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2.4

The time-dependent Schrödinger equation

Stationary states are only particular solutions. The general one is a linear superposition of them — and once the coefficients are fixed, the entire future is fixed with them.

Recommended first

After this section you should be able to

  • Write the complete time evolution of an arbitrary initial state
  • Compute the expansion coefficients cₙ from an initial wavefunction and say what |cₙ|² means
  • Explain why the Schrödinger equation is first order in time, and what follows from that

The previous section produced a family of particular solutions, Ψn=ψne−iEnt/ℏ\Psi_n=\psi_n\ee^{-\ii E_nt/\hbar}. But physically you cannot expect an initial state to be exactly one of them. Real initial states are arbitrary — “the particle is confined to the left half of the well”, say. What then?

The answer is hidden in one property of the equation: linearity.

Superposition and the general solution

The Schrödinger equation is linear: if Ψ1\Psi_1 and Ψ2\Psi_2 are solutions, so is c1Ψ1+c2Ψ2c_1\Psi_1+c_2\Psi_2. So we may stack infinitely many stationary states:

Ψ(x,t)=∑ncn ψn(x) e−iEnt/ℏ(2.4.1)\Psi(x,t)=\sum_n c_n\,\psi_n(x)\,\ee^{-\ii E_n t/\hbar}\tag{2.4.1}

This is the general solution. Fix the coefficients cnc_n and you know Ψ\Psi at every future time.

Finding the coefficients

The stationary states ψn\psi_n form a complete orthonormal basis (proved in section 3.5; used here on trust):

∫−∞∞ψm∗(x)ψn(x) dx=δmn(2.4.2)\int_{-\infty}^{\infty}\psi_m^*(x)\psi_n(x)\,\dd x=\delta_{mn}\tag{2.4.2}
Extracting cₙ with orthogonalitybasic~4 min

At t=0t=0,

Ψ(x,0)=∑ncnψn(x)(2.4.3)\Psi(x,0)=\sum_n c_n\psi_n(x)\tag{2.4.3}

Multiply both sides by ψm∗(x)\psi_m^*(x) and integrate over all space:

∫ψm∗Ψ(x,0) dx=∑ncn∫ψm∗ψn dx=∑ncnδmn=cm(2.4.4)\int\psi_m^*\Psi(x,0)\,\dd x=\sum_n c_n\int\psi_m^*\psi_n\,\dd x=\sum_n c_n\delta_{mn}=c_m\tag{2.4.4}

so that

cn=∫−∞∞ψn∗(x) Ψ(x,0) dx(2.4.5)c_n=\int_{-\infty}^{\infty}\psi_n^*(x)\,\Psi(x,0)\,\dd x\tag{2.4.5}

This manoeuvre is called the Fourier trick. It works entirely because orthogonality annihilates every term in the sum but one. In Dirac notation it becomes cn=⟨ψn|Ψ(0)⟩c_n=\braket{\psi_n}{\Psi(0)}, which reads simply as “take the component” — exactly isomorphic to projecting a three-dimensional vector onto a basis vector.

The coefficients also satisfy

∑n∣cn∣2=1(2.4.6)\sum_n|c_n|^2=1\tag{2.4.6}
Proof that Σ|cₙ|² = 1advanced~3 min
1=∫∣Ψ∣2dx=∫(∑mcmψm) ⁣∗(∑ncnψn)dx=∑m,ncm∗cn∫ψm∗ψn dx(2.4.7)1=\int|\Psi|^2\dd x=\int\left(\sum_m c_m\psi_m\right)^{\!*}\left(\sum_n c_n\psi_n\right)\dd x =\sum_{m,n}c_m^*c_n\int\psi_m^*\psi_n\,\dd x\tag{2.4.7}=∑m,ncm∗cn δmn=∑n∣cn∣2(2.4.8)=\sum_{m,n}c_m^*c_n\,\delta_{mn}=\sum_n|c_n|^2\tag{2.4.8}

Note that the result is time-independent: including the time factors sends cn→cne−iEnt/ℏc_n\to c_n\ee^{-\ii E_nt/\hbar}, which leaves the moduli untouched. So if the initial state is normalised, it stays normalised forever — the same conclusion reached in section 2.2 via the continuity equation, obtained here with far less work.

A complete example

Example: an initially localised state in an infinite welladvanced~7 min

Take an infinite square well of width LL, whose stationary states are (see section 2.7)

ψn(x)=2Lsin⁡nπxL,En=n2π2ℏ22mL2(2.4.9)\psi_n(x)=\sqrt{\frac{2}{L}}\sin\frac{n\pi x}{L},\qquad E_n=\frac{n^2\pi^2\hbar^2}{2mL^2}\tag{2.4.9}

Let the initial state be an equal-weight superposition of the two lowest states:

Ψ(x,0)=12[ψ1(x)+ψ2(x)](2.4.10)\Psi(x,0)=\frac{1}{\sqrt2}\left[\psi_1(x)+\psi_2(x)\right]\tag{2.4.10}

so c1=c2=1/2c_1=c_2=1/\sqrt2 and everything else vanishes. Evolving,

Ψ(x,t)=12[ψ1e−iE1t/ℏ+ψ2e−iE2t/ℏ](2.4.11)\Psi(x,t)=\frac{1}{\sqrt2}\left[\psi_1\ee^{-\ii E_1t/\hbar}+\psi_2\ee^{-\ii E_2t/\hbar}\right]\tag{2.4.11}

and the probability density is

∣Ψ∣2=12[ψ12+ψ22+2ψ1ψ2cos⁡(E2−E1)tℏ](2.4.12)|\Psi|^2=\frac12\left[\psi_1^2+\psi_2^2+2\psi_1\psi_2\cos\frac{(E_2-E_1)t}{\hbar}\right]\tag{2.4.12}

Look carefully at the cross term. The first two terms do not move; the third oscillates at angular frequency

ω21=E2−E1ℏ=3π2ℏ2mL2(2.4.13)\omega_{21}=\frac{E_2-E_1}{\hbar}=\frac{3\pi^2\hbar}{2mL^2}\tag{2.4.13}

so the probability density sloshes from side to side inside the well with period T=2πℏ/(E2−E1)T=2\pi\hbar/(E_2-E_1).

Three things worth noting:

  1. The frequency depends only on the level difference, not on where the energy zero is. Shift every EnE_n by a constant and nothing physical changes — the choice of zero really is arbitrary.
  2. If only one cnc_n is non-zero (a pure stationary state) there is no cross term and ∣Ψ∣2|\Psi|^2 is static.
  3. This ω21\omega_{21} is exactly the frequency at which the system absorbs or emits a photon (section 7.5). Bohr’s frequency condition ΔE=ℏω\Delta E=\hbar\omega has appeared on its own.

Why first order in time

Schrödinger actually wrote down a relativistic second-order equation first — later called the Klein–Gordon equation — and retreated to the non-relativistic first-order form only after finding that it gave the wrong fine structure for hydrogen.

The picture

Being first order in time buys two crucial properties:

  • Probability conservation works. The conserved quantity of the second-order equation is not positive definite and cannot be read as a probability — this is precisely the “negative probability” difficulty of the Klein–Gordon equation.
  • The initial condition is just Ψ(x,0)\Psi(x,0), which makes the notion of a state very clean.

The price is that the equation is second order in space and first order in time, so space and time are not on the same footing — unacceptable relativistically, and eventually fixed by the Dirac equation.

The mathematics

iℏ∂Ψ∂t=H^Ψ(2.4.14)\ii\hbar\frac{\partial\Psi}{\partial t}=\hat H\Psi\tag{2.4.14}

Formal solution:

Ψ(t)=U^(t)Ψ(0),U^(t)=e−iH^t/ℏ(2.4.15)\Psi(t)=\hat U(t)\Psi(0),\qquad \hat U(t)=\ee^{-\ii\hat Ht/\hbar}\tag{2.4.15}

U^\hat U is unitary (U^†U^=1\hat U^\dagger\hat U=1), which is the operator version of probability conservation. See section 3.9.

The relativistic generalisation is chapter 12.

Key formulas

  • Time-dependent Schrödinger equation

    iℏ∂Ψ∂t=H^Ψ\ii\hbar\frac{\partial\Psi}{\partial t}=\hat H\Psi

    First order in time: Ψ(x,0) determines the future uniquely

  • General solution

    Ψ(x,t)=∑ncnψn(x) e−iEnt/ℏ\Psi(x,t)=\sum_n c_n\psi_n(x)\,\ee^{-\ii E_n t/\hbar}

    Each component turns its phase at its own energy

  • Expansion coefficients

    cn=∫ψn∗(x) Ψ(x,0) dxc_n=\int\psi_n^*(x)\,\Psi(x,0)\,\dd x

    The Fourier trick; |cₙ|² is the probability of measuring Eₙ

  • Bohr frequency

    ωmn=Em−Enℏ\omega_{mn}=\frac{E_m-E_n}{\hbar}

    Oscillation frequency of the cross term, and of the emitted photon

Self-check3 questions

  1. 1.

    You know a particle's wavefunction Ψ(x,0) at t = 0. What extra information is needed to predict Ψ at t > 0?

  2. 2.

    Shift every energy level by a constant E₀ (that is, choose a different zero of energy). Which quantities change? (Select all that apply.)

    Select all that apply

  3. 3.

    An infinite well starts in Ψ(x,0) = (ψ₁ + ψ₂)/√2. How does |Ψ(x,t)|² behave?

What comes next

We now have the full picture of the evolution. The next section lays out the difference between stationary and non-stationary states completely — and lets you watch it in a simulation.

Section 12 of 106 · use ← → to turn the page