Exercises — Wave function ψ — probability density - ψ - ²

Answer: $|\psi {|}^2$.

• $\psi$ — can be negative or complex, and a probability can never be negative. ✗
• ${\psi }^2$ — if $\psi$ is complex, ${\psi }^2$ is still complex (e.g. $i^2=-1$). ✗
• $|\psi |$ — non-negative, so it looks fine, but diffraction experiments (amplitudes add, then square) match $|\psi {|}^2$, not $|\psi |$. ✗
• $|\psi {|}^2={\psi }^{\ast }\psi =a^2+b^2$ — real, $\ge 0$, and matches experiment. ✓

$|\psi {|}^{2}dx$ is a pure probability (dimensionless). Since $dx$ has units of metres (m), $|\psi {|}^2$ must have units ${\text{m}}^{-1}$ (probability per metre). Because $|\psi {|}^2=(\psi {)}^2$ in magnitude, $\psi$ has units $({\text{m}}^{-1}{)}^{1/2}={\text{m}}^{-1/2}$.

What: demand total probability $=1$. Why: the particle must lie somewhere in the box. ${\int }_0^L|A{|}^{2}dx=|A{|}^{2}L=1\Rightarrow A=\frac{1}{\sqrt{L}}.$ A flat $|\psi {|}^2=1/L$ means equally likely anywhere in the box — the uniform distribution.

$P={\int }_0^{L/4}\frac{1}{L}dx=\frac{1}{L}\cdot \frac{L}{4}=\frac{1}{4}.$ A quarter of the box holds a quarter of the probability — exactly what "uniform" should give.

What: multiply by the conjugate. Why: ${\psi }^{\ast }$ flips $i\to -i$ in the phase. $|\psi {|}^2={\psi }^{\ast }\psi =(A^{\ast }{e}^{-ikx}{e}^{-|x|/L}\left)\right(A{e}^{ikx}{e}^{-|x|/L}).$ The phases combine: $e^{-ikx}{e}^{ikx}=e^0=1$. So $|\psi {|}^2=|A{|}^2{e}^{-2|x|/L}.$ The oscillating factor $e^{ikx}$ (which carries the momentum, via the de Broglie Wavelength) is invisible to probability density.

Normalize. We need ${\int }_0^L{A}^2{\sin }^2\frac{\pi x}{L}dx=1$. Use ${\sin }^2\theta =\frac{1}{2}(1-\cos 2\theta )$: ${\int }_0^L{\sin }^2\frac{\pi x}{L}dx=\frac{1}{2}{\int }_0^L(1-\cos \frac{2\pi x}{L})dx=\frac{1}{2}(L-0)=\frac{L}{2}.$ (The cosine integrates to zero over a full period.) So $A^2\cdot \frac{L}{2}=1\Rightarrow A=\sqrt{2/L}$.

Forecast (symmetry). ${\sin }^2\frac{\pi x}{L}$ is a hump symmetric about the box centre $x=L/2$. So the left half and right half must carry equal probability $\Rightarrow P=\frac{1}{2}$.

Verify by integrating. $P=\frac{2}{L}{\int }_0^{L/2}{\sin }^2\frac{\pi x}{L}dx=\frac{2}{L}\cdot \frac{1}{2}[x-\frac{L}{2\pi }\sin \frac{2\pi x}{L}{]}_0^{L/2}=\frac{1}{L}\cdot \frac{L}{2}=\frac{1}{2}.✓$

$P=\frac{2}{L}{\int }_0^{L/4}{\sin }^2\frac{\pi x}{L}dx=\frac{2}{L}\cdot \frac{1}{2}[x-\frac{L}{2\pi }\sin \frac{2\pi x}{L}{]}_0^{L/4}.$ At $x=L/4$: the bracket is $\frac{L}{4}-\frac{L}{2\pi }\sin \frac{\pi }{2}=\frac{L}{4}-\frac{L}{2\pi }$. $P=\frac{1}{L}(\frac{L}{4}-\frac{L}{2\pi })=\frac{1}{4}-\frac{1}{2\pi }\approx \mathrm{0.0908.}$ Note it is less than $\frac{1}{4}$ — the hump pushes probability toward the centre, starving the edges.

Normalize. $|\psi {|}^2=|A{|}^2{e}^{-2|x|/L}$. Use the even symmetry to fold the integral: ${\int }_{-\infty }^{\infty }|A{|}^2{e}^{-2|x|/L}dx=2|A{|}^2{\int }_0^{\infty }{e}^{-2x/L}dx=2|A{|}^2\cdot \frac{L}{2}=|A{|}^{2}L.$ Set $=1\Rightarrow A=1/\sqrt{L}$.

Probability within one decay length. $P(|x|\le L)=\frac{2}{L}{\int }_0^L{e}^{-2x/L}dx=\frac{2}{L}\cdot \frac{L}{2}(1-e^{-2})=1-e^{-2}\approx \mathrm{0.8647.}$ So about $86$ of the probability lives inside $\pm L$ — the tails are thin, as a square-integrable $\psi$ demands.

Compute the density of ${\psi }_2$: $|{\psi }_2{|}^2=(e^{i\pi /3}A\sin \frac{\pi x}{L}{)}^{\ast }\left(e^{i\pi /3}A\sin \frac{\pi x}{L}\right)=e^{-i\pi /3}{e}^{i\pi /3}|A{|}^2{\sin }^2\frac{\pi x}{L}=|A{|}^2{\sin }^2\frac{\pi x}{L}=|{\psi }_1{|}^2.$ Since $|e^{i\theta }|=1$ for any real $\theta$, a global phase $e^{i\theta }$ multiplies $|\psi {|}^2$ by $1$. All probabilities — and all Expectation Values $⟨x⟩=\int x|\psi {|}^{2}dx$ — are unchanged. Principle: a global phase is physically invisible.

(a) Normalize. The density is piecewise constant, so the integral is just height² × width: ${\int }_0^d{\alpha }^{2}dx+{\int }_d^{2d}{\beta }^{2}dx={\alpha }^{2}d+{\beta }^{2}d=1\Rightarrow d({\alpha }^2+{\beta }^2)=1.$

(b) Equally likely regions. The probability of the left region is ${\alpha }^{2}d$, of the right is ${\beta }^{2}d$. Equal $\Rightarrow {\alpha }^{2}d={\beta }^{2}d=\frac{1}{2}$, so $\alpha =\beta =\frac{1}{\sqrt{2d}}.$ (Check: $d({\alpha }^2+{\beta }^2)=d(\frac{1}{2d}+\frac{1}{2d})=1.✓$)

(c) Expectation value. With $|\psi {|}^2=\frac{1}{2d}$ on $[0,2d]$ (uniform over the whole $2d$): $⟨x⟩={\int }_0^{2d}x\frac{1}{2d}dx=\frac{1}{2d}\cdot \frac{(2d{)}^2}{2}=\frac{1}{2d}\cdot 2d^2=d.$ The mean position is the midpoint $x=d$ — exactly the symmetry centre. Sensible.

Multiply the two spreads: $\Delta x\Delta p=\Delta x\cdot \frac{\hslash }{2\Delta x}=\frac{\hslash }{2}.$ The product equals exactly $\hslash /2$, so the Gaussian sits on the equality edge of $\Delta x\Delta p\ge \hslash /2$ — it is the minimum-uncertainty state. Any other shape spreads $|\psi {|}^2$ and its Fourier partner wider, pushing the product above $\hslash /2$.

L1 ::: Recognise that only $|\psi {|}^2$ is a valid density, and know its units. L2 ::: Apply normalization ($÷\sqrt{N}$) and strip a global phase. L3 ::: Integrate ${\sin }^2$ and use symmetry; equal lengths ≠ equal probability unless flat. L4 ::: Handle decaying tails and prove global-phase invariance. L5 ::: Build superpositions, compute $⟨x⟩$, and connect to the uncertainty bound.