Schrödinger equation — time-dependent, time-independent: Visual walkthrough

Step 0 — The three characters we need

Before any physics, meet the actors. Everything below is built from just these.

Definition The complex wiggle

$e^{i θ}$ — why a rotating arrow A real sine wave only goes up and down. But a quantum wave needs to store two things at once: how tall it is and where it is in its cycle (its phase). The cleanest object that carries both is a little arrow of length 1 that spins around a circle.

We write that spinning unit arrow as $e^{i θ}$ . Here $i$ is the "quarter-turn" number ( $i^{2} = - 1$ , i.e. two quarter-turns = a half-turn = flip to negative), and $θ$ is the angle the arrow has turned through. Its shadow on the horizontal axis is $cos θ$ ; its shadow on the vertical axis is $sin θ$ . So $e^{i θ} = cos θ + i sin θ .$ Why this and not plain $sin$ ? Because — as you'll see in Steps 2–3 — differentiating a spinning arrow just spins it a bit more, which turns calculus into simple multiplication. That is the whole reason quantum mechanics is written with $i$ .

Step 1 — Write the one wave we already understand

Read the exponent piece by piece, right where each symbol sits:

$Ψ (x, t) = A unit arrow, spinning e^{i (k x position phase - ω t time phase)}$

$A$ — the amplitude, sets the arrow's length (how "loud" the wave is).
$k x$ — as you walk right by $x$ , the arrow has already spun $k$ radians per metre. This paints the ripples in space.
$- ω t$ — as time passes, the arrow spins backwards by $ω$ radians per second. The minus sign makes the whole pattern drift to the right (crest moves to keep $k x - ω t$ fixed).

Step 2 — Ask "how fast does the arrow spin in time?" → get $E$

Differentiating a spinning arrow just multiplies it by "the spin rate times $i$ ": $\frac{\partial Ψ}{\partial t} = pull down the spin rate (- iω) Ψ.$

Now swap $ω$ for energy using $E = ℏ ω$ , i.e. $ω = E /ℏ$ :

\;\;\Longrightarrow\;\; \boxed{\;i\hbar\frac{\partial \Psi}{\partial t}=E\,\Psi\;}$$ Term by term in the boxed line: - $\dfrac{\partial \Psi}{\partial t}$ — the measured spin-rate of the wave in time. - Multiplying by $i\hbar$ **cancels** the $-i/\hbar$ that popped out, leaving a clean $E$. - Result: *"apply $i\hbar\,\partial_t$ to the wave and it hands you back $E$ times the same wave."* The operation **is** energy in disguise. --- ## Step 3 — Ask "how curved is the arrow in space?" → get $p^2$ > [!intuition] WHAT > Now the **space** derivative. One space derivative reports the wave's *slope*; a **second** space derivative $\dfrac{\partial^2}{\partial x^2}$ reports its *curvature* — how sharply it bends. > [!intuition] WHY twice, and why space > Momentum lives in $k$ (via $p=\hbar k$), and $k$ is the *space-ripple* rate — so we differentiate in $x$. We take it **twice** because kinetic energy needs $p^2$, and each space derivative brings down one factor of $k$. Two derivatives → $k^2$ → $p^2$. $$\frac{\partial^2 \Psi}{\partial x^2}=(ik)^2\Psi=\underbrace{-k^2}_{\text{two quarter-turns} = \text{flip}}\Psi.$$ The $i^2=-1$ is why curvature comes with a **minus**: a wave curves *back toward zero*, opposite to its own height. Swap $k$ for momentum via $k=p/\hbar$: $$\frac{\partial^2\Psi}{\partial x^2}=-\frac{p^2}{\hbar^2}\Psi \;\;\Longrightarrow\;\; \boxed{\;-\hbar^2\frac{\partial^2\Psi}{\partial x^2}=p^2\,\Psi\;}$$ Term by term: - $\dfrac{\partial^2\Psi}{\partial x^2}$ — the curvature of the ripples. - Multiplying by $-\hbar^2$ undoes the $-1/\hbar^2$, leaving a clean $p^2$. - Reading: *"the more sharply the wave curves, the more momentum (squared) it carries."* Sharp bends = fast particle. --- ## Step 4 — Feed both into energy conservation > [!intuition] WHAT > Now the one line of **real physics**: for a free particle all the energy is kinetic, > $$E=\frac{p^2}{2m},$$ > where $m$ is the particle's mass. (Small $p$, big $m$ → small energy: heavy slow things carry little kinetic energy.) > [!intuition] WHY this step is the heart > Steps 2 and 3 gave us two *machines*: one that produces $E\Psi$, one that produces $p^2\Psi$. Energy conservation is the sentence that **relates** those two quantities. Plug the machines into that sentence and the wave's private details ($A$, exact $k$) drop out — leaving a rule true for *every* such wave. Multiply $E=\dfrac{p^2}{2m}$ on the right by $\Psi$, then substitute the boxes from Steps 2–3: $$\underbrace{i\hbar\frac{\partial\Psi}{\partial t}}_{=\,E\Psi\ \text{(Step 2)}} \;=\; \frac{1}{2m}\underbrace{\left(-\hbar^2\frac{\partial^2\Psi}{\partial x^2}\right)}_{=\,p^2\Psi\ \text{(Step 3)}} \;=\; -\frac{\hbar^2}{2m}\frac{\partial^2\Psi}{\partial x^2}.$$ --- ## Step 5 — Let the particle feel a force: add $V(x)$ > [!intuition] WHAT > A real particle sits in a landscape of hills and valleys — a **potential energy** $V(x)$ (high where it's hard to be, low where it's easy). Total energy is now kinetic **plus** potential: > $$E=\frac{p^2}{2m}+V(x).$$ > [!intuition] WHY just add a term > Energy conservation still holds; we only enriched what energy *is*. Repeating Step 4 with the extra $V(x)$ term (which just multiplies $\Psi$) glues one more piece onto the right-hand side. > [!formula] Time-Dependent Schrödinger Equation (TDSE) > $$\boxed{\;i\hbar\frac{\partial \Psi}{\partial t} > = \underbrace{-\frac{\hbar^2}{2m}\frac{\partial^2\Psi}{\partial x^2}}_{\text{kinetic energy}} > + \underbrace{V(x)\,\Psi}_{\text{potential energy}}\;}$$ > The bracket $\hat H=-\dfrac{\hbar^2}{2m}\partial_x^2+V(x)$ is the ==[[Hamiltonian Operator|Hamiltonian]]== — "energy as a recipe of derivatives." Compactly: $i\hbar\,\partial_t\Psi=\hat H\Psi$. --- ## Step 6 — Degenerate/edge cases: check the equation didn't break A derivation is only trustworthy if it survives the corners. Three to test. > [!intuition] Case A — $V=0$ everywhere (the free particle we started from) > The $V\Psi$ term vanishes and we're back to Step 4. Plugging in the plane wave gives $\hbar\omega=\dfrac{\hbar^2k^2}{2m}$, i.e. $E=\dfrac{p^2}{2m}$. ✓ It reproduces exactly the physics we built it from — internal consistency. > [!intuition] Case B — constant potential $V=V_0$ (a flat plateau, not zero) > Then $E=\dfrac{p^2}{2m}+V_0$. The wave still travels, but slower for the same $E$: the ripples get *longer* ($k$ smaller) because kinetic energy $E-V_0$ shrank. A constant $V_0$ just **shifts the energy zero** — physics is unchanged. This is why only *differences* in potential matter. > [!intuition] Case C — $\omega$ or $k$ equal to zero (no wiggle) > If $k=0$ the wave is flat in space: zero curvature → zero momentum → a particle at rest. If $\omega=0$ it never ticks in time → zero energy. Both derivatives correctly return zero. The equation degrades gracefully; it does not blow up. --- ## Step 7 — Freezing time out: TDSE → TISE > [!intuition] WHAT > When the landscape $V(x)$ doesn't change with time, we guess the wave **factors** into a fixed *shape* times a pure *phase-ticking*: > $$\Psi(x,t)=\psi(x)\,\phi(t).$$ > [!intuition] WHY this is allowed > A machine whose rules never change in time can't mix "where the ripples are" with "when we look." So the space-shape $\psi(x)$ and the time-clock $\phi(t)$ live independent lives — the hallmark of a **separable** problem. Substitute and divide by $\psi\phi$. The left becomes a function of $t$ only, the right a function of $x$ only: $$\underbrace{i\hbar\frac{1}{\phi}\frac{d\phi}{dt}}_{\text{time only}} =\underbrace{\frac{1}{\psi}\hat H\psi}_{\text{space only}}=E.$$ A thing depending only on $t$ can equal a thing depending only on $x$ **for all** $x,t$ only if both are the same constant — and that constant has units of energy, so it *is* $E$. - Time half: $i\hbar\,\dot\phi=E\phi \Rightarrow \phi(t)=e^{-iEt/\hbar}$ — the arrow spins forever at rate $E/\hbar$. - Space half: > [!formula] Time-Independent Schrödinger Equation (TISE) > $$\boxed{\;-\frac{\hbar^2}{2m}\frac{d^2\psi}{dx^2}+V(x)\,\psi=E\,\psi\;}\qquad\text{i.e. }\hat H\psi=E\psi.$$ > An ==eigenvalue equation==: only special shapes $\psi$ (with special energies $E$) survive. Full state: $\Psi(x,t)=\psi(x)\,e^{-iEt/\hbar}$, so $|\Psi|^2=|\psi(x)|^2$ **never moves** — a [[Energy Quantization|stationary state]]. See [[Particle in a Box]] for the first place these special energies appear. --- ## The one-picture summary > [!recall]- Feynman retelling — the whole walkthrough in plain words > Start with the one wave we already trust: a free particle drawn as a little arrow spinning as you move along and as time passes. De Broglie tells us the *fineness* of the ripples is the particle's momentum, and the *speed of ticking* is its energy. > > Now ask the arrow two questions. "How fast do you tick in time?" — the answer hands us the energy. "How sharply do you bend in space?" — the answer, done twice, hands us the momentum-squared. Two questions, two machines. > > Then we say the one true physics sentence: energy = kinetic + potential. We pour our two machines into that sentence, the wave's private details wash out, and what remains is the Schrödinger equation — a rule true for *any* wave, not just our starting one. > > Finally, when the landscape sits still, we split the wave into a frozen shape times a spinning clock. The clock always spins at rate energy/ℏ; the shape must solve a special equation that only accepts certain energies. That "only certain energies" is where quantum steps — discrete energy levels — are born. > [!mnemonic] The four fishing hooks > **Time-derivative fishes $E$. Space-derivative-twice fishes $p^2$. Energy conservation ties them. A potential adds the last knot.** --- ## Connections - [[2.3.09 Schrödinger equation — time-dependent, time-independent (Hinglish)|Parent topic]] - [[de Broglie Hypothesis]] — gives $p=\hbar k$, $E=\hbar\omega$, the whole starting point. - [[Wavefunction and Born Interpretation]] — why $|\Psi|^2$ is what we watch. - [[Hamiltonian Operator]] — the $\hat H$ recipe assembled in Step 5. - [[Particle in a Box]] · [[Quantum Harmonic Oscillator]] — the TISE solved in real landscapes. - [[Energy Quantization]] — special energies from special shapes. - [[Heisenberg Uncertainty Principle]] — sharp ripples vs. wide spread, the same wave nature. --- #flashcards/physics

Step 0 — The three characters we need

Step 1 — Write the one wave we already understand

Step 2 — Ask "how fast does the arrow spin in time?" → get E

Step 2 — Ask "how fast does the arrow spin in time?" → get $E$