From Quantum State to the Schrödinger Equation
Physics ·The two-path interference argument constructed a state that predicts path and detector probabilities. We now ask a new question:
Given the state now, what rule predicts the state later?
1. What experiment produces a time-update problem?
The previous post defined
\[\Omega:=\{A,B\}, \qquad \mathcal H:=\mathbb C^\Omega :=\{\psi\mid\psi:\Omega\to\mathbb C\}.\]For the ordered path labels \((A,B)\), define
\[\begin{aligned} \operatorname{coord}_{(A,B)}&:\mathcal H\longrightarrow\mathbb C^2,\\ \operatorname{coord}_{(A,B)}(\psi) &:= \begin{pmatrix} \psi(A)\\ \psi(B) \end{pmatrix}. \end{aligned}\]After choosing path \(A\) as the phase reference, its two-path state satisfies
\[\begin{aligned} \psi_\theta(A) &:=\sqrt{P(X=A\mid C=C_p,\theta)} =\frac1{\sqrt2},\\ \psi_\theta(B) &:=\sqrt{P(X=B\mid C=C_p,\theta)}\,e^{i\theta} =\frac{e^{i\theta}}{\sqrt2}. \end{aligned}\]The parameter \(\theta\) labels relative phase, not time. To create a time-update experiment, drive the phase adjuster at a constant measured angular frequency \(\omega\). Choose \(t=0\) when \(\theta=0\) and choose the sign convention
\[\theta(t):=-\omega t.\]Repeated trials then give
\[P(D_0\mid C=C_0,t) =\cos^2\left(\frac{\omega t}{2}\right).\]The state at time \(t\) is
\[\psi(t):=\psi_{\theta(t)},\]so after an additional interval \(\tau\),
\[\begin{aligned} \psi(t+\tau)(A)&=\psi(t)(A),\\ \psi(t+\tau)(B)&=e^{-i\omega\tau}\psi(t)(B). \end{aligned}\]Using the same linearity postulate as for the beam splitter, extend the observed component update to an arbitrary input state:
\[\begin{aligned} U_\omega(\tau)&:\mathcal H\longrightarrow\mathcal H,\\ \bigl(U_\omega(\tau)\psi\bigr)(A)&:=\psi(A),\\ \bigl(U_\omega(\tau)\psi\bigr)(B)&:=e^{-i\omega\tau}\psi(B),\\ \psi(t+\tau)&=U_\omega(\tau)\psi(t). \end{aligned}\]The original experiment has now supplied one example of a time-update map. The general problem is to determine which properties any isolated-system update
\[U(\tau):\mathcal H\longrightarrow\mathcal H\]must have.
1.1. What must the update preserve?
Define the inner product and norm on \(\mathcal H\) by
\[\langle\phi,\psi\rangle :=\sum_{X\in\Omega}\overline{\phi(X)}\psi(X), \qquad \lVert\psi\rVert^2:=\langle\psi,\psi\rangle.\]The squared norm is total probability. An isolated-system update must therefore satisfy
\[\lVert U(\tau)\psi\rVert^2 =\lVert\psi\rVert^2\]for every \(\psi\in\mathcal H\). Amplitudes also obey linear superposition:
\[U(\tau)(a\psi+b\phi) =aU(\tau)\psi+bU(\tau)\phi, \qquad a,b\in\mathbb C.\]If the physical rule does not change with the clock’s starting time, successive intervals compose as
\[I:\mathcal H\longrightarrow\mathcal H, \qquad I\psi:=\psi,\]and
\[U(0)=I, \qquad U(\tau_2)U(\tau_1)=U(\tau_1+\tau_2).\]Finally, assume that \(U(\tau)\) changes differentiably with \(\tau\). These assumptions state the prediction problem precisely; they do not yet specify a particular system’s dynamics.
2. What generates a continuous probability-preserving update?
Define the instantaneous generator
\[A:\mathcal H\longrightarrow\mathcal H, \qquad A:=\lim_{\tau\to0}\frac{U(\tau)-I}{\tau}.\]Differentiability then gives the short-interval expansion
\[U(\tau)=I+\tau A+o(\tau).\]Here \(o(\tau)\) is an operator remainder for which every matrix entry of \(o(\tau)/\tau\) tends to zero as \(\tau\to0\).
For any linear operator \(L:\mathcal H\to\mathcal H\), its adjoint \(L^\dagger\) is defined by
\[\langle\phi,L\psi\rangle =\langle L^\dagger\phi,\psi\rangle\]for all \(\phi,\psi\in\mathcal H\). By the polarization identity, a linear norm-preserving map also preserves the inner product:
\[\langle U(\tau)\phi,U(\tau)\psi\rangle =\langle\phi,\psi\rangle.\]Using the definition of the adjoint,
\[\langle U(\tau)^\dagger U(\tau)\phi,\psi\rangle =\langle\phi,\psi\rangle\]for all \(\phi,\psi\in\mathcal H\), so
\[U(\tau)^\dagger U(\tau)=I.\]Substitute the short-interval expansion:
\[\begin{aligned} U(\tau)^\dagger U(\tau) &=\left(I+\tau A^\dagger+o(\tau)\right) \left(I+\tau A+o(\tau)\right)\\ &=I+\tau\left(A^\dagger+A\right)+o(\tau). \end{aligned}\]Equality with \(I\) requires
\[A^\dagger=-A.\]Thus the time generator is anti-Hermitian. The experiment, however, labels its oscillations by the real number \(\omega\). To represent such real frequencies by a Hermitian operator, define
\[G:\mathcal H\longrightarrow\mathcal H, \qquad G:=iA.\]Then
\[G^\dagger =(iA)^\dagger =-iA^\dagger =iA =G,\]so \(G\) is Hermitian and
\[A=-iG.\]From \(\psi(t+\tau)=U(\tau)\psi(t)\),
\[\begin{aligned} \lim_{\tau\to0} \frac{\psi(t+\tau)-\psi(t)}{\tau} &=A\psi(t)\\ &=-iG\psi(t). \end{aligned}\]Therefore
\[i\frac{d\psi(t)}{dt}=G\psi(t).\]For an operator \(L:\mathcal H\to\mathcal H\), define its matrix in the ordered coordinates \((A,B)\) by
\[\operatorname{coord}_{(A,B)}(L\psi) =[L]_{(A,B)}\operatorname{coord}_{(A,B)}(\psi).\]For the driven two-path example,
\[[G_\omega]_{(A,B)} = \begin{pmatrix} 0&0\\ 0&\omega \end{pmatrix},\]which produces
\[[U_\omega(\tau)]_{(A,B)} =e^{-i\tau[G_\omega]_{(A,B)}} = \begin{pmatrix} 1&0\\ 0&e^{-i\omega\tau} \end{pmatrix}.\]The abstract derivation has therefore recovered the update already measured in the motivating experiment.
3. Why identify the generator with energy?
Norm preservation tells us that \(G\) is Hermitian, but it does not identify the physical quantity represented by its eigenvalues. That identification comes from experiment.
For a state with a single measured oscillation frequency \(\omega\), experiments give
\[E=\hbar\omega,\]where the universal constant \(\hbar>0\) converts angular frequency into energy. Define
\[\widehat H:\mathcal H\longrightarrow\mathcal H, \qquad \widehat H:=\hbar G.\]Because \(G^\dagger=G\),
\[\widehat H^\dagger=\widehat H.\]If \(\psi_E\in\mathcal H\) satisfies
\[G\psi_E=\omega\psi_E,\]then
\[\widehat H\psi_E =\hbar\omega\psi_E =E\psi_E.\]The measured energies are therefore the eigenvalues of \(\widehat H\). Multiplying the generator equation by \(\hbar\) gives
\[i\hbar\frac{d\psi(t)}{dt} =\widehat H\psi(t).\]This is the abstract Schrödinger equation. The operator \(\widehat H\) is the Hamiltonian, or energy operator.
For an energy eigenstate,
\[\psi_E(t)=e^{-iEt/\hbar}\psi_E(0),\]and therefore
\[\lVert\psi_E(t)\rVert^2 =\lVert\psi_E(0)\rVert^2.\]4. Why can the imaginary unit not simply be omitted?
Suppose the same Hermitian energy operator instead generated
\[\hbar\frac{d\psi(t)}{dt} =\widehat H\psi(t).\]For \(\widehat H\psi_E=E\psi_E\), the solution would be
\[\psi_E(t)=e^{Et/\hbar}\psi_E(0),\]so
\[\lVert\psi_E(t)\rVert^2 =e^{2Et/\hbar}\lVert\psi_E(0)\rVert^2.\]Except when \(E=0\), total probability would change with time. The factor \(-i\) instead turns the Hermitian energy operator into the anti-Hermitian generator required by norm preservation:
\[\left(-\frac{i}{\hbar}\widehat H\right)^\dagger =\frac{i}{\hbar}\widehat H =-\left(-\frac{i}{\hbar}\widehat H\right).\]The imaginary unit changes real exponential growth into phase rotation:
\[e^{Et/\hbar} \quad\longrightarrow\quad e^{-iEt/\hbar}.\]5. How does the spatial Schrödinger equation arise?
For a particle whose possible positions form \(\mathbb R^3\), define the position-state space
\[\mathcal H_{\mathrm{pos}} :=L^2(\mathbb R^3,\mathbb C) :=\left\{ \psi\ \middle|\ \psi:\mathbb R^3\to\mathbb C, \ \int_{\mathbb R^3}|\psi(\mathbf x)|^2\,\mathrm d^3\mathbf x<\infty \right\}.\]At time \(t\), the state is the wavefunction
\[\psi(t)\in\mathcal H_{\mathrm{pos}}, \qquad \psi(\mathbf x,t):=\psi(t)(\mathbf x).\]Let the result of a position measurement be the random vector \(\mathbf X\in\mathbb R^3\). For every measurable region \(R\subseteq\mathbb R^3\), the position probability rule is
\[P(\mathbf X\in R\mid t) :=\int_R|\psi(\mathbf x,t)|^2\,\mathrm d^3\mathbf x.\]A normalized state satisfies
\[\int_{\mathbb R^3}|\psi(\mathbf x,t)|^2\,\mathrm d^3\mathbf x=1.\]To apply the abstract time-update equation, we must determine the spatial action of \(\widehat H\).
5.1. What operator represents momentum?
Matter-wave experiments relate measured momentum \(\mathbf p\) to measured wave vector \(\mathbf k\):
\[\mathbf p=\hbar\mathbf k.\]Write \(\mathbf x=(x_1,x_2,x_3)\) and define
\[\nabla :=\begin{pmatrix} \partial/\partial x_1\\ \partial/\partial x_2\\ \partial/\partial x_3 \end{pmatrix}, \qquad \nabla^2 :=\sum_{j=1}^3\frac{\partial^2}{\partial x_j^2}.\]An idealized plane wave is
\[\psi_{\mathbf k}(\mathbf x):=e^{i\mathbf k\cdot\mathbf x}.\]Its spatial derivative satisfies
\[\nabla\psi_{\mathbf k} =i\mathbf k\psi_{\mathbf k}.\]Therefore the linear operator
\[\widehat{\mathbf p}:=-i\hbar\nabla\]has the experimentally required value on every plane wave:
\[\widehat{\mathbf p}\,\psi_{\mathbf k} =\hbar\mathbf k\psi_{\mathbf k} =\mathbf p\,\psi_{\mathbf k}.\]5.2. What operator represents kinetic and potential energy?
For a nonrelativistic free particle of measured mass \(m>0\), experiment gives
\[K=\frac{|\mathbf p|^2}{2m}.\]Using the momentum operator,
\[\widehat K :=\frac{\widehat{\mathbf p}^{\,2}}{2m} =-\frac{\hbar^2}{2m}\nabla^2.\]If the particle’s potential-energy function is
\[V:\mathbb R^3\longrightarrow\mathbb R,\]define its action on a wavefunction by
\[(\widehat V\psi)(\mathbf x) :=V(\mathbf x)\psi(\mathbf x).\]The nonrelativistic energy operator is
\[\widehat H :=\widehat K+\widehat V =-\frac{\hbar^2}{2m}\nabla^2+\widehat V.\]Substitution into the abstract equation gives
\[i\hbar\frac{\partial\psi(\mathbf x,t)}{\partial t} = \left[ -\frac{\hbar^2}{2m}\nabla^2 +V(\mathbf x) \right]\psi(\mathbf x,t).\]Every term now has a defined role:
| Term | Role |
|---|---|
| \(\psi(\mathbf x,t)\) | Position-state amplitude at \(\mathbf x\) and time \(t\) |
| \(\lvert\psi(\mathbf x,t)\rvert^2\) | Position probability density |
| \(\partial\psi/\partial t\) | Instantaneous time change of the state |
| \(\widehat H\) | Energy operator; \(-i\widehat H/\hbar\) is the time generator |
| \(-\hbar^2\nabla^2/(2m)\) | Nonrelativistic kinetic-energy operator |
| \(V(\mathbf x)\) | Position-dependent potential energy |
| \(\hbar\) | Conversion factor from angular frequency to energy |
| \(i\) | Converts the Hermitian energy operator into an anti-Hermitian time generator |
6. Do we take the real part at the end?
No. A quantum state’s relative phase affects later probabilities through interference.
For example,
\[\psi_1:=\frac1{\sqrt2} \begin{pmatrix}1\\1\end{pmatrix}, \qquad \psi_2:=\frac1{\sqrt2} \begin{pmatrix}1\\i\end{pmatrix}.\]Both states give path probabilities \(1/2\) and \(1/2\), but their relative phases differ. The beam-splitter map therefore gives different detector probabilities. Replacing \(\psi_2\) by its real part would discard information required for that prediction.
Real measurement probabilities instead arise from the squared-amplitude rule. If a selected measurement outcome is represented by a normalized \(\phi\in\mathcal H\) and the system state is the normalized \(\psi\in\mathcal H\), then
\[P(\phi\mid\psi) :=|\langle\phi,\psi\rangle|^2.\]For a Hermitian observable \(\widehat A:\mathcal H\to\mathcal H\), its expected value in state \(\psi\) is the real scalar
\[\langle\widehat A\rangle_\psi :=\langle\psi,\widehat A\psi\rangle.\]7. Could the same theory use only real numbers?
Yes. Replace each complex amplitude by its two real components and represent multiplication by \(i\) with
\[J:= \begin{pmatrix} 0&-1\\ 1&0 \end{pmatrix}, \qquad J^2=-I_2.\]Here \(I_2\) is the \(2\times2\) identity matrix.
This gives an equivalent real description with twice as many scalar components and a distinguished linear operation \(J\). The rotation structure remains; only its notation changes.
Therefore the strongest conclusion is:
Continuous, reversible, probability-preserving quantum evolution requires a phase-rotation structure. Complex numbers provide its economical representation.
The symbol \(i\) is not removed at the end of a quantum calculation. It represents the rotation that converts energy into norm-preserving time evolution rather than exponential growth.