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Wyatt's Notes — University

Wave Functions and the Schrodinger Equation

3.1 Wave Functions

In the position representation, the state is described by a wave function ψ(r,t)\psi(\mathbf{r}, t) Where ∣ψ(r,t)∣2|\psi(\mathbf{r}, t)|^2 is the probability density:

P(r∈[r,r+dr])=∣ψ(r,t)∣2 d3rP(\mathbf{r} \in [\mathbf{r}, \mathbf{r} + d\mathbf{r}]) = |\psi(\mathbf{r}, t)|^2\, d^3\mathbf{r}

Normalisation: ∫−∞∞∣ψ(r,t)∣2 d3r=1\int_{-\infty}^{\infty} |\psi(\mathbf{r}, t)|^2\, d^3\mathbf{r} = 1.

3.2 Time-Dependent Schrodinger Equation

iℏ∂ψ∂t=H^ψ=(−ℏ22m∇2+V(r,t))ψi\hbar \frac{\partial \psi}{\partial t} = \hat{H}\psi = \left(-\frac{\hbar^2}{2m}\nabla^2 + V(\mathbf{r}, t)\right)\psi

3.3 Time-Independent Schrodinger Equation

For time-independent potentials V(r)V(\mathbf{r})Separate variables: ψ(r,t)=ϕ(r)e−iEt/ℏ\psi(\mathbf{r}, t) = \phi(\mathbf{r}) e^{-iEt/\hbar}:

H^ϕ=Eϕi.e.,−ℏ22m∇2ϕ+Vϕ=Eϕ\hat{H}\phi = E\phi \quad \mathrm{i}.e., \quad -\frac{\hbar^2}{2m}\nabla^2\phi + V\phi = E\phi

This is an eigenvalue problem: EE is the energy eigenvalue, ϕ\phi is the energy eigenstate.

Properties of energy eigenstates:

  1. Orthogonality. If H^ϕn=Enϕn\hat{H}\phi_n = E_n\phi_n and H^ϕm=Emϕm\hat{H}\phi_m = E_m\phi_m with En≠EmE_n \neq E_m then ∫ϕn∗ϕm dx=0\int \phi_n^*\phi_m\,dx = 0 (since H^\hat{H} is Hermitian).

  2. Completeness. The energy eigenstates form a complete basis: any state can be expanded as ψ(x,0)=∑ncnϕn(x)\psi(x,0) = \sum_n c_n\phi_n(x) where cn=∫ϕn∗(x)ψ(x,0) dxc_n = \int \phi_n^*(x)\psi(x,0)\,dx.

  3. Stationary states. If ψ(x,0)=ϕn(x)\psi(x,0) = \phi_n(x)Then ψ(x,t)=ϕn(x)e−iEnt/ℏ\psi(x,t) = \phi_n(x)e^{-iE_nt/\hbar}. The probability density ∣ψ∣2=∣ϕn∣2|\psi|^2 = |\phi_n|^2 is time-independent.

  4. Reality of ϕ\phi. If V(x)V(x) is real and there is no magnetic field, ϕn(x)\phi_n(x) can be chosen to be real. This is because if ϕn\phi_n is a solution, so is ϕn∗\phi_n^*And degenerate solutions can be combined into real linear combinations.

3.4 Probability Current

The probability current density is

J=ℏ2mi(ψ∗∇ψ−ψ∇ψ∗)\mathbf{J} = \frac{\hbar}{2mi}(\psi^* \nabla\psi - \psi \nabla\psi^*)

It satisfies the continuity equation: ∂∣ψ∣2∂t+∇⋅J=0\frac{\partial |\psi|^2}{\partial t} + \nabla \cdot \mathbf{J} = 0 Expressing conservation of probability.

Derivation of the continuity equation. Start with the Schrodinger equation and its complex conjugate:

iℏ∂ψ∂t=−ℏ22m∇2ψ+Vψi\hbar \frac{\partial \psi}{\partial t} = -\frac{\hbar^2}{2m}\nabla^2\psi + V\psi

−iℏ∂ψ∗∂t=−ℏ22m∇2ψ∗+Vψ∗-i\hbar \frac{\partial \psi^*}{\partial t} = -\frac{\hbar^2}{2m}\nabla^2\psi^* + V\psi^*

Multiply the first by ψ∗\psi^* and the second by ψ\psiThen subtract:

iℏ ⁣(ψ∗∂ψ∂t+ψ∂ψ∗∂t)=−ℏ22m ⁣(ψ∗∇2ψ−ψ∇2ψ∗)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-hand side is iℏ ∂∣ψ∣2/∂ti\hbar\,\partial|\psi|^2/\partial t. The right-hand side is a divergence:

∇⋅(ψ∗∇ψ−ψ∇ψ∗)=ψ∗∇2ψ−ψ∇2ψ∗\nabla \cdot (\psi^*\nabla\psi - \psi\nabla\psi^*) = \psi^*\nabla^2\psi - \psi\nabla^2\psi^*

Therefore:

iℏ∂∣ψ∣2∂t=−ℏ22m∇⋅(ψ∗∇ψ−ψ∇ψ∗)i\hbar\frac{\partial |\psi|^2}{\partial t} = -\frac{\hbar^2}{2m}\nabla \cdot (\psi^*\nabla\psi - \psi\nabla\psi^*)

Dividing by iℏi\hbar:

∂∣ψ∣2∂t+∇⋅[ℏ2mi(ψ∗∇ψ−ψ∇ψ∗)]=0\frac{\partial |\psi|^2}{\partial t} + \nabla \cdot \left[\frac{\hbar}{2mi}(\psi^*\nabla\psi - \psi\nabla\psi^*)\right] = 0

∂∣ψ∣2∂t+∇⋅J=0■\frac{\partial |\psi|^2}{\partial t} + \nabla \cdot \mathbf{J} = 0 \qquad \blacksquare

3.5 Unitarity of Time Evolution

Theorem 3.1. Time evolution governed by the Schrodinger equation with a Hermitian Hamiltonian Is unitary, and therefore preserves the norm of the state vector.

Proof. The time evolution operator U^(t,t0)\hat{U}(t, t_0) is defined by:

∣ψ(t)⟩=U^(t,t0)∣ψ(t0)⟩|\psi(t)\rangle = \hat{U}(t, t_0)|\psi(t_0)\rangle

For a time-independent Hamiltonian:

U^(t,t0)=exp⁡ ⁣(−iH^(t−t0)ℏ)\hat{U}(t, t_0) = \exp\!\left(-\frac{i\hat{H}(t - t_0)}{\hbar}\right)

To prove unitarity, we show U^†U^=I^\hat{U}^\dagger \hat{U} = \hat{I}:

U^†=exp⁡ ⁣(iH^†(t−t0)ℏ)=exp⁡ ⁣(iH^(t−t0)ℏ)\hat{U}^\dagger = \exp\!\left(\frac{i\hat{H}^\dagger(t - t_0)}{\hbar}\right) = \exp\!\left(\frac{i\hat{H}(t - t_0)}{\hbar}\right)

Since H^=H^†\hat{H} = \hat{H}^\dagger (Hermitian). Therefore:

U^†U^=exp⁡ ⁣(iH^(t−t0)ℏ)exp⁡ ⁣(−iH^(t−t0)ℏ)=I^\hat{U}^\dagger \hat{U} = \exp\!\left(\frac{i\hat{H}(t - t_0)}{\hbar}\right)\exp\!\left(-\frac{i\hat{H}(t - t_0)}{\hbar}\right) = \hat{I}

Since commuting operators satisfy eAe−A=Ie^A e^{-A} = I.

Consequence. Norm preservation:

⟨ψ(t)∣ψ(t)⟩=⟨ψ(t0)∣U^†U^∣ψ(t0)⟩=⟨ψ(t0)∣ψ(t0)⟩\langle\psi(t)|\psi(t)\rangle = \langle\psi(t_0)|\hat{U}^\dagger\hat{U}|\psi(t_0)\rangle = \langle\psi(t_0)|\psi(t_0)\rangle

Total probability is conserved under time evolution. ■\blacksquare

Composing evolutions. For successive time intervals, the evolution operator composes as:

U^(t2,t0)=U^(t2,t1) U^(t1,t0)\hat{U}(t_2, t_0) = \hat{U}(t_2, t_1)\,\hat{U}(t_1, t_0)

This composition law, combined with unitarity, is the group structure underlying quantum dynamics. For a time-dependent Hamiltonian, the evolution operator is given by Dyson”s time-ordered exponential:

U^(t,t0)=Texp⁡ ⁣(−iℏ∫t0tH^(t′) dt′)\hat{U}(t, t_0) = \mathcal{T}\exp\!\left(-\frac{i}{\hbar}\int_{t_0}^{t}\hat{H}(t')\,dt'\right)

Where T\mathcal{T} denotes time ordering (later times appear to the left).

For a time-dependent Hamiltonian, the evolution operator satisfies iℏ ∂U^/∂t=H^(t)U^i\hbar\,\partial\hat{U}/\partial t = \hat{H}(t)\hat{U} With U^(t0,t0)=I^\hat{U}(t_0, t_0) = \hat{I}. Unitarity still holds: d(U^†U^)/dt=0d(\hat{U}^\dagger\hat{U})/dt = 0 since H^(t)=H^†(t)\hat{H}(t) = \hat{H}^\dagger(t).

3.6 Normalisation of Wave Functions

A physically valid wave function must satisfy ∫∣ψ∣2 dx=1\int |\psi|^2\,dx = 1. This determines the normalisation Constant.

Example 3.1. Normalise the wave function ψ(x)=Ae−α∣x∣\psi(x) = Ae^{-\alpha|x|} for −∞<x<∞-\infty \lt x \lt \infty Where α>0\alpha \gt 0.

Solution

∫−∞∞∣A∣2e−2α∣x∣ dx=2∣A∣2∫0∞e−2αx dx=2∣A∣2⋅12α=∣A∣2α=1\int_{-\infty}^{\infty} |A|^2 e^{-2\alpha|x|}\,dx = 2|A|^2 \int_0^{\infty} e^{-2\alpha x}\,dx = 2|A|^2 \cdot \frac{1}{2\alpha} = \frac{|A|^2}{\alpha} = 1

Therefore ∣A∣=α|A| = \sqrt{\alpha}And we choose A=αA = \sqrt{\alpha}:

ψ(x)=α e−α∣x∣\psi(x) = \sqrt{\alpha}\,e^{-\alpha|x|}

To find ⟨x⟩\langle x \rangle:

⟨x⟩=α∫−∞∞x e−2α∣x∣ dx=0\langle x \rangle = \alpha \int_{-\infty}^{\infty} x\,e^{-2\alpha|x|}\,dx = 0

By symmetry (the integrand is odd). For ⟨x2⟩\langle x^2 \rangle:

⟨x2⟩=2α∫0∞x2e−2αx dx=2α⋅2(2α)3=12α2\langle x^2 \rangle = 2\alpha \int_0^{\infty} x^2 e^{-2\alpha x}\,dx = 2\alpha \cdot \frac{2}{(2\alpha)^3} = \frac{1}{2\alpha^2}

So Δx=⟨x2⟩−⟨x⟩2=1/(2 α)\Delta x = \sqrt{\langle x^2 \rangle - \langle x \rangle^2} = 1/(\sqrt{2}\,\alpha).

The uncertainty product for this state is σx σp=ℏ/(22)\sigma_x\,\sigma_p = \hbar/(2\sqrt{2})Which is larger Than the minimum ℏ/2\hbar/2Showing this is not a minimum-uncertainty state.

Example 3.2. Normalise ψ(x)=Axe−αx2\psi(x) = Axe^{-\alpha x^2} for −∞<x<∞-\infty \lt x \lt \infty.

Solution

∫−∞∞∣A∣2x2e−2αx2 dx=∣A∣2⋅14απ2α=1\int_{-\infty}^{\infty} |A|^2 x^2 e^{-2\alpha x^2}\,dx = |A|^2 \cdot \frac{1}{4\alpha}\sqrt{\frac{\pi}{2\alpha}} = 1

Using the Gaussian integral ∫−∞∞x2e−ax2 dx=12aπa\int_{-\infty}^{\infty} x^2 e^{-ax^2}\,dx = \frac{1}{2a}\sqrt{\frac{\pi}{a}}. Therefore:

A=2α(2απ)1/4A = 2\sqrt{\alpha}\left(\frac{2\alpha}{\pi}\right)^{1/4}

3.7 Time-Dependent Perturbation Theory

When the Hamiltonian has a time-dependent perturbation, H^(t)=H^0+V^(t)\hat{H}(t) = \hat{H}_0 + \hat{V}(t)The Transition probability from initial state ∣i⟩|i\rangle to final state ∣f⟩|f\rangle (with Ei≠EfE_i \neq E_f) is computed in the interaction picture.

First-order transition amplitude. If the system starts in ∣i⟩|i\rangle at t=0t = 0The probability Amplitude for being in ∣f⟩|f\rangle at time tt is, to first order:

cf(t)=−iℏ∫0t⟨f∣V^(t′)∣i⟩ eiωfit′ dt′c_f(t) = -\frac{i}{\hbar}\int_0^t \langle f | \hat{V}(t') | i \rangle\, e^{i\omega_{fi}t'}\,dt'

Where ωfi=(Ef−Ei)/ℏ\omega_{fi} = (E_f - E_i)/\hbar is the Bohr frequency.

Constant perturbation. If V^(t)=V^0\hat{V}(t) = \hat{V}_0 (constant) for 0<t<T0 \lt t \lt T:

cf(T)=−iℏVfi∫0Teiωfit′ dt′=−Vfiℏωfi ⁣(eiωfiT−1)c_f(T) = -\frac{i}{\hbar}V_{fi}\int_0^T e^{i\omega_{fi}t'}\,dt' = -\frac{V_{fi}}{\hbar\omega_{fi}}\!\left(e^{i\omega_{fi}T} - 1\right)

The transition probability is:

Pi→f(T)=∣Vfi∣2ℏ2 sin⁡2(ωfiT/2)(ωfi/2)2P_{i \to f}(T) = \frac{|V_{fi}|^2}{\hbar^2}\,\frac{\sin^2(\omega_{fi}T/2)}{(\omega_{fi}/2)^2}

This function is sharply peaked around ωfi=0\omega_{fi} = 0 (resonance), with width Δω∼2π/T\Delta\omega \sim 2\pi/T.

Interpretation. As T→∞T \to \inftyThe function sin⁡2(ωfiT/2)/(ωfi/2)2→2πT δ(ωfi)\sin^2(\omega_{fi}T/2)/(\omega_{fi}/2)^2 \to 2\pi T\,\delta(\omega_{fi}) So transitions occur only when energy is conserved (Ef=EiE_f = E_i). For finite TTEnergy conservation Is approximate to within ΔE∼ℏ/T\Delta E \sim \hbar/TA manifestation of the time-energy uncertainty Relation.

Fermi’s Golden Rule. For a transition to a continuum of final states with density of states ρ(Ef)\rho(E_f)The transition rate (probability per unit time) is:

Γi→f=2πℏ∣⟨f∣V^∣i⟩∣2 ρ(Ef)\Gamma_{i \to f} = \frac{2\pi}{\hbar}|\langle f | \hat{V} | i \rangle|^2\,\rho(E_f)

This is one of the most important results in quantum mechanics, with applications to spontaneous Emission, scattering theory, and condensed matter physics.

Sudden and adiabatic approximations.

  • Sudden approximation. If the Hamiltonian changes rapidly compared to the system’s natural timescale ∼ℏ/ΔE\sim \hbar/\Delta EThe state does not have time to adjust: ∣ψafter⟩=∣ψbefore⟩|\psi_{\mathrm{after}\rangle = |\psi_{\mathrm{before}\rangle}}. The probability of finding the system in the new nn-th eigenstate is Pn=∣⟨nnew∣ψbefore⟩∣2P_n = |\langle n_{\mathrm{new}|\psi_{\mathrm{before}\rangle|^2}}.

  • Adiabatic theorem. If the Hamiltonian changes slowly enough (specifically, if ∣⟨m∣∂H^/∂t∣n⟩∣/(ℏωmn2)≪1|\langle m|\partial\hat{H}/\partial t|n\rangle|/(\hbar\omega_{mn}^2) \ll 1 for all m≠nm \neq n), the system remains in an instantaneous eigenstate without transitions. The adiabatic condition requires the rate of change to be much slower than the energy gap divided by ℏ\hbar.

Harmonic perturbation. For a sinusoidal perturbation V^(t)=V^1 e−iωt+V^1† eiωt\hat{V}(t) = \hat{V}_1\,e^{-i\omega t} + \hat{V}_1^\dagger\,e^{i\omega t} The first-order transition rate from ∣i⟩|i\rangle to ∣f⟩|f\rangle is significant only when ω≈ωfi\omega \approx \omega_{fi} (absorption) or ω≈−ωfi\omega \approx -\omega_{fi} (stimulated emission). The transition probability for Resonant absorption (ω≈ωfi\omega \approx \omega_{fi}) is:

Pi→f(t)=∣⟨f∣V^1∣i⟩∣2ℏ2 sin⁡2((ω−ωfi)t/2)(ω−ωfi)2/4P_{i\to f}(t) = \frac{|\langle f|\hat{V}_1|i\rangle|^2}{\hbar^2}\,\frac{\sin^2((\omega - \omega_{fi})t/2)}{(\omega - \omega_{fi})^2/4}

In the long-time limit, this reduces to Fermi’s Golden Rule with the replacement Vfi→⟨f∣V^1∣i⟩V_{fi} \to \langle f|\hat{V}_1|i\rangle.