Separation of variables and the constant E
Guess a product, divide, and the constant is E
Almost every problem in this course opens with the same three lines: guess a product, divide by it, set both sides equal to E. Learn why those lines are legal, what they quietly assume about V, and why the solutions they hand you are the building blocks of the answer rather than the answer itself.
Build the model
One guess, one division, and a constant that turns out to be the energy.
1The big idea
The Schrödinger equation iħ ∂Ψ/∂t = −(ħ²/2m) ∂²Ψ/∂x² + V(x)Ψ is one partial differential equation in two variables, and separation of variables is a bet that some of its solutions factor as Ψ(x, t) = ψ(x) f(t). Substitute and divide by ψf: every t now sits on the left, in iħ f′/f, and every x on the right, in −(ħ²/2m)ψ″/ψ + V(x). A function of t alone can equal a function of x alone at every x and every t only if both are the same constant. Call it E, because it carries joules and because the right-hand equation says Ĥψ = Eψ. The time equation then integrates in one line to f(t) = e(−iEt/ħ): a pure phase turning at ω = E/ħ, which leaves |Ψ|² and every expectation value frozen, and which conserves probability only if E is real. The bet has two costs. It needs V to keep x and t apart: a static trap passes, and so does an offset U(t) that is the same everywhere, but a driving force whose potential energy is −F₀x cos Ωt ties x to t, and then there are no stationary states at all. And even when V is static the products are special: the equation is linear, so sums of them with different E also solve it, and those sums are not products. The general solution is such a sum. Separation supplies its terms, each carrying its own clock.
2In one sentence
Separation of variables looks for solutions of the Schrödinger equation of the form Ψ(x, t) = ψ(x) f(t); when V depends on x alone this forces f(t) = e(−iEt/ħ), where the real constant E is the energy of the state.
Example. An electron in a separated state with E = 2.00 eV has a time factor turning at ω = E/ħ = 3.04 × 10¹⁵ rad s⁻¹: Re Ψ at any point repeats every h/E = 2.07 fs, while |Ψ(x, t)|² = |ψ(x)|² does not change at all.
3Guess a product and divide by it
Take a potential that depends on position only, V = V(x), and try Ψ(x, t) = ψ(x) f(t). Each partial derivative now acts on one factor: ∂Ψ/∂t = ψ f′ and ∂²Ψ/∂x² = ψ″ f, so the Schrödinger equation becomes iħ ψ f′ = −(ħ²/2m) ψ″ f + V ψ f. Divide through by ψf. The left side becomes iħ f′/f, which contains no x; the right side becomes −(ħ²/2m) ψ″/ψ + V(x), which contains no t. Check the dimensions before going on: ħ f′/f is J s × s⁻¹ = J, and ħ²ψ″/(mψ) is J² s² ÷ (kg m²) = J, the same as V. The division is illegal at a node where ψ = 0, but the equations it leads to contain no division once multiplied back out, and continuity carries them across the node.
4A function of t that equals a function of x is a constant
Call the left side T(t) and the right side X(x). The equation says T(t) = X(x) for every x and every t. Hold x fixed at some x₀ and let t run: X(x₀) is one number, so T(t) cannot change. Hold t fixed and let x run: by the same argument X(x) cannot change. Both are therefore constant, and the same constant, since they are equal. Name it E. The argument leans entirely on V being free of t. Apply a uniform driving force F₀ cos Ωt, at a drive frequency Ω unrelated to E/ħ, and V gains −F₀x cos Ωt: the right side now changes with t at fixed x, the argument collapses, and there is no separation constant. An offset U(t) that is the same at every x is the one harmless exception, because it can be moved across to the time side, where it only adds to the phase.
5The time equation integrates in one line
The left side gives iħ df/dt = E f, whose solution is f(t) = f(0) e(−iEt/ħ). Fold the constant f(0) into ψ, and the separated solution is Ψ(x, t) = ψ(x) e(−iEt/ħ). The factor has modulus 1 and phase −Et/ħ, turning at ω = E/ħ: for E = 1.00 eV that is 1.52 × 10¹⁵ rad s⁻¹, one turn every h/E = 4.14 fs. For a real ψ, Re Ψ = ψ cos(Et/ħ) and Im Ψ = −ψ sin(Et/ħ): one fixed shape scaled up and down, with nodes that never move. Because |f| = 1, a ψ normalised once stays normalised, ∫|Ψ|² dx = ∫|ψ|² dx = 1 at every t, and |Ψ|² = |ψ|² makes every expectation value time-independent. That is what a stationary state is: Ψ keeps turning, and nothing measurable moves.
6Why the constant deserves the name energy
Three separate facts earn E its name. Its unit is the joule, as the division showed. The spatial equation reads Ĥψ = Eψ, so the Hamiltonian returns ψ multiplied by E: for a normalised state ⟨H⟩ = ∫Ψ* ĤΨ dx = E and ⟨H²⟩ = E², the spread in energy is zero, and every energy measurement returns E. And the phase rate ω = E/ħ is the Planck–Einstein relation E = ħω: the equation was built to respect it for free plane waves, and separation shows it holding for the separated states of every static V. E must also be real: a trial E = E₀ + iη makes the density |ψ|² e(2ηt/ħ), so the total probability would grow or decay, which a real V forbids. The classical echo: a Hamiltonian with no explicit t conserves energy, and the same static V is what makes the separation constant a conserved energy here.
7Only differences in E can be seen
Add a constant V₀ to the potential everywhere. The spatial equation is satisfied by the same ψ with E raised to E + V₀, and the solution gains a factor e(−iV₀t/ħ) common to every x. Lower the zero of energy by 10.0 eV and each separated solution simply turns 1.52 × 10¹⁶ rad s⁻¹ faster, while no density and no expectation value changes. So the absolute rotation rate of a single stationary state is a bookkeeping choice, not a measurement. What survives a change of reference is a difference: two separated solutions with energies E₁ and E₂ keep a relative phase advancing at (E₂ − E₁)/ħ whatever V₀ is. Spectral lines measure exactly these differences, which is why an atom's photon frequencies are fixed while its absolute energies depend on where the zero is put.
8Special solutions, not the general one
The equation is linear, so c₁ψ₁ e(−iE₁t/ħ) + c₂ψ₂ e(−iE₂t/ħ) also solves it. It is not a product. Pull out the first time factor and what remains, c₁ψ₁ + c₂ψ₂ e(−iΔEt/ħ) with ΔE = E₂ − E₁, still changes its shape in x as t runs, because ψ₁ and ψ₂ are different functions, so no single ψ(x) and f(t) reproduce it and the state has no single E. Its density moves. For an electron in a 1.00 nm infinite well, E₁ = 0.376 eV and E₂ = 1.504 eV, and an equal mix sloshes from one side of the well to the other and back every h/(E₂ − E₁) = 3.67 fs. A general start state Ψ(x,0) is expanded as Σ cₙψₙ(x), and each term is then given its own clock e(−iEₙt/ħ). That is the real use of separation: it does not solve the time-dependent problem, it supplies the pieces every solution is assembled from.
Key equations
What each relation says, and the units it is written in.
Product ansatz
Ψ(x, t) = ψ(x) f(t)
A guess, not a theorem. When it works, one PDE in two variables becomes two ODEs, but it can only find solutions that happen to be products.
ψ in m⁻¹⁄² for a normalised state on a line; f dimensionless; x in m, t in s
Separated equation
iħ f′(t)/f(t) = −(ħ²/2m) ψ″(x)/ψ(x) + V(x) = E
The left side depends on t only and the right side on x only, so both must equal one constant E. That step needs a V that does not mix x with t.
ħ = 1.055 × 10⁻³⁴ J s; m in kg; V and E in J; holds wherever ψf ≠ 0
Time factor
f(t) = e(−iEt/ħ)
All the time dependence is a phase turning at ω = E/ħ, so |Ψ(x, t)|² = |ψ(x)|² and every expectation value is constant: a stationary state.
E in J, t in s, phase Et/ħ in rad; |f| = 1; any constant prefactor is folded into ψ
Spatial equation
−(ħ²/2m) ψ″ + V(x) ψ = E ψ
It reads Ĥψ = Eψ: ψ is an eigenfunction of the Hamiltonian with eigenvalue E, which is why the separation constant is the energy.
the same E as in the time factor; every term in J m⁻¹⁄²
Assembling the general solution
Ψ(x, t) = Σₙ cₙ ψₙ(x) e(−iEₙt/ħ)
Linearity makes the sum a solution. With two or more distinct Eₙ it is not a product, and its density moves at the differences (Eₙ − Eₘ)/ħ.
cₙ dimensionless, fixed by the start state Ψ(x,0) = Σₙ cₙ ψₙ(x); ψₙ in m⁻¹⁄², Eₙ in J
Why E must be real
|Ψ(x, t)|² = |ψ(x)|² e(2ηt/ħ)
A complex E would make the total probability grow or decay exponentially. Models of decay use E = E₀ − iΓ/2 on purpose, giving a lifetime ħ/Γ.
for a trial E = E₀ + iη, with E₀ and η in J; a real V conserves ∫|Ψ|² dx, so η = 0
Interactive lab
Change one variable. Predict first, then read the result.
Controls
Try this. Keep p = 0 and drag t: Re Ψ and Im Ψ trade places inside one fixed envelope and the density never moves. Set p = 0.5 and drag again: the sum is not a product, its density sloshes, and ⟨x⟩ swings between 0.32 L and 0.68 L once every h/(E₂ − E₁), which is 3.67 fs at L = 1.0 nm.
Live results
- GROUND LEVEL E₁
- 0.376 eV
- PHASE PERIOD h/E₁
- 11.00 fs
- BEAT PERIOD h/(E₂ − E₁)
- 3.67 fs
- ⟨x⟩ / L
- 0.500
Catch the common trap
The mistake this lesson is built to prevent.
The tempting conclusion is that separation of variables has solved the Schrödinger equation, so every particle is in a state of definite energy E and its |Ψ|² never changes. It has not. The product form was an assumption, and it catches only the solutions that happen to be products. Test a free Gaussian ψ = exp[−x²/(2a²)] with V = 0: −(ħ²/2m)ψ″/ψ = (ħ²/2ma²)(1 − x²/a²) changes with x, so no single E exists, and a free Gaussian packet really does spread. A general state is a sum of separated solutions, each turning at its own rate Eₙ/ħ, and its density moves at the differences. A definite energy and a frozen density go together, but only for the special states the ansatz returns; every other state is a superposition of them.
Test it: what changes?
- Ψ is no longer a product ψ(x)f(t)
- Density sloshes and ⟨x⟩ oscillates
- Density repeats every h/(E₂ − E₁), whatever the mix
- No single energy E describes the state
- Every Eₙ rises by V₀; each ψₙ unchanged
- Each phase turns V₀/ħ faster
- No probability density changes
- Beat rate (E₂ − E₁)/ħ unchanged
Practice & worked examples
Three tiers. Try each before opening the solution.
EasyExample 1
An electron is in a separated state of a static trap with E = 1.20 eV, with energies measured from V = 0 far from the trap. A colleague measures from a reference 5.00 eV lower, so her potential is V + 5.00 eV everywhere. Find the angular frequency of the time factor in each description, the phase by which the two wavefunctions differ after 1.00 fs, and whether any probability differs.
View solution →
- Her spatial equation is −(ħ²/2m)ψ″ + (V + V₀)ψ = E′ψ with V₀ = 5.00 eV. The same ψ satisfies it with E′ = E + V₀ = 6.20 eV, because the extra V₀ψ is matched on both sides.
- With ħ = 6.582 × 10⁻¹⁶ eV s: ω = E/ħ = 1.20 eV ÷ 6.582 × 10⁻¹⁶ eV s = 1.82 × 10¹⁵ rad s⁻¹, and ω′ = 6.20 eV ÷ 6.582 × 10⁻¹⁶ eV s = 9.42 × 10¹⁵ rad s⁻¹.
- Her solution is Ψ′ = ψ e(−iE′t/ħ) = Ψ e(−iV₀t/ħ). After t = 1.00 fs the extra phase is V₀t/ħ = (5.00 eV × 1.00 × 10⁻¹⁵ s) ÷ 6.582 × 10⁻¹⁶ eV s = 7.60 rad, the same at every x.
- Probabilities: |Ψ′|² = |Ψ|² × |e(−iV₀t/ħ)|² = |Ψ|², so both descriptions give the same normalised density and the same expectation values. The 7.60 rad is a global phase that no measurement can detect.
Answer. ω = 1.82 × 10¹⁵ rad s⁻¹ and ω′ = 9.42 × 10¹⁵ rad s⁻¹; the wavefunctions differ by a global phase of 7.60 rad after 1.00 fs, and no probability differs.
MediumExample 2
For a free electron (V = 0), test two candidate spatial factors: φ(x) = A cos(kx) with k = 5.00 nm⁻¹, and χ(x) = B exp[−x²/(2a²)] with a = 0.200 nm. Which can be multiplied by a single f(t) to solve the Schrödinger equation, and for that one, what are E and the angular frequency ω of its time factor?
View solution →
- With V = 0 the separated equation needs −(ħ²/2m)ψ″/ψ to be one number at every x. Its prefactor is ħ²/2m = (1.0546 × 10⁻³⁴ J s)² ÷ (2 × 9.109 × 10⁻³¹ kg) = 6.104 × 10⁻³⁹ J m².
- For φ: φ″ = −k²φ, so the ratio is ħ²k²/2m at every x, a true constant. With k = 5.00 × 10⁹ m⁻¹, E = 6.104 × 10⁻³⁹ J m² × 2.50 × 10¹⁹ m⁻² = 1.526 × 10⁻¹⁹ J = 0.952 eV.
- Then ω = E/ħ = 1.526 × 10⁻¹⁹ J ÷ 1.0546 × 10⁻³⁴ J s = 1.45 × 10¹⁵ rad s⁻¹. Classical check: cos(kx) is an equal mix of momenta +ħk and −ħk, and the magnitude |p| = ħk = 5.27 × 10⁻²⁵ kg m s⁻¹ gives a speed v = |p|/m = 5.79 × 10⁵ m s⁻¹, and ½mv² = 0.952 eV, the same energy.
- For χ: χ″ = (x²/a⁴ − 1/a²)χ, so the ratio is (ħ²/2ma²)(1 − x²/a²). With ħ²/2ma² = 6.104 × 10⁻³⁹ J m² ÷ 4.00 × 10⁻²⁰ m² = 1.526 × 10⁻¹⁹ J = 0.952 eV, it is +0.952 eV at x = 0, zero at x = a, and −2.86 eV at x = 2a.
- The ratio for χ changes with x, so no constant E exists and χ(x)f(t) fails for every choice of f. At x = 0 it agrees with φ because the Gaussian curves like cos(x/a) near its centre, but nowhere else. Limiting check: as a grows the ratio flattens towards zero, the E of a constant ψ at rest.
Answer. Only φ works: E = ħ²k²/2m = 0.952 eV and ω = 1.45 × 10¹⁵ rad s⁻¹. The Gaussian's ratio runs from +0.952 eV at x = 0 to −2.86 eV at x = 2a, so it is not a separated solution.
HardExample 3
An electron is in a separated state of a static trap with E = 1.50 eV. From t = 0 the whole apparatus is driven so that the potential energy at every point gains the same U(t) = U₀ sin(Ωt), with U₀ = 0.200 eV and Ω = 2π × 1.00 × 10¹⁴ rad s⁻¹. Show that separation still works, find f(t), find the largest extra phase and when it first occurs, and explain why a uniform force with potential energy −F₀x sin(Ωt) cannot be treated the same way.
View solution →
- Substitute Ψ = ψ(x)f(t) into iħ ∂Ψ/∂t = −(ħ²/2m) ∂²Ψ/∂x² + [V(x) + U(t)]Ψ and divide by ψf. Moving U(t) to the left gives iħ f′/f − U(t) = −(ħ²/2m)ψ″/ψ + V(x): t only on the left, x only on the right, so both equal one constant, and the right side is the undriven trap's equation. ψ and E = 1.50 eV are unchanged.
- The time equation is iħ f′ = [E + U(t)] f, so f(t) = exp[−(i/ħ)(Et + ∫₀ᵗ U dt′)] = exp[−iEt/ħ − i(U₀/ħΩ)(1 − cos Ωt)]. Limiting check: as U₀ → 0 this returns to e(−iEt/ħ).
- |f| = 1 still, so |Ψ|² = |ψ|² and the norm stays 1 at every t. A potential that is the same everywhere exerts no force, and in this state the drive changes nothing measurable.
- The extra phase is −(U₀/ħΩ)(1 − cos Ωt), largest in size when cos Ωt = −1, first at t = π/Ω = 5.00 fs. With ħΩ = 6.582 × 10⁻¹⁶ eV s × 6.283 × 10¹⁴ s⁻¹ = 0.4136 eV, its size there is 2U₀/ħΩ = 0.400 eV ÷ 0.4136 eV = 0.967 rad.
- With −F₀x sin(Ωt) instead, dividing by ψf leaves −F₀x sin(Ωt) on the right, a term that changes with t at fixed x. It cannot be moved left either, because the left must stay free of x. No separation constant exists. A uniform force is not a uniform potential: it pushes the electron, and its density does move.
Answer. f(t) = exp[−iEt/ħ − i(U₀/ħΩ)(1 − cos Ωt)] with |Ψ|² unchanged; the extra phase reaches 0.967 rad at t = 5.00 fs. The force term −F₀x sin(Ωt) ties x to t, so separation fails.
Quick check
Choose an answer, then check it. Explain to yourself why the others fail.
Key takeaway
Each separated state is ψ(x) e(−iEt/ħ); general states are sums.
Separate only when V keeps x and t apart: a static V works, and so does a uniform offset U(t); a driving term −F₀x cos Ωt does not. For static V, solve Ĥψ = Eψ, attach e(−iEt/ħ) with E real, and build general states as Σₙ cₙψₙ e(−iEₙt/ħ).