Exercises — Schrödinger equation — time-dependent, time-independent

(a) has ${\partial }_t$ and $\Psi (x,t)$ → it is the TDSE. The bracket is $\hat{H}$, the Hamiltonian. (b) has an ordinary derivative $d^2/d{x}^2$, only $\psi (x)$, and a constant $E$ on the right → it is the TISE, the eigenvalue equation $\hat{H}\psi =E\psi$. What told us: a time derivative ⇒ TDSE; a bare number $E$ multiplying $\psi$ ⇒ TISE.

Only $|\Psi {|}^2$ is observable ($\Psi$ and $\psi$ are complex, not directly measurable). $|\Psi {|}^2={\Psi }^{\ast }\Psi =\left({\psi }^{\ast }{e}^{+iEt/\hslash }\right)\left(\psi e^{-iEt/\hslash }\right)=|\psi {|}^2{e}^0=|\psi {|}^2.$ The two phase factors are complex conjugates; their product is $e^{iEt/\hslash -iEt/\hslash }=e^0=1$. So $|\Psi {|}^2=|\psi (x){|}^2$ has no $t$ — that is why it is called a stationary state.

Plug numbers with $n=1$: $E_1=\frac{{\pi }^2(1.055\times 10^{-34}{)}^2}{2(9.11\times 10^{-31})(0.50\times 10^{-9}{)}^2}.$ Numerator: ${\pi }^2(1.113\times 10^{-68})=1.099\times 10^{-67}$. Denominator: $2(9.11\times 10^{-31})(2.5\times 10^{-19})=4.555\times 10^{-49}$. $E_1=2.41\times 10^{-19}\text{J}=\frac{2.41\times 10^{-19}}{1.602\times 10^{-19}}=1.51\text{eV}.$ Why this matters: a room-temperature electron trapped in an atom-sized box has energies of order 1 eV — matching real atomic/optical scales.

Since $E_n\propto n^2$: $E_3/E_1=3^2/1^2=9$. $E_3=9E_1=9(1.51)=13.6\text{eV}$, so $E_3-E_1=13.6-1.51=12.1\text{eV}$. Why: only the $n^2$ factor changes; everything else cancels in the ratio.

Normalisation demands total probability $1$: ${\int }_0^a|C{|}^{2}dx=|C{|}^{2}a=1\Rightarrow C=\frac{1}{\sqrt{a}}.$ Units check first (crucial): $|\psi {|}^{2}dx$ is a pure probability (dimensionless), and $dx$ has units of metres, so $|\psi {|}^2$ must carry ${\text{m}}^{-1}$ and hence $\psi$ carries ${\text{m}}^{-1/2}$. That is exactly what $1/\sqrt{a}$ gives. Numerically: $C=\frac{1}{\sqrt{2.0\times 10^{-9}\text{m}}}=2.24\times 10^4{\text{m}}^{-1/2}.$ Why: $|\psi {|}^2$ is a flat block of height $C^2$ (units ${\text{m}}^{-1}$) over width $a$ (units m); its area $C^{2}a$ is dimensionless and forced to $1$. Dropping the units of $a$ would give a meaningless bare number for $C$.

(a) ${\psi }^{\prime \prime }=-k^2\psi$ ⇒ $-\frac{{\hslash }^2}{2m}(-k^2)\psi =\frac{{\hslash }^2{k}^2}{2m}\psi$. Yes, $E={\hslash }^2{k}^2/2m=p^2/2m>0$ with $p=\hslash k$. ✓ (b) ${\psi }^{\prime \prime }=-k^2\psi$ too ⇒ same $E>0$. Yes ✓ (a standing combination of $e^{\pm ikx}$, i.e. equal and opposite momenta $\pm \hslash k$). (c) ${\psi }^{\prime \prime }=+k^2\psi$ ⇒ $E=-\frac{{\hslash }^2{k}^2}{2m}<0$. This is a valid eigenfunction but with negative $E$ and it blows up as $x\to +\infty$, so it cannot be normalised for a free particle. Not physical here. Why this matters: the sign of ${\psi }^{\prime \prime }/\psi$ decides everything. Negative (${\psi }^{\prime \prime }=-k^2\psi$) ⇒ oscillation ⇒ real wavenumber $k$ ⇒ real momentum $p=\hslash k$ ⇒ a genuine travelling/standing wave. Positive (${\psi }^{\prime \prime }=+k^2\psi$) ⇒ exponential growth/decay ⇒ imaginary $k$ ⇒ no real momentum ⇒ this only survives inside a classically forbidden barrier. The figure below draws both, and marks how the momentum arrow $p=\hslash k$ attaches only to the oscillating case.

$n=0$ gives $\sin (0)=0$ everywhere ⇒ $\psi \equiv 0$ ⇒ $\int |\psi {|}^2=0\ne 1$: no particle exists, cannot be normalised. Negative $n$: $\sin (-n\pi x/L)=-\sin (n\pi x/L)$. This is just $-1$ times the $+n$ state; multiplying a wavefunction by a constant phase gives the same physical state ($|\Psi {|}^2$ unchanged). So $n=1,2,3,\dots$ enumerate all distinct states. Why: physical content lives in $|\psi {|}^2$, which is blind to an overall sign or phase.

Attach the time factor: ${\Psi }_1(x,t)=\sqrt{2/L}\sin (\pi x/L)e^{-i{E}_{1}t/\hslash }$. LHS (time derivative pulls down $-i{E}_1/\hslash$): $i\hslash {\partial }_t{\Psi }_1=i\hslash \left(-\frac{i{E}_1}{\hslash }\right){\Psi }_1=E_1{\Psi }_1.$ RHS ($V=0$ inside; ${\psi }_1^{\prime \prime }=-(\pi /L{)}^2{\psi }_1$ so $\hat{H}{\psi }_1=\frac{{\hslash }^2}{2m}(\pi /L{)}^2{\psi }_1=E_1{\psi }_1$): $\hat{H}{\Psi }_1=E_1{\Psi }_1.$ LHS $=$ RHS. ✓ Why this matters: it shows that solving the TISE and multiplying by $e^{-iEt/\hslash }$ automatically solves the TDSE — that is the whole point of separation of variables.

Write $\Psi =\frac{1}{\sqrt{2}}({\psi }_1{e}^{-i{E}_{1}t/\hslash }+{\psi }_2{e}^{-i{E}_{2}t/\hslash })$ with ${\psi }_1,{\psi }_2$ real. Then $|\Psi {|}^2=\frac{1}{2}({\psi }_1^2+{\psi }_2^2+2{\psi }_1{\psi }_2\cos [(E_2-E_1)t/\hslash ]).$ The cross term keeps a $\cos ({\omega }_{21}t)$: the probability cloud sloshes back and forth. A single energy eigenstate is stationary; a mix of two is not. Why this matters: all real dynamics (an electron oscillating, emitting light) comes from superposing states of different energy. Same-energy phases cancel in $|\Psi {|}^2$; different-energy phases don't.

(a) $E_2=4E_1=6.05\text{eV}$, so $E_2-E_1=3E_1=4.54\text{eV}$. Convert: $4.54\times 1.602\times 10^{-19}=7.27\times 10^{-19}\text{J}$. (b) $\nu =\frac{7.27\times 10^{-19}}{6.626\times 10^{-34}}=1.097\times 10^{15}\text{Hz}$. (c) $\lambda =\frac{3.0\times 10^8}{1.097\times 10^{15}}=2.73\times 10^{-7}\text{m}=273\text{nm}$ (ultraviolet). Why this matters: the box model predicts a real spectral line in the UV — quantum jumps between allowed levels create the discrete colours atoms emit.

$\frac{E_{n+1}-E_n}{E_n}=\frac{(n+1{)}^2-n^2}{n^2}=\frac{2n+1}{n^2}.$ At $n=1$: $\frac{3}{1}=3$ (300% jump — very quantum, levels far apart). At $n=100$: $\frac{201}{10000}=0.0201$ (about 2%). As $n\to \infty$, $\frac{2n+1}{n^2}\to 0$: neighbouring levels merge into a near-continuum. Why this matters: this is the correspondence principle — quantum discreteness becomes invisible at large quantum numbers, recovering classical physics. Ties directly to Energy Quantization.

Quantization in the box came from boundary conditions $\psi (0)=\psi (L)=0$, which forced $kL=n\pi$ — discrete $k$. A free particle has no confining walls ⇒ no boundary condition selecting $k$ ⇒ any real $k$ (hence any $E={\hslash }^2{k}^2/2m\ge 0$) is allowed ⇒ a continuous spectrum. Why: discreteness is not built into the Schrödinger equation — it is imposed by confinement. Remove the box, remove the quantization.