=2),

[ E_2=L(2)(3)=6L. ]

Hence the energy difference is

[ \Delta E=E_2-E_1=6L-2L=4L. ]

[ \boxed{\Delta E=4L} ]

So the energy difference between the stationary states (l=1) and (l=2) is (\mathbf{4L}).

For the hydrogen-atom wave function
[ \psi(r,\theta,\phi)=A e^{-r/a_0} ] the normalization condition is [ \int |\psi|^2,dV=1. ]

In spherical polar coordinates, [ dV=r^2\sin\theta,dr,d\theta,d\phi. ]

So, [ \int_0^\infty\int_0^\pi\int_0^{2\pi} A^2 e^{-2r/a_0}r^2\sin\theta,dr,d\theta,d\phi=1. ]

The angular integrals give [ \int_0^\pi \sin\theta,d\theta\int_0^{2\pi}d\phi =2\times 2\pi=4\pi. ]

Thus, [ 4\pi A^2\int_0^\infty r^2 e^{-2r/a_0},dr=1. ]

Using the standard integral [ \int_0^\infty r^n e^{-\alpha r},dr=\frac{n!}{\alpha^{n+1}}, ] with (n=2) and (\alpha=\frac{2}{a_0}), [ \int_0^\infty r^2 e^{-2r/a_0},dr =\frac{2!}{(2/a_0)^3} =\frac{2a_0^3}{8} =\frac{a_0^3}{4}. ]

Therefore, [ 4\pi A^2\left(\frac{a_0^3}{4}\right)=1 ] [ \pi A^2 a_0^3=1 ] [ A^2=\frac{1}{\pi a_0^3}. ]

Hence, [ \boxed{A=\frac{1}{\sqrt{\pi a_0^3}}} ] or equivalently, [ \boxed{A=\left(\frac{1}{\pi a_0^3}\right)^{1/2}}. ]

So the normalized hydrogen wave function is
[ \psi(r,\theta,\phi)=\frac{1}{\sqrt{\pi a_0^3}}e^{-r/a_0}. ]

The momentum operator in one dimension is
[ \hat p=-i\hbar\frac{\partial}{\partial x}. ]

Given
[ \Psi(x)=N\exp\left[-\frac{x^2}{2\sigma^2}-ikx\right], ] the normalized probability density is
[ |\Psi(x)|^2=N^2\exp\left[-\frac{x^2}{\sigma^2}\right]. ]

Now, [ \frac{\partial \Psi}{\partial x}

\Psi\left[-\frac{x}{\sigma^2}-ik\right]. ]

Therefore, [ \hat p\Psi

-i\hbar\frac{\partial \Psi}{\partial x}

-i\hbar\left[-\frac{x}{\sigma^2}-ik\right]\Psi

\left(\frac{i\hbar x}{\sigma^2}-\hbar k\right)\Psi. ]

Hence the expectation value of momentum is [ \langle p\rangle

\int_{-\infty}^{\infty}\Psi^*\hat p\Psi,dx

\int_{-\infty}^{\infty}|\Psi|^2\left(\frac{i\hbar x}{\sigma^2}-\hbar k\right)dx. ]

The first term contains the odd function (x|\Psi|^2), so its integral over a symmetric interval is zero: [ \int_{-\infty}^{\infty}\frac{i\hbar x}{\sigma^2}|\Psi|^2dx=0. ]

Thus, [ \langle p\rangle

-\hbar k\int_{-\infty}^{\infty}|\Psi|^2dx. ]

Since the wave function is normalized, [ \int_{-\infty}^{\infty}|\Psi|^2dx=1. ]

Therefore, [ \boxed{\langle p\rangle=-\hbar k} ]

So the expectation value of momentum is (\boxed{-\hbar k}).