University Physics IV · The Schrödinger Equation · 6.3
Separation of Variables & the Time-Independent Equation
The time-dependent equation is a partial differential equation in two variables, and you will not integrate it by hand. This topic buys the trick that makes it tractable — split x from t — then charges you honestly: it works only for a static potential, and it delivers only the states whose density never moves.
Build the model
Connect the measurement to the mechanism.
The time-dependent Schrödinger equation is a partial differential equation in x and t at once, and nothing about it invites a general solution. Separation of variables is the standard move: guess that some solutions factorise, Ψ(x, t) = ψ(x)φ(t), substitute, and divide through by the product. If — and only if — V carries no explicit time dependence, every surviving term on one side depends on t alone and every term on the other on x alone, so both sides must equal the same constant.
Comparing the resulting time factor with the plane wave, whose phase turns as e(−iωt) with E = ħω, identifies that constant as the energy. The time equation then integrates in one line to φ(t) = e(−iEt/ħ), a pure phase of unit modulus whenever E is real, and what is left of the spatial part is the eigenvalue problem Ĥψ = Eψ. What this costs is scope.
You have not solved the time-dependent equation; you have found its stationary states, the special solutions whose probability density never moves. A general state is a superposition of them carrying different energies, and a superposition satisfies the time-dependent equation but not Ĥψ = Eψ, because there is no single E to put on the right. And if V itself depends on time — a driven atom, a wall that moves — the division that produced the constant never comes off, and the method is simply unavailable.
- Simple definition
- Separation of variables writes a solution as a function of position alone multiplied by a function of time alone, which reduces the time-dependent Schrödinger equation to the eigenvalue equation Ĥψ = Eψ — but only when the potential has no explicit time dependence.
- Example
- For an electron in the n = 1 state of a 1.00 nm infinite well, E₁ = h²/8mL² = 6.02 × 10⁻²⁰ J = 0.376 eV, so the factor e(−iEt/ħ) turns through a full 2π every h/E₁ = 1.10 × 10⁻¹⁴ s while |Ψ|² never changes at all.
It selects the stationary states out of all the solutions; it is not the general solution.
ψ in m(−1/2) for a 1-D state, φ dimensionless; the split is a guess, tested by substituting it.
Only a time-independent V lets the x-side be free of t. This single line is the whole method — and its whole restriction.
Left side a function of t only, right side of x only, so both equal one constant E, in joules.
Unit modulus is why an eigenstate's density is frozen; a complex E would drain the norm instead.
E in J, ħ = 1.055 × 10⁻³⁴ J s; the phase turns through 2π in a time h/E.
An eigenvalue problem: the boundary conditions, not the differential equation, decide which E survive.
Ĥ = p̂²/2m + V(x); with ψ in m(−1/2), every term carries J m(−1/2).
Every expectation value of a time-independent operator is then a constant — hence the name stationary.
True for a single eigenstate only, and only because e(−iEt/ħ) cancels its own conjugate.
Solves the time-dependent equation; obeys Ĥψ = Eψ only when exactly one cₙ is non-zero.
cₙ = ∫ψₙ*(x) Ψ(x, 0) dx, dimensionless, with Σ|cₙ|² = 1.
The equation you have to solve, and why it resists
The time-dependent Schrödinger equation, iħ ∂Ψ/∂t = −(ħ²/2m) ∂²Ψ/∂x² + V(x, t)Ψ, is a partial differential equation: one unknown function of two variables, differentiated once in t and twice in x. For an arbitrary V there is no general closed form, and even where one exists you would not reach it by direct attack. Separation of variables sidesteps the problem rather than solving it. Ask a narrower question — are there any solutions of the special product form Ψ(x, t) = ψ(x)φ(t), in which the x-dependence and the t-dependence never mix? — and the answer comes in three lines. The narrower question is worth asking because the special solutions turn out to form a complete set: once you have all of them, every other solution is a superposition of them. But keep the logic straight. The ansatz is a guess about the form of a solution, justified only by substituting it and finding that it works. Nothing so far promises that such a solution exists, and nothing claims a general Ψ has this form.
Substitute, divide, and watch the variables come apart
Put Ψ = ψφ into the equation. The time derivative touches only φ and the space derivative only ψ, so iħ ψ φ′ = −(ħ²/2m) ψ″ φ + V ψφ. Divide every term by the product ψφ — legitimate wherever it is non-zero — and you are left with iħ φ′/φ = −(ħ²/2m) ψ″/ψ + V. Now read the two sides. The left is a function of t and of nothing else. The right is a function of x and of nothing else, provided V = V(x). A function of t alone can equal a function of x alone only if neither actually varies: hold x fixed and the right side is a fixed number, so the left must equal that number for every t, and conversely. Both sides equal one constant. If instead V = V(x, t) — a laser field switched on, a well wall in motion — the right-hand side still contains t after the division, the argument collapses at this exact line, and no product solution exists. That is the entire restriction on the method, and it is worth noticing that it shows up here, at the start, rather than at the end.
Naming the constant: it has to be the energy
Call the constant E. Nothing in the algebra says what it is; the identification comes from outside. Two arguments agree. First, the time equation is iħφ′ = Eφ, so φ ∝ e(−iEt/ħ), and comparing this with the free-particle plane wave e(i(kx − ωt)) gives E = ħω — the Planck relation the whole equation was built on. Second, the spatial equation reads Ĥψ = Eψ with Ĥ = −(ħ²/2m)d²/dx² + V the operator for total energy, so for normalised ψ the mean energy is ⟨H⟩ = ∫ψ*Ĥψ dx = E∫|ψ|² dx = E. The variance goes the same way: ⟨H²⟩ = ∫ψ*Ĥ(Eψ) dx = E², so σH² = E² − E² = 0. A separable state has a sharp energy, not a spread of energies, and E is its value. That is the strongest justification for the name — not that the units work out, but that an energy measurement on such a state returns E every single time.
A pure phase is what makes a state stationary
Integrating iħφ′ = Eφ gives φ(t) = e(−iEt/ħ), any constant of integration being absorbed into ψ. For real E this has modulus exactly one: e(−iEt/ħ) × e(+iEt/ħ) = 1. So |Ψ(x, t)|² = |ψ(x)|² |φ(t)|² = |ψ(x)|², frozen for all time — and with it ⟨x⟩, ⟨p⟩ and the expectation of every operator carrying no explicit time dependence. This is why the separable solutions are called stationary states, and it is a strong physical claim: an electron in the n = 2 level of an atom is not orbiting, and its probability cloud is not sloshing. The phase, meanwhile, is very much alive. It turns through 2π in a time h/E, which for the 0.376 eV ground state of a 1.00 nm well is 1.10 × 10⁻¹⁴ s. No measurement on this state can see that rotation, because a global phase cancels out of every expectation value. Only the relative phase between two terms of a superposition is observable — which is exactly what makes superpositions the interesting case.
Boundary conditions, not the ODE, quantise E
What remains is −(ħ²/2m)ψ″ + Vψ = Eψ, an ordinary differential equation in x, second order and linear. As an ODE it is unfussy: choose any real E and any V and solutions exist. Quantisation therefore does not come from the equation. It comes from insisting that ψ be an admissible wavefunction — normalisable, continuous, with ψ′ continuous wherever V is finite, and satisfying whatever the boundaries demand. In an infinite well of width L those conditions force ψ(0) = ψ(L) = 0, which kills every solution except sin(nπx/L) and leaves Eₙ = n²π²ħ²/2mL² = n²h²/8mL². For an electron in L = 1.00 nm the ladder is 0.376 eV, 1.504 eV, 3.384 eV — discrete because the box refuses everything else, not because the differential equation did. Ask for E = 0.5 eV and a mathematical solution still exists; it simply fails to vanish at both walls. Reading the spatial part as an eigenvalue problem is what makes the finite well, the harmonic oscillator and the hydrogen atom one technique instead of three.
A superposition solves one equation and not the other
Because Ĥ is linear, any sum Ψ(x, t) = Σ cₙ ψₙ(x) e(−iEₙt/ħ) also solves the time-dependent equation, with the coefficients fixed once and for all by the initial state through cₙ = ∫ψₙ*(x)Ψ(x,0) dx. That sum is the general solution, and it is the reason separating was worth the trouble. But it is not an eigenstate. Feed it to Ĥψ = Eψ and the left side returns Σ cₙ Eₙ ψₙ e(−iEₙt/ħ), which is not E times the state for any single E unless exactly one cₙ is non-zero. Concretely, an equal mix of n = 1 and n = 2 in that 1.00 nm well carries a cross term in |Ψ|² that oscillates at the Bohr frequency (E₂ − E₁)/h = 1.128 eV ÷ h = 2.73 × 10¹⁴ Hz: the density genuinely moves, so no time-independent equation can describe it. The rule to carry away is to write Ĥψ = Eψ only for a state you have already committed to being an energy eigenstate. Everything else evolves.
Change one variable at a time
Make the relationship visible.
Set c = 0 and sweep θ: nothing moves, because a single eigenstate is stationary whatever its phase does. Now set c = 0.70 and sweep θ again — the density sloshes and ⟨x⟩ swings across the well once per beat. Superposition, not the eigenvalue equation, is what makes anything happen.
WEIGHT ON ψ₂0.49
MEAN ENERGY2.47 E₁
MEAN POSITION0.320 L
INTERFERENCE TERM1.00
Live interpretationWEIGHT ON ψ₂: 0.49. MEAN ENERGY: 2.47 E₁. MEAN POSITION: 0.320 L. INTERFERENCE TERM: 1.00
Catch the common trap
Explain before calculating.
An electron in a 1.00 nm infinite well is prepared in the normalised state Ψ(x, 0) = (ψ₁ + ψ₂)/√2, where ψ₁ and ψ₂ are the first two energy eigenfunctions. Which statement about this state is correct?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyA particle is in the separable state Ψ(x, t) = ψ(x) e(−iEt/ħ) with real E = 3.00 eV. Show that its probability density carries no time dependence, and find how long the phase factor takes to complete one full turn.
- Form the density: |Ψ|² = Ψ*Ψ = ψ*(x)e(+iEt/ħ) · ψ(x)e(−iEt/ħ). For real E the two exponentials multiply to e⁰ = 1.
- So |Ψ(x, t)|² = |ψ(x)|², independent of t — and with it every expectation value of an operator carrying no explicit t. That is exactly what stationary means.
- The phase returns to its starting value when Et/ħ = 2π, so t = 2πħ/E = h/E.
- E = 3.00 eV = 3.00 × 1.602 × 10⁻¹⁹ = 4.807 × 10⁻¹⁹ J, giving t = 6.626 × 10⁻³⁴ ÷ 4.807 × 10⁻¹⁹ = 1.38 × 10⁻¹⁵ s.
Answer|Ψ|² = |ψ|² for every t, so the state is stationary; the phase completes one turn in h/E = 1.38 × 10⁻¹⁵ s, a rotation no measurement on this state alone can detect.
MediumSubstitute Ψ(x, t) = ψ(x)φ(t) into iħ ∂Ψ/∂t = −(ħ²/2m) ∂²Ψ/∂x² + V Ψ and carry the separation through to the two ordinary equations. Then state precisely what changes if the potential is instead V(x, t) = V₀(x) + qE₀ x cos(ωt).
- Differentiate the product: ∂Ψ/∂t = ψ φ′ and ∂²Ψ/∂x² = ψ″ φ, so the equation becomes iħ ψ φ′ = −(ħ²/2m) ψ″ φ + V ψ φ.
- Divide every term by ψφ wherever it is non-zero: iħ φ′/φ = −(ħ²/2m) ψ″/ψ + V(x).
- The left side depends only on t, the right only on x. Fix x and the right side is a number, so the left equals that same number for every t. Both sides equal one constant; call it E.
- Time equation: iħ φ′ = Eφ, so φ(t) = e(−iEt/ħ). Space equation: −(ħ²/2m) ψ″ + Vψ = Eψ, which is Ĥψ = Eψ.
- With V = V₀(x) + qE₀ x cos(ωt), the same division leaves iħ φ′/φ = −(ħ²/2m) ψ″/ψ + V₀(x) + qE₀ x cos(ωt). The right-hand side still contains t, so the argument that both sides must be constant never starts. A driven system has no stationary states, and the separation is unavailable, not merely harder.
Answeriħ φ′/φ = E = −(ħ²/2m) ψ″/ψ + V, giving φ = e(−iEt/ħ) and Ĥψ = Eψ. With the driving term, t survives on the x-side, both sides cannot be one constant, and no product solution exists.
HardAn electron sits in a 1.00 nm infinite square well, so Eₙ = n²h²/8mL². It is prepared as Ψ(x, 0) = (√3 ψ₁ + ψ₃)/2. Check the normalisation, write Ψ(x, t), find ⟨E⟩, find the frequency at which |Ψ|² oscillates, and say whether this state satisfies Ĥψ = Eψ.
- The ladder: E₁ = h²/8mL² = (6.626 × 10⁻³⁴)² ÷ (8 × 9.109 × 10⁻³¹ × (1.00 × 10⁻⁹)²) = 4.390 × 10⁻⁶⁷ ÷ 7.288 × 10⁻⁴⁸ = 6.02 × 10⁻²⁰ J = 0.376 eV, so E₃ = 9E₁ = 3.384 eV.
- Normalisation: the ψₙ are orthonormal, so |c₁|² + |c₃|² = 3/4 + 1/4 = 1. The state is already normalised.
- Each term carries its own phase, never a shared one: Ψ(x, t) = (√3/2) ψ₁ e(−iE₁t/ħ) + (1/2) ψ₃ e(−iE₃t/ħ).
- Mean energy: ⟨E⟩ = Σ |cₙ|² Eₙ = 0.75 × 0.376 + 0.25 × 3.384 = 0.282 + 0.846 = 1.128 eV, which is 3E₁.
- The cross term in |Ψ|² carries the relative phase e(−i(E₃−E₁)t/ħ), so the density oscillates at f = (E₃ − E₁)/h = 8E₁/h = 8 × 6.02 × 10⁻²⁰ ÷ 6.626 × 10⁻³⁴ = 7.27 × 10¹⁴ Hz, a period of 1.37 fs.
- Test the eigenvalue equation: Ĥ acting on the state gives (√3/2)E₁ψ₁e(−iE₁t/ħ) + (1/2)E₃ψ₃e(−iE₃t/ħ). Since E₁ ≠ E₃ this is not any constant times Ψ.
Answer⟨E⟩ = 3E₁ = 1.13 eV; |Ψ|² oscillates at 7.27 × 10¹⁴ Hz, a period of 1.37 fs; and no — Ĥψ = Eψ fails because E₁ ≠ E₃. Only the time-dependent equation holds.