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| Figure 1: A two-state system (qubit) |
The time evolution of a quantum system, where the unitary operator \(U\) transforms the state of the system at time \(0\) to time \(t\), is defined as:
\[\psi(t) = U(t)\psi(0) \tag{1} \]
Operator \(U\), where the Hamiltonian \(H\) encodes the system's energy levels (spectrum), is defined as:[2]
\[U(t) = e^{−iHt} \tag{2} \]
Intuition: Time evolution is phase accumulation: in the energy basis, each component of \(\psi\) picks up a phase \(e^{-iE_k t}\). For each component, visualize a clock hand (or arrow) that spins at a rate set by its energy.
Differentiating and using \(\frac{d}{dt}U(t) = -iHU(t)\) gives the Schrödinger equation:[3]
\[i \frac{d}{dt}\psi(t) = H \psi(t) \tag{3} \]
Intuition: Visualize each energy component of \(\psi(t)\) as a clock hand rotating on the complex plane. Its time derivative is the tangent to that rotating clock hand (i.e., the clock hand rotated 90° clockwise and scaled by its energy). Multiplying by \(i\) rotates that tangent back to the radial direction. In the energy basis, applying \(H\) to \(\psi(t)\) simply scales each energy component by its corresponding energy level. Thus \(i \frac{d}{dt}\psi(t)\) (tangent rotated back) and \(H \psi(t)\) (component-wise scaling) are the same vector, giving the Schrödinger equation.
A trivial one-state system (no phase change)
Consider a one-dimensional Hilbert space, e.g., the spin degree of freedom of a spin-0 particle. Set \(H=0\) and \(\psi(0)=1\). Then
\[\psi(t)=e^{-i\cdot 0\cdot t}\,1 = 1 \]
for all \(t\) (as illustrated in Figure 2). For a one-state system, measurement always gives the same result with probability \(1\) (per the Born Rule, \(|1|^2 = 1\)). For a spin-0 particle, the measured spin is always \(0\).
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| Figure 2: \(\psi(t) = 1\) horizontal-axis: time | lateral axes: complex amplitude |
A one-state system with a 1Hz phase (global phase change only)
\(H = \theta\) and \(\psi(t)=e^{-i \cdot \theta \cdot t}\) where \(\theta\) is a scalar similarly represents a single physical state.
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| Figure 3: \(\psi(t) = e^{-i\cdot\tau\cdot t}\) |
The amplitudes at selected times per Figure 3 for \(\theta = \tau\) (where \(\theta\) is the energy eigenvalue and \(\tau \equiv 2\pi\), representing the radians in one cycle):
\[t=0,1: 1, \quad t=0.25: -i,\]
\[t=0.5: -1, \quad t=0.75: i.\]
This system can equally be used to model the spin degree of freedom of a spin-0 particle since the global phase is unobservable in quantum systems. The squared magnitude of any point on this curve is \(1\) (which, per the Born rule, represents the probability of measuring that state), thus the measured spin will again be \(0\) with certainty. Note that the oscillating state phase becomes relevant for multi-state systems which we will consider next.
A superposed two-state system (qubit)
Now consider a two-dimensional Hilbert space comprising a ground state (with \(0\) energy) and an excited state (with \(\tau\) energy) in equal superposition.[4][5] For example, a spin-½ particle with up (ground) and down (excited) states.
Energy basis \(\{|E_0\rangle,|E_1\rangle\}\) with
\[H_{energy} = \begin{bmatrix}0 & 0\\0 & \tau\end{bmatrix}, \quad \psi_{energy}(0) = \dfrac{1}{\sqrt2} \begin{bmatrix}1\\1\end{bmatrix} \]
Time evolution:
\[\begin{align}\psi_{energy}(t) &= U_{energy}(t)\psi_{energy}(0) \\ &= e^{−iH_{energy}t} \frac{1}{\sqrt2} \begin{bmatrix}1\\1\end{bmatrix} \\ &= \begin{bmatrix} 1 & 0 \\ 0 & e^{−i \tau t}\end{bmatrix} \frac{1}{\sqrt2} \begin{bmatrix}1\\1\end{bmatrix} \\ &= \frac{1}{\sqrt2} \begin{bmatrix}1 \\ e^{−i \tau t}\end{bmatrix} \end{align} \]
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| Figure 4: \(|\psi_{energy}(t)⟩ = \frac{1}{\sqrt2}(|E_0⟩ + e^{-i \tau t}|E_1⟩)\) |
The (superposed) amplitudes at selected times per Figure 4:
\[t=0,1: \dfrac{1}{\sqrt2}\begin{bmatrix}1\\1\end{bmatrix}, \quad t=0.25: \dfrac{1}{\sqrt2}\begin{bmatrix}1\\-i\end{bmatrix},\]
\[t=0.5: \dfrac{1}{\sqrt2}\begin{bmatrix}1\\-1\end{bmatrix}, \quad t=0.75: \dfrac{1}{\sqrt2}\begin{bmatrix}1\\i\end{bmatrix}.\]
These amplitudes are the linear combination of the amplitudes for the one-state systems in Figures 2 and 3 (normalized such that the square of the magnitudes equals 1).
Since the magnitude of the individual amplitudes for each state are \(\tfrac{1}{\sqrt2}\) at all points, the probability of measuring a particular state is \(\tfrac{1}{2}\) (per the Born rule, \((\tfrac{1}{\sqrt2})^2\)).
Time (Hadamard/DFT-2) basis. Define \(|T_0\rangle, |T_1\rangle\) via the Hadamard:
\[\begin{align}\psi_{time}(t) &= \mathcal{F}^{-1} \psi_{energy}(t) \\ &= \frac{1}{\sqrt2} \begin{bmatrix}1 & 1\\1 & -1\end{bmatrix} \frac{1}{\sqrt2}\begin{bmatrix} 1 \\ e^{−i \tau t} \end{bmatrix} \\ &= \frac{1}{2}\begin{bmatrix}1 + e^{−i \tau t} \\ 1 - e^{−i \tau t}\end{bmatrix} \end{align} \]
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| Figure 5: \(|\psi_{time}(t)⟩ = \frac{1}{2}(1 + e^{-i \tau t})|T_0⟩ + \frac{1}{2}(1 - e^{-i \tau t})|T_1⟩ \) |
The (superposed) amplitudes at selected times per Figure 5:[6]
\[t=0,1: \begin{bmatrix}1\\0\end{bmatrix}, \quad t=0.25: \dfrac{1}{2} \begin{bmatrix}1 - i\\1 + i\end{bmatrix},\]
\[t=0.5: \begin{bmatrix}0\\1\end{bmatrix}, \quad t=0.75: \dfrac{1}{2} \begin{bmatrix}1 + i\\1 - i\end{bmatrix}.\]
A measurement in the time basis (with eigenstates \(|T_0\rangle, |T_1\rangle\)) at physical time \(t=0\) would return the eigenstate \(|T_0\rangle\) (e.g., the spin-½ particle ground state in the time basis) with certainty and at physical time \(t=0.25\) would return either eigenstate with equal probability.
Wrapping up
We started with the basic rule for how quantum systems change over time and built up from the simplest case: a system that never changes to one that oscillates and then to a two-state system (a qubit). An energy level is analogous to a tiny clock hand spinning at its own speed. With just one clock, nothing interesting happens, but with two clocks spinning at different rates, their positions combine to create interference patterns. These two-clock systems, or qubits, are the building blocks for quantum computing.
Related posts
Footnotes
[1] Units, \(\tau\), and representations
Using Planck units with \(\hbar=1\) and representing the radians in one cycle with \(\tau\equiv 2\pi\), then
\[h=\tau,\qquad E=hf=\tau f = \omega,\]
referencing Planck's constant (\(h\)), energy (\(E\)), frequency (\(f\)) and angular frequency (\(\omega\)).
A quantum state is a complex wavefunction written here either in the energy basis or in a convenient “time (Fourier)” basis. In a one-state system these coincide (so any Heisenberg trade-off is vacuous). In finite-dimensional quantum systems, a “time basis” is not the eigenbasis of a fundamental time operator. It is a complementary basis obtained by a unitary transformation (Hadamard/DFT in 2D). Such systems satisfy SU(2) (spin-½) commutation relations, e.g. \([Sx,Sy]=iSz\), but not canonical commutation, such as \([E,T]=iI\). Here, \(Sx\) corresponds to energy, \(Sy\) to time and \(Sz\) to the relative phase between energy and time.
[2] Exponentiating the Hamiltonian
One-state system. With \(E_f=\tau f\) (the energy at a specific frequency),
\[H=E_f,\qquad U(t)=e^{-iHt}=e^{-iE_f t}.\]
Two-state system. With energy basis \(\{|E_0\rangle,|E_1\rangle\}\)),
\[H=\begin{bmatrix}E_0&0\\[2pt]0&E_1\end{bmatrix},\qquad U(t)=e^{-iHt}=\begin{bmatrix}e^{-iE_0 t}&0\\[2pt]0&e^{-iE_1 t}\end{bmatrix}.\]
Eigen/evolution relations (operator form):
\[H|E_n\rangle=E_n|E_n\rangle,\qquad U(t)|E_n\rangle=e^{-iE_n t}|E_n\rangle\quad(n=0,1).\]
[3] Deriving the Schrödinger equation
Starting from \(\psi(t)=U(t)\psi(0)\) with \(U(t)=e^{-iHt}\) and time-independent \(H\),
\[\begin{align}i \frac{d}{dt}\psi(t) &= i \frac{d}{dt}e^{−iHt}\psi(0) \\ &= i \cdot -iH e^{−iHt}\psi(0) \\ &= H \psi(t). \end{align} \]
[4] Transforming between the time and energy representations
The time and energy representations of a quantum system are related via a unitary Fourier transform:
\[\psi_{energy} = \mathcal{F} \psi_{time} , \quad \psi_{time} = \mathcal{F}^{-1} \psi_{energy} .\]
One state system: the Identity matrix,
\[\mathcal{F}_1 = \mathcal{F}_1^{-1} = \begin{bmatrix}1\end{bmatrix} . \]
Two state system: the Hadamard/DFT-2,
\[\mathcal{F}_2 = \mathcal{F}_2^{-1} = \frac{1}{\sqrt2} \begin{bmatrix}1 & 1\\1 & -1\end{bmatrix} . \]
[5] Energy and time bases
In the two-state system with \(\hbar=1\) and \(\tau\equiv 2\pi\) we set
\[ H_{\text{energy}}=\begin{bmatrix}0&0\\[2pt]0&\tau\end{bmatrix}.\]
So \(|E_0\rangle, |E_1\rangle\) satisfy
\[H_{\text{energy}}|E_0\rangle=0\,|E_0\rangle,\qquad H_{\text{energy}}|E_1\rangle=\tau\,|E_1\rangle,\]
and, in the energy basis coordinates,
\[|E_0\rangle = \begin{bmatrix}1\\0\end{bmatrix},\qquad |E_1\rangle = \begin{bmatrix}0\\1\end{bmatrix}.\]
We also define the time (Hadamard/DFT-2) basis by
\[|T_0\rangle=\tfrac{1}{\sqrt2}(|E_0\rangle+|E_1\rangle)=\tfrac{1}{\sqrt2}\begin{bmatrix}1\\[2pt]1\end{bmatrix},\qquad |T_1\rangle=\tfrac{1}{\sqrt2}(|E_0\rangle-|E_1\rangle)=\tfrac{1}{\sqrt2}\begin{bmatrix}1\\[2pt]-1\end{bmatrix}.\]
[6] Transform-then-evolve (and other alternatives)
A second method for calculating \( \psi_{time}(t) \) is to transform \(H_{energy} \) and \(\psi_{energy}(0)\) to the time basis first and then evolve the state.
\[\begin{align}H_{time} &= \mathcal{F}^{-1} H_{energy} \mathcal{F} \\ &= \frac{1}{\sqrt2} \begin{bmatrix}1 & 1\\1 & -1\end{bmatrix} \begin{bmatrix}0 & 0\\0 & \tau\end{bmatrix} \frac{1}{\sqrt2} \begin{bmatrix}1 & 1\\1 & -1\end{bmatrix} \\ &= \begin{bmatrix}\pi & -\pi\\-\pi & \pi\end{bmatrix} \end{align} \]
\[\begin{align}\psi_{time}(0) &= \mathcal{F}^{-1} \psi_{energy}(0) \\ &= \frac{1}{\sqrt2} \begin{bmatrix}1 & 1\\1 & -1\end{bmatrix} \frac{1}{\sqrt2} \begin{bmatrix}1\\1\end{bmatrix} \\ &= \begin{bmatrix}1\\0\end{bmatrix} \end{align} \]
\[\begin{align}\psi_{time}(t) &= U_{time}(t)\psi_{time}(0) \\ &= e^{−iH_{time}t} \begin{bmatrix}1\\0\end{bmatrix} \\ &= \frac{1}{2} \begin{bmatrix}1 + e^{−i \tau t} & 1 - e^{−i \tau t} \\ 1 - e^{−i \tau t} & 1 + e^{−i \tau t}\end{bmatrix} \begin{bmatrix}1\\0\end{bmatrix} \\ &= \frac{1}{2} \begin{bmatrix}1 + e^{−i \tau t} \\ 1 - e^{−i \tau t}\end{bmatrix} \end{align} \]
A third method for calculating \( \psi_{time}(t) \) is via the Schrödinger equation.
\[i \frac{d}{dt}\psi(t) = H \psi(t), \quad H_{energy} = \begin{bmatrix}0 & 0\\0 & \tau\end{bmatrix}, \quad \psi_{energy}(0) = \dfrac{1}{\sqrt2} \begin{bmatrix}1\\1\end{bmatrix}.\]
Because the Hamiltonian is diagonal, the Schrödinger equation decouples into independent equations for each component. For the first component:
\[ i \frac{d}{dt}\psi_0(t) = 0 \cdot \psi_0(t) = 0 \]
\[ \frac{d}{dt}\psi_0(t) = 0 \implies \psi_0(t) = \psi_0(0) = \frac{1}{\sqrt2} \]
For the second component:
\[ i \frac{d}{dt}\psi_1(t) = \tau \cdot \psi_1(t) \]
\[\psi_1(t) = \psi_1(0) e^{−i \tau t} = \frac{1}{\sqrt2} e^{−i \tau t}\]
Therefore, the solution in the energy basis is:
\[\psi_{energy}(t) = \frac{1}{\sqrt2} \begin{bmatrix}\psi_0(t) \\ \psi_1(t) \end{bmatrix} = \frac{1}{\sqrt2} \begin{bmatrix}1 \\ e^{−i \tau t}\end{bmatrix}\]
Finally, transform \(\psi_{energy}(t)\) to the time basis:
\[\begin{align}\psi_{time}(t) &= \mathcal{F}^{-1} \psi_{energy}(t) \\ &= \frac{1}{\sqrt2} \begin{bmatrix}1 & 1\\1 & -1\end{bmatrix} \frac{1}{\sqrt2}\begin{bmatrix} 1 \\ e^{−i \tau t} \end{bmatrix} \\ &= \frac{1}{2}\begin{bmatrix}1 + e^{−i \tau t} \\ 1 - e^{−i \tau t}\end{bmatrix} \end{align} \]
