Particle in a one-dimensional box (infinite potential well)

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

$$V(x)= \begin{cases} 0, & 0<x<L\\[2mm] \infty, & x\le 0 \text{ or } x\ge L \end{cases}$$

Since the potential is infinite outside the box, the wave function must vanish there:

$$\psi(x)=0 \quad \text{for } x\le 0,\ x\ge L$$

Inside the box, (V(x)=0), so the time-independent Schrödinger equation is

$$\frac{d^2\psi}{dx^2}+\frac{2m}{\hbar^2}E\psi=0$$

Let

$$k^2=\frac{2mE}{\hbar^2}$$

Then

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

The general solution is

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

Boundary conditions

At (x=0),

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

So

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

At (x=L),

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

For a nontrivial solution, (A\ne 0). Hence

$$\sin kL=0$$ $$kL=n\pi,\qquad n=1,2,3,\dots$$

Therefore,

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

Energy eigenvalues

Using (k^2=\dfrac{2mE}{\hbar^2}),

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

or, since (\hbar=\dfrac{h}{2\pi}),

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

Thus the energy of the particle is quantized. The lowest energy, called the zero-point energy, is

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

The particle cannot have zero energy inside the box.

Normalization and wave function

The wave function is

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

Normalize it:

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

Using

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

we get

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

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

$$\psi_n(x)=0 \quad \text{outside the box.}$$

Final results

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

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

These wave functions are orthonormal:

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

The energy levels are discrete, and the spacing between successive levels increases as (n) increases.

Physical Implications of the Schrödinger Equation

1. The Equation Itself

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

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

and for stationary states $\Psi(\mathbf{r},t)=\psi(\mathbf{r})e^{-iEt/\hbar}$ it reduces to the time-independent form

$-\frac{\hbar^2}{2m}\nabla^2\psi + V\psi = E\psi$

2. Physical Implications

(a) Wave–particle duality and the wave nature of matter. The equation treats a material particle as being described by a wave function $\Psi$. It is the quantitative expression of de Broglie's hypothesis: matter propagates as a wave, and interference and diffraction effects observed for electrons, neutrons and atoms follow directly from it.

(b) The wave function $\Psi$ as a probability amplitude. $\Psi$ is not directly measurable. Its physical meaning is given by Born's interpretation: $|\Psi(\mathbf{r},t)|^2\,d^3r$ is the probability of finding the particle in the volume element $d^3r$ about $\mathbf{r}$ at time $t$. Thus the theory is inherently statistical — it predicts probabilities, not exact trajectories. This is why classical determinism (a definite position and momentum at every instant) is replaced by a probability distribution.

(c) Normalisation and conservation of probability. Since total probability must be unity,

$\int |\Psi|^2\,d^3r = 1$

The equation guarantees this permanently: it leads to the continuity equation

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

so probability is neither created nor destroyed — it only flows, exactly like a conserved fluid. This is the quantum analogue of charge conservation.

(d) Quantisation of energy and other observables. Because $\Psi$ must be single-valued, finite and continuous, acceptable solutions exist only for certain discrete values of $E$. Energy levels of atoms, molecules and nuclei therefore come out automatically — the equation explains quantisation rather than assuming it. The same applies to angular momentum, which appears in units of $\hbar$.

(e) The uncertainty principle. The spread of $|\Psi|^2$ in position and in momentum are related by $\Delta x\,\Delta p_x \ge \hbar/2$. The uncertainty relation is thus a direct consequence of the wave nature of $\Psi$, not an extra postulate.

(f) Superposition and interference. The equation is linear, so any linear combination of solutions is again a solution. This gives rise to superposition of states, interference terms in $|\Psi|^2$, and the possibility of wave packets. A localised particle is a superposition (packet) of many momentum eigenstates, which spreads with time — the wave packet dispersion.

(g) Stationary states and discrete spectra. For bound systems the allowed states have definite energies and time dependence only through the phase factor $e^{-iEt/\hbar}$. Transitions between such states give spectral lines of frequency $\nu = (E_2-E_1)/h$, in agreement with the observed line spectra of hydrogen and other atoms.

(h) Tunnelling and classically forbidden regions. Where $E<V$, the solutions are exponentially decaying rather than zero, so there is a finite probability of penetration through a barrier. This explains alpha decay, tunnel diodes, field emission and scanning tunnelling microscopy — effects impossible in classical mechanics.

(i) Correspondence with classical mechanics. For large quantum numbers and macroscopic systems, the average of quantum predictions reproduces Newton's laws (Ehrenfest's theorem). Thus classical mechanics appears as a limiting case, satisfying Bohr's correspondence principle.

(j) Zero-point energy. Confinement of a particle in a finite region forces a minimum kinetic energy; a particle can never be completely at rest in a potential well. This zero-point energy is experimentally confirmed, e.g. in liquid helium.

3. Conclusion

The Schrödinger equation is the fundamental equation of non-relativistic quantum mechanics. Its physical implications are: matter behaves as a wave described by a probability amplitude; measurable quantities are obtained as expectation values; energy and angular momentum are quantised; probability is conserved; and phenomena such as tunnelling, interference and zero-point energy — all foreign to classical physics — follow naturally from it. It has been verified by an enormous range of experiments and forms the basis of atomic, molecular, nuclear and solid-state physics.

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OR

Radial Part of the Schrödinger Equation for the Hydrogen Atom and Its Solution

1. The Radial Equation

For a central potential $V(r)$, write $\Psi(r,\theta,\phi)=R(r)\,Y_l^m(\theta,\phi)$. Substituting into the time-independent Schrödinger equation in spherical polar coordinates and separating the angular part gives the radial equation

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

For hydrogen, $V(r)=-\dfrac{e^2}{4\pi\varepsilon_0 r}$. Putting $u(r)=rR(r)$ reduces it to the one-dimensional form

$-\frac{\hbar^2}{2m}\frac{d^2u}{dr^2} + \left[-\frac{e^2}{4\pi\varepsilon_0 r}+\frac{l(l+1)\hbar^2}{2mr^2}\right]u = Eu$

The term $\dfrac{l(l+1)\hbar^2}{2mr^2}$ is the centrifugal barrier; it keeps the electron away from the nucleus for $l\neq 0$.

2. Solution

(a) Asymptotic behaviour. As $r\to\infty$, the equation becomes $u'' = -\dfrac{2mE}{\hbar^2}u$. For bound states $E<0$, so $u \sim e^{-\kappa r}$ with

$\kappa = \frac{\sqrt{-2mE}}{\hbar}$

As $r\to 0$, the centrifugal term dominates and $u \sim r^{\,l+1}$.

(b) Trial solution. Write

$u(r) = r^{\,l+1}e^{-\kappa r}\,v(r)$

Substitution gives a differential equation for $v(r)$ which is solved by a power series $v(r)=\sum_{k}a_k r^k$. The recursion relation between successive coefficients is

$a_{k+1} = \frac{2\kappa(k+l+1)-2m e^2/(4\pi\varepsilon_0\hbar^2)}{(k+1)(k+2l+2)}\,a_k$

(c) Termination and quantisation. For the wave function to remain finite at infinity the series must terminate, i.e. there must be an integer $n_r$ such that $a_{n_r+1}=0$. This requires

$\frac{2m e^2}{4\pi\varepsilon_0\hbar^2} = 2\kappa\,(n_r+l+1)$

Defining the principal quantum number $n = n_r + l + 1$ ($n = 1,2,3,\dots$), we get

$\kappa = \frac{1}{n a_0},\qquad a_0 = \frac{4\pi\varepsilon_0\hbar^2}{m e^2} = 0.529\ \text{Å}$

(d) Energy eigenvalues. Using $\kappa = \sqrt{-2mE}/\hbar$,

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

This exactly reproduces the Bohr energy levels and the observed hydrogen spectrum.

(e) Radial wave functions. The terminating series is the associated Laguerre polynomial $L_{n+l}^{2l+1}(2\kappa r)$, so

$R_{nl}(r) = N_{nl}\left(\frac{2r}{na_0}\right)^{l} e^{-r/na_0}\, L_{n+l}^{2l+1}\!\left(\frac{2r}{na_0}\right)$

with normalisation constant

$N_{nl} = \sqrt{\left(\frac{2}{na_0}\right)^3\frac{(n-l-1)!}{2n\,[(n+l)!]^3}}$

The first few are:

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

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

$R_{21} = \frac{1}{\sqrt{24}\,a_0^{3/2}}\frac{r}{a_0}e^{-r/2a_0}$

3. Discussion of the Solution

1. Quantisation: Acceptable (finite, single-valued, normalisable) solutions exist only for $n=1,2,3,\dots$; hence energy is quantised as $E_n=-13.6/n^2$ eV. The ground state is $n=1$, $l=0$.
2. Degeneracy: $E_n$ depends only on $n$, not on $l$ or $m$. For a given $n$, $l=0,1,\dots,n-1$ and $m=-l,\dots,+l$, giving $n^2$-fold degeneracy (ignoring spin).
3. Nodes: $R_{nl}$ has $n-l-1$ radial nodes, so the total number of nodes is $n-1$.
4. Behaviour near origin: $R_{nl}\propto r^{l}$, so only $s$-states ($l=0$) have non-zero probability density at the nucleus.
5. Behaviour at large $r$: $R_{nl}\sim e^{-r/na_0}$; the exponential decay confirms the electron is bound, and the Bohr radius $a_0$ sets the size of the atom.
6. Expectation values: $\langle r\rangle = \dfrac{n^2a_0}{Z}\left[1+\tfrac12\left(1-\dfrac{l(l+1)}{n^2}\right)\right]$; for the ground state $\langle r\rangle = 1.5a_0$, and the most probable radius is $a_0$.
7. Agreement with experiment: The predicted spectral series (Lyman, Balmer, Paschen, etc.) and ionisation energy of 13.6 eV agree with observation, confirming the theory.

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

1. Setting up the uncertainty relation

For an electron confined to a region of size $r$ (the radius of the orbit), 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 $\hbar = \dfrac{h}{2\pi}$.

2. Total energy of the electron

The electron of charge $-e$ moves in the Coulomb field of a proton of charge $+e$ at a distance $r$. Its total energy is

$E = \underbrace{\frac{p^2}{2m}}_{\text{kinetic}} \;-\; \underbrace{\frac{e^2}{4\pi\varepsilon_0 r}}_{\text{potential}}$

Substituting $p \approx \dfrac{\hbar}{r}$,

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

3. Minimising the energy to find the radius

For the stable (ground-state) orbit, $E$ must be a minimum. Therefore

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

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

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

This is exactly the radius of the first Bohr orbit (the Bohr radius, $r_1 = 0.53\ \text{Å}$).

4. Ground-state energy

Put this value of $r$ back into $E(r)$. From the minimisation condition,

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

so the kinetic energy is

$K = \frac{\hbar^2}{2mr^2} = \frac{1}{2}\cdot\frac{e^2}{4\pi\varepsilon_0 r} = \frac{|V|}{2}$

Hence

$E = \frac{|V|}{2} - |V| = -\frac{|V|}{2} = -\frac{e^2}{8\pi\varepsilon_0 r}$

Substituting $r = \dfrac{4\pi\varepsilon_0\hbar^2}{me^2}$,

$E = -\frac{e^2}{8\pi\varepsilon_0}\cdot\frac{me^2}{4\pi\varepsilon_0\hbar^2}$

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

5. Numerical value

With $m = 9.1\times10^{-31}\ \text{kg}$, $e = 1.6\times10^{-19}\ \text{C}$, $\varepsilon_0 = 8.85\times10^{-12}\ \text{C}^2\text{N}^{-1}\text{m}^{-2}$ and $\hbar = 1.055\times10^{-34}\ \text{J s}$,

$E = -\frac{(9.1\times10^{-31})(1.6\times10^{-19})^4}{32\pi^2(8.85\times10^{-12})^2(1.055\times10^{-34})^2} \approx -2.18\times10^{-18}\ \text{J}$

$E = \frac{-2.18\times10^{-18}}{1.6\times10^{-19}} \approx -13.6\ \text{eV}$

Result: The uncertainty principle alone gives the Bohr radius $r = \dfrac{4\pi\varepsilon_0\hbar^2}{me^2} = 0.53\ \text{Å}$ and the ground-state energy of hydrogen $E = -\dfrac{me^4}{32\pi^2\varepsilon_0^2\hbar^2} = -13.6\ \text{eV}$, in complete agreement with the Bohr theory.

Ehrenfest’s Theorem

Statement:
For a quantum mechanical system described by the wavefunction $\psi(\mathbf r,t)$, the expectation values of position and momentum operators obey the classical equations of motion:

$$\frac{d}{dt}\langle x\rangle=\frac{\langle p_x\rangle}{m}, \qquad \frac{d}{dt}\langle p_x\rangle=-\left\langle \frac{\partial V}{\partial x}\right\rangle=\langle F_x\rangle .$$

More generally, for any operator $A$,

$$\frac{d}{dt}\langle A\rangle=\frac{1}{i\hbar}\langle [A,H]\rangle+\left\langle \frac{\partial A}{\partial t}\right\rangle .$$

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Proof

Let the expectation value of an operator $A$ be

$$\langle A\rangle=\int \psi^* A\psi\,d^3r .$$

Differentiating with respect to time,

$$\frac{d}{dt}\langle A\rangle = \int \frac{\partial \psi^*}{\partial t}A\psi\,d^3r + \int \psi^* \frac{\partial A}{\partial t}\psi\,d^3r + \int \psi^* A\frac{\partial \psi}{\partial t}\,d^3r .$$

Using the time-dependent Schrödinger equations,

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

we get

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

Substituting,

$$\frac{d}{dt}\langle A\rangle = \frac{i}{\hbar}\int (H\psi)^*A\psi\,d^3r + \left\langle \frac{\partial A}{\partial t}\right\rangle - \frac{i}{\hbar}\int \psi^* A H\psi\,d^3r .$$

Since $H$ is Hermitian,

$$\int (H\psi)^*A\psi\,d^3r=\int \psi^* H A\psi\,d^3r .$$

Therefore,

$$\frac{d}{dt}\langle A\rangle = \frac{i}{\hbar}\int \psi^*(HA-AH)\psi\,d^3r + \left\langle \frac{\partial A}{\partial t}\right\rangle .$$

But $HA-AH=[H,A]=-[A,H]$. Hence,

$$\boxed{ \frac{d}{dt}\langle A\rangle = \frac{1}{i\hbar}\langle [A,H]\rangle + \left\langle \frac{\partial A}{\partial t}\right\rangle }$$

This is the general form of Ehrenfest’s theorem.

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1. Position–momentum relation

Take $A=x$. Then $\dfrac{\partial x}{\partial t}=0$, and

$$[x,H]=\left[x,\frac{p_x^2}{2m}+V(x)\right] = \frac{1}{2m}[x,p_x^2].$$

Using $[x,p_x]=i\hbar$,

$$[x,p_x^2]=2i\hbar p_x .$$

Thus,

$$[x,H]=\frac{i\hbar}{m}p_x .$$

Therefore,

$$\frac{d}{dt}\langle x\rangle = \frac{1}{i\hbar}\left\langle \frac{i\hbar}{m}p_x\right\rangle = \frac{\langle p_x\rangle}{m}.$$

So,

$$\boxed{\frac{d}{dt}\langle x\rangle=\frac{\langle p_x\rangle}{m}}$$

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2. Momentum–force relation

Take $A=p_x$. Then $\dfrac{\partial p_x}{\partial t}=0$, and

$$[p_x,H]=[p_x,V(x)].$$

Using the operator relation

$$[p_x,f(x)]=-i\hbar\frac{\partial f}{\partial x},$$

we get

$$[p_x,V(x)]=-i\hbar\frac{\partial V}{\partial x}.$$

Hence,

$$\frac{d}{dt}\langle p_x\rangle = \frac{1}{i\hbar}\left\langle -i\hbar\frac{\partial V}{\partial x}\right\rangle = -\left\langle \frac{\partial V}{\partial x}\right\rangle .$$

Since $F_x=-\dfrac{\partial V}{\partial x}$,

$$\boxed{\frac{d}{dt}\langle p_x\rangle=\langle F_x\rangle}$$

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Physical Significance

Ehrenfest’s theorem shows that the average motion of a quantum mechanical wave packet follows the classical Newtonian equations of motion. Thus, classical mechanics emerges as a limiting case of quantum mechanics for large systems.

Free particle: solution of the Schrödinger wave equation

A free particle is one on which no force acts, so its potential energy is zero everywhere:

[ V(x)=0 ]

1. Schrödinger wave equation

The time-dependent Schrödinger equation in one dimension is

[ i\hbar \frac{\partial \Psi}{\partial t}

-\frac{\hbar^2}{2m}\frac{\partial^2 \Psi}{\partial x^2} + V(x)\Psi ]

For a free particle, (V(x)=0), so

[ i\hbar \frac{\partial \Psi}{\partial t}

-\frac{\hbar^2}{2m}\frac{\partial^2 \Psi}{\partial x^2} ]

2. Separation of variables

Let

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

where (E) is the total energy of the particle. Substituting this into the free-particle Schrödinger equation gives the time-independent Schrödinger equation:

[ -\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 ]

3. Solution of the differential equation

Put

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

Then the equation becomes

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

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

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

or, equivalently,

[ \psi(x)=C e^{ikx}+D e^{-ikx} ]

where (A,B,C,D) are arbitrary constants determined by boundary conditions.

4. Energy of the free particle

From

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

we get

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

Since the free particle is not confined to any finite region, there is no boundary condition that restricts (k). Therefore (k) can take any real value, and the energy (E) is continuous, not quantised.

5. Complete time-dependent wave function

Using (\Psi(x,t)=\psi(x)e^{-iEt/\hbar}), the complete wave function is

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

where

[ \omega=\frac{E}{\hbar} ]

Thus the free-particle wave function is a superposition of plane waves travelling in the (+x) and (-x) directions.

6. Extension to three dimensions

In three dimensions, the free-particle Schrödinger equation is

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

Its plane-wave solution is

[ \psi(\mathbf r)=A e^{i\mathbf k\cdot \mathbf r} ]

and the energy is

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

where

[ k^2=k_x^2+k_y^2+k_z^2 ]

Important note

A free-particle wave function extends over all space and cannot be normalised in the usual way, because

[ \int_{-\infty}^{\infty}|\Psi|^2dx ]

diverges. Such states are normalised in a box or treated as plane waves with continuous energy. Thus the free particle has a continuous energy spectrum with energy

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

and wave function

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

Probability Density and Probability Current Density

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 represent a physical quantity. Instead, the physically meaningful quantity is the probability density, defined as

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

The quantity $\rho(\mathbf{r}, t)\,d^3r$ gives the probability of finding the particle in the volume element $d^3r$ around the point $\mathbf{r}$ at time $t$. Because the particle must be found somewhere in space, the probability density satisfies the normalization condition

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

Probability Current Density

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

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

where $m$ is the mass of the particle and $\hbar$ is the reduced Planck's constant. The quantity $\mathbf{J}\cdot d\mathbf{S}$ gives the probability flowing per unit time through the surface element $d\mathbf{S}$.

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Derivation of the Equation of Continuity

The time-dependent Schrödinger equation and its complex conjugate are

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

Multiply equation (1) by $\Psi^*$ from the left:

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

Multiply equation (2) by $\Psi$ from the left:

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

Subtracting equation (4) from equation (3), the potential terms $V\Psi^*\Psi$ cancel:

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

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

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

Dividing throughout by $i\hbar$:

$$\frac{\partial \rho}{\partial t} = -\frac{\hbar}{2mi}\left(\Psi^* \nabla^2 \Psi - \Psi \nabla^2 \Psi^*\right) \tag{5}$$

Now consider the divergence of the probability current density:

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

The first and third terms cancel, leaving

$$\nabla \cdot \mathbf{J} = \frac{\hbar}{2mi}\left(\Psi^* \nabla^2 \Psi - \Psi \nabla^2 \Psi^*\right) \tag{6}$$

Comparing equations (5) and (6), we obtain

$$\frac{\partial \rho}{\partial t} = -\,\nabla \cdot \mathbf{J}$$

or

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

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Conclusion

This is the equation of continuity for probability in quantum mechanics. It has exactly the same form as the continuity equation of fluid flow or of electric charge conservation. It expresses the conservation of probability: the rate of decrease of probability density in any volume equals the net probability current flowing out through the surface bounding that volume. In integral form,

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

Thus, if the wave function vanishes at infinity (as required for a normalizable state), the total probability $\int \rho\, d^3r$ remains constant in time, confirming that probability is conserved.

(a) Particle velocity equals the group velocity of a wave packet.

A wave packet is a superposition of plane waves:

$\psi(x,t)=\int A(k)\,e^{i(kx-\omega t)}\,dk,\qquad v_g=\frac{d\omega}{dk}.$

For a matter wave, $p=\hbar k$ and $E=\hbar\omega$. For a free particle $E=\dfrac{p^2}{2m}=\dfrac{\hbar^2k^2}{2m}$, so

$\omega=\frac{\hbar k^2}{2m}\;\Rightarrow\; v_g=\frac{d\omega}{dk}=\frac{\hbar k}{m}=\frac pm=v_{\text{particle}} .$

The group velocity of the packet equals the velocity of the particle.

OR: Degeneracy. When two or more linearly independent eigenfunctions have the same energy eigenvalue, the level is degenerate, and the number of such eigenfunctions is its degree of degeneracy.

For a hydrogen-like atom the energy depends only on the principal quantum number $n$. For a given $n$, $l=0,1,\dots,n-1$, and for each $l$ there are $2l+1$ values of $m=-l,\dots,+l$. Hence

$\sum_{l=0}^{n-1}(2l+1)=n^2$

eigenfunctions $\psi_{nlm}$ share the energy $E_n$ ($2n^2$ if the two spin states are included).

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(b) The momentum operator is Hermitian.

With $\hat p=-i\hbar\,\dfrac{d}{dx}$ and square-integrable $\phi,\psi$, integration by parts gives

$\langle\phi|\hat p\psi\rangle=\int_{-\infty}^{\infty}\phi^*\left(-i\hbar\frac{d\psi}{dx}\right)dx=-i\hbar\Big[\phi^*\psi\Big]_{-\infty}^{\infty}+i\hbar\int_{-\infty}^{\infty}\frac{d\phi^*}{dx}\,\psi\,dx .$

The boundary term vanishes because $\phi,\psi\to0$ at infinity. Then

$\langle\phi|\hat p\psi\rangle=\int_{-\infty}^{\infty}\left(-i\hbar\frac{d\phi}{dx}\right)^{*}\psi\,dx=\langle\hat p\phi|\psi\rangle ,$

so $\hat p^\dagger=\hat p$ and the momentum operator is Hermitian.

OR: Ball of mass 10 g in a box of length 10 cm. Here $m=0.01$ kg and $L=0.1$ m. For a particle in a box $E_n=\dfrac{n^2h^2}{8mL^2}$, so

$E_1=\frac{(6.626\times10^{-34})^2}{8(0.01)(0.1)^2}\approx5.5\times10^{-64}\ \text{J}\approx3.4\times10^{-45}\ \text{eV}.$

At room temperature $kT\approx4.1\times10^{-21}$ J, so the ball occupies a very high level:

$n\approx\sqrt{\frac{kT}{E_1}}\approx2.7\times10^{21}.$

The spacing between neighbouring levels is

$\Delta E\approx2nE_1\approx3\times10^{-42}\ \text{J}\approx2\times10^{-23}\ \text{eV},$

far too small to resolve. The de Broglie wavelength is

$\lambda=\frac{h}{\sqrt{2mkT}}\approx7\times10^{-23}\ \text{m},$

which is vastly smaller than $L=0.1$ m. Hence the energy levels form a quasi-continuum and the quantised energy states cannot be observed for the ball. This is the classical limit.

(a) General results for two-particle bound states.

For particles of masses $m_1,m_2$ interacting through a potential $V(\mathbf r)$ that depends only on $\mathbf r=\mathbf r_1-\mathbf r_2$, the motion separates into centre-of-mass and relative motion:

$\Psi(\mathbf r_1,\mathbf r_2)=e^{i\mathbf K\cdot\mathbf R}\,\phi(\mathbf r),\qquad \mathbf R=\frac{m_1\mathbf r_1+m_2\mathbf r_2}{m_1+m_2}.$

• The centre of mass moves like a free particle of mass $M=m_1+m_2$, with energy $E_{cm}=\hbar^2K^2/2M$.
• The relative motion is that of a single particle of reduced mass $\mu=\dfrac{m_1m_2}{m_1+m_2}$ in the potential $V(r)$: $\left[-\frac{\hbar^2}{2\mu}\nabla^2+V(r)\right]\phi=E_{rel}\,\phi .$
• The total energy is $E=E_{cm}+E_{rel}$. Bound states have $E_{rel}<0$ and discrete values.
• For a central potential $\phi=R(r)Y_{lm}(\theta,\varphi)$ and each level is $(2l+1)$-fold degenerate in $m$.
• The binding energy depends on $\mu$, not on the individual masses. For hydrogen $E_n=-\dfrac{\mu e^4}{2(4\pi\epsilon_0)^2\hbar^2n^2}$. In positronium $\mu=m_e/2$, so its levels are half those of hydrogen, and the finite proton mass causes the isotope shift between H and D.

(b) Virtual binding.

A virtual state is a state in which two particles almost, but not quite, form a bound system. The interaction is attractive, but the well is slightly too shallow to hold a bound state. For an s-wave square well of depth $V_0$ and range $b$, a bound state exists only if $V_0>\dfrac{\pi^2\hbar^2}{8\mu b^2}$.

If $V_0$ falls just short of this value:

• There is no normalisable bound-state energy level.
• The scattering length is large and negative, and the low-energy cross section is very large.
• The outside wave function resembles a bound state but grows as $e^{+\kappa r}$ instead of decaying, a pole of the scattering amplitude on the negative imaginary $k$ axis.

An example is the singlet ($S=0$) neutron–proton state, with $a_s\approx-23.7$ fm. It is only virtually bound, whereas the triplet state is the bound deuteron.

(c) Conditions on the wave function. The wave function $\psi$ must be:

1. Single-valued, so that the probability at each point is unique.
2. Finite everywhere, and square-integrable ($\int|\psi|^2d\tau$ finite, normalisable); for a bound state $\psi\to0$ at infinity.
3. Continuous everywhere.
4. Such that its first derivative $d\psi/dx$ is continuous wherever the potential is finite (at an infinite wall only $\psi$ is continuous, with $\psi=0$ inside the wall).

(d) Meaning and importance of parity.

The parity operator $\hat\Pi$ inverts the space coordinates: $\hat\Pi\psi(\mathbf r)=\psi(-\mathbf r)$. Since $\hat\Pi^2=1$ its eigenvalues are $\pm1$: an even wave function has parity $+1$ and an odd one has parity $-1$. For spherical harmonics the parity is $(-1)^l$.

Importance:

• If $V(\mathbf r)=V(-\mathbf r)$, then $[\hat H,\hat\Pi]=0$, so parity is a constant of motion and stationary states have definite parity. In a symmetric 1D well the states alternate even, odd, even, and so on, starting with the even ground state.
• It simplifies calculations, since integrals of odd functions over symmetric limits vanish.
• It gives selection rules: electric dipole transitions need a change of parity (Laporte rule, $\Delta l=\pm1$).
• Parity is conserved in strong and electromagnetic interactions, but not in weak interactions.

Given: Mass of particle, (m = 1\ \text{gm} = 1 \times 10^{-3}\ \text{kg})

Spring is stretched by (x = 1\ \text{cm} = 1 \times 10^{-2}\ \text{m}) by force (F = 0.1\ \text{N})

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{10000}=100\ \text{rad/s} ]

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

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

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

Given:
Side of cubical box, (a = 1,\text{Å} = 1 \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 an electron in a 3-D infinite potential cubical box, the energy is

[ E_{n_x n_y n_z}=\frac{h^2}{8ma^2}\left(n_x^2+n_y^2+n_z^2\right) ]

The lowest energy occurs for

[ n_x=n_y=n_z=1 ]

So,

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

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

Substituting values,

[ E_{\min}=\frac{3(6.626\times10^{-34})^2}{8(9.11\times10^{-31})(1\times10^{-10})^2} ]

[ E_{\min}=\frac{3(4.390\times10^{-67})}{8(9.11\times10^{-31})(1\times10^{-20})} ]

[ E_{\min}=\frac{1.317\times10^{-66}}{7.288\times10^{-50}} ]

[ E_{\min}=1.807\times10^{-17},\text{J} ]

In electron volt,

[ E_{\min}=\frac{1.807\times10^{-17}}{1.602\times10^{-19}},\text{eV} ]

[ E_{\min}\approx 112.8,\text{eV} ]

[ \boxed{E_{\min}\approx 1.81\times10^{-17},\text{J}\approx 112.8,\text{eV}} ]

So, the lowest energy of the electron confined in a cubical box of side (1,\text{Å}) is about 112.8 eV.

Given: Two operators (A) and (B) satisfy the commutation relation [ [A, B] = \alpha B ] where (\alpha) is a number (a c-number, i.e. commutes with (A) and (B)).

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

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Step 1: Define the operator function

Let [ f(\lambda) = e^{\lambda A} B e^{-\lambda A} ] where (\lambda) is a real parameter. We want the value at (\lambda = 1).

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Step 2: Differentiate with respect to (\lambda)

[ \frac{df}{d\lambda} = \frac{d}{d\lambda}\left(e^{\lambda A} B e^{-\lambda A}\right) ]

Using the product rule and the fact that (\frac{d}{d\lambda}e^{\lambda A} = A,e^{\lambda A} = e^{\lambda A}A),

[ \frac{df}{d\lambda} = A e^{\lambda A} B e^{-\lambda A} + e^{\lambda A} B \left(-A e^{-\lambda A}\right) ]

[ = A,f(\lambda) - e^{\lambda A} B A e^{-\lambda A} ]

Now write the second term as (e^{\lambda A} B A e^{-\lambda A} = e^{\lambda A}(BA)e^{-\lambda A}). Since (BA = AB - [A,B] = AB - \alpha B),

[ \frac{df}{d\lambda} = A f(\lambda) - e^{\lambda A}(AB - \alpha B)e^{-\lambda A} ]

[ = A f(\lambda) - e^{\lambda A}A B e^{-\lambda A} + \alpha, e^{\lambda A} B e^{-\lambda A} ]

[ = A f(\lambda) - A f(\lambda) + \alpha f(\lambda) ]

[ \boxed{\frac{df}{d\lambda} = \alpha, f(\lambda)} ]

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Step 3: Solve the differential equation

[ \frac{df}{f} = \alpha, d\lambda ]

Integrating,

[ \ln f(\lambda) = \alpha \lambda + C ]

[ f(\lambda) = C' e^{\alpha \lambda} ]

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Step 4: Apply the boundary condition

At (\lambda = 0),

[ f(0) = e^{0} B e^{0} = B ]

So (C' = B). Hence,

[ f(\lambda) = B, e^{\alpha \lambda} ]

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Step 5: Put (\lambda = 1)

[ f(1) = e^{A} B e^{-A} = B, e^{\alpha} ]

[ \boxed{e^{A} B e^{-A} = e^{\alpha} B} ]

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Hence proved.

This is the operator form of the Baker–Campbell–Hausdorff (BCH) relation for the special case ([A,B] = \alpha B), and it is the key identity used in the theory of Lie algebras and in quantum mechanics (e.g. displacement and squeezing operators).

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

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

where (I) is the moment of inertia of the molecule about its centre of mass.

For (l=1),

[ E_1=\frac{\hbar^2}{2I}(1)(2)=\frac{\hbar^2}{I} ]

For (l=2),

[ E_2=\frac{\hbar^2}{2I}(2)(3)=\frac{3\hbar^2}{I} ]

Therefore, the energy difference is

[ \Delta E=E_2-E_1=\frac{3\hbar^2}{I}-\frac{\hbar^2}{I} ]

[ \boxed{\Delta E=\frac{2\hbar^2}{I}} ]

In terms of Planck’s constant (h), since (\hbar=\frac{h}{2\pi}),

[ \Delta E=\frac{2}{I}\left(\frac{h}{2\pi}\right)^2 ]

[ \boxed{\Delta E=\frac{h^2}{2\pi^2 I}} ]

Thus, the energy difference between the stationary states (l=1) and (l=2) is (\frac{2\hbar^2}{I}).

For a hydrogen-atom wave function depending only on (r),

[ \psi(r,\theta,\phi)=Ae^{-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 ]

Separate the integrals:

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

Now,

[ \int_0^\pi \sin\theta,d\theta=2 ]

[ \int_0^{2\pi}d\phi=2\pi ]

So the angular part gives

[ 2\times 2\pi=4\pi ]

Also,

[ \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,

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

[ A^2\pi a_0^3=1 ]

[ A^2=\frac{1}{\pi a_0^3} ]

Hence,

[ \boxed{A=\frac{1}{\sqrt{\pi a_0^3}}} ]

So the normalized hydrogen wave function is

[ \boxed{\psi(r,\theta,\phi)=\frac{1}{\sqrt{\pi a_0^3}}e^{-r/a_0}} ]

The momentum expectation value is

[ \langle p\rangle=\int_{-\infty}^{\infty}\Psi^*(x)\left(-i\hbar\frac{\partial}{\partial x}\right)\Psi(x),dx ]

Given

[ \Psi(x)=N\exp\left[-\frac{x^2}{2\alpha^2}-ikx\right] ]

First compute the derivative:

[ \frac{\partial \Psi}{\partial x} =\Psi(x)\frac{d}{dx}\left[-\frac{x^2}{2\alpha^2}-ikx\right] ]

[ \frac{\partial \Psi}{\partial x} =\Psi(x)\left[-\frac{x}{\alpha^2}-ik\right] ]

Therefore,

[ -i\hbar\frac{\partial \Psi}{\partial x} =-i\hbar \Psi(x)\left[-\frac{x}{\alpha^2}-ik\right] ]

[ -i\hbar\frac{\partial \Psi}{\partial x} =\Psi(x)\left[\frac{i\hbar x}{\alpha^2}-\hbar k\right] ]

Now,

[ \langle p\rangle =\int_{-\infty}^{\infty}\Psi^*(x)\Psi(x) \left[\frac{i\hbar x}{\alpha^2}-\hbar k\right]dx ]

Since

[ \Psi^*(x)\Psi(x)=|N|^2\exp\left[-\frac{x^2}{\alpha^2}\right] ]

is an even function of (x), the term containing (x) gives zero:

[ \int_{-\infty}^{\infty} x\exp\left[-\frac{x^2}{\alpha^2}\right]dx=0 ]

Thus,

[ \langle p\rangle =-\hbar k\int_{-\infty}^{\infty}|\Psi(x)|^2dx ]

Using normalization,

[ \int_{-\infty}^{\infty}|\Psi(x)|^2dx=1 ]

Hence,

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

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