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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]
\[U(t) = e^{−iHt} \tag{2} \]
Time evolution 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]
\[i \frac{d}{dt}\psi(t) = H \psi(t) \tag{3} \]
Schrödinger equation 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 particle with a frequency of 0Hz. Set \(H=0\) and \(\psi(0)=1\). Then
\[\psi(t)=e^{-i\cdot 0\cdot t}\cdot 1 = 1 \]
for all \(t\) (as illustrated in Figure 2). For a one-state system, there is only one possible measurement outcome, namely, that single state.
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| Figure 2: \(\psi(t) = 1\) (0Hz) horizontal-axis: time | lateral axes: complex amplitude |
0Hz system intuition: We are, in effect, modeling a clock whose hand does not rotate (0Hz). Hence the graph shows the amplitude fixed in its initial phase over time. Note, however, that we cannot subsequently measure the angle (phase) of the clock hand, or even its frequency (more on this below).
A one-state system with a 1Hz phase (global phase change only)
Now consider a particle with a frequency of 1Hz. Set \(H = \theta\) and \(\psi(0)=1\). Then
\[\psi(t)=e^{-i\cdot \theta \cdot t} \cdot 1 = e^{-i\cdot \theta \cdot t} \]
where \(\theta\) is a scalar.
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| Figure 3: \(\psi(t) = e^{-i\cdot\tau\cdot t}\) (1Hz) |
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):
\[t=0,1: 1, \quad t=0.25: -i,\]
\[t=0.5: -1, \quad t=0.75: i.\]
As noted before: for a one-state system, there is only one possible measurement outcome, namely, that single state. Since the global phase (including both the degree of rotation around the complex plane and its rate of change) is unobservable in quantum systems, this system is indistinguishable from the earlier 0Hz system. The different oscillations become relevant for multi-state systems which we will consider next.
1Hz system intuition: Now we're modeling a clock whose hand rotates once per second (1Hz). Hence the graph shows the amplitude rotating in a unit circle from 1 through -i, -1 and i back to 1. This is visualized as a corkscrew along the time axis. As noted earlier, we cannot measure the angle of the clock hand or its frequency. Only relative phase (between two or more clocks) can be measured. Each clock is, in effect, enclosed in its own black box with no way to experimentally distinguish them.
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 (0Hz ground) and down (1Hz excited) states.
Hilbert space intuition: Consider the familiar 2D Euclidean plane that represent lengths in an x and y direction, and also angles and magnitude. A 2D Hilbert space allows the x and y coordinates to be complex numbers. This retains the familiar geometry but adds structure for phases and interference. A vector in a 2D Hilbert space (i.e., a qubit) can be visualized as a point on the surface of a Bloch sphere where the latitude represents proximity to the poles (0 and 1) and the longitude represents the relative phase between them.
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: \(|\psi_{energy}(t)⟩ = \frac{1}{\sqrt2}(|E_0⟩ + e^{-i \tau t}|E_1⟩)\) (0Hz+1Hz) |
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).
Vector intuition: The vectors have two elements, associated with the two component states. Reading
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, \((\tfrac{1}{\sqrt2})^2\)).
Born rule intuition: The two states can be represented as the adjacent and opposite sides of a right-angled triangle, with the hypotenuse of unit length \(1\). Per the Pythagorean Theorem, if the magnitudes of the two states are \(0.6\) and \(0.8\), then the probability of measuring those states per the Born rule is \(0.6^2 = 0.36\) and \(0.8^2 = 0.64\) respectively (with \(0.36 + 0.64 = 1\)). In the main qubit example, the triangle side lengths are each \(\tfrac{1}{\sqrt2}\) whose squares are each \(\tfrac{1}{2}\) (summing to 1). A measurement is a projection onto one of the sides per the Born probabilities. Thus, in the main qubit example, there is an equal probability of measuring either state (i.e., 0Hz or 1Hz, ground or excited, 0 or 1).
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 \(|T_0\rangle\) (i.e., 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.
Qubit intuition: The two independent clocks discussed earlier can be superposed in a single two-state system (a qubit). A measurement projects onto just one of the clocks (0Hz or 1Hz, ground or excited, 0 or 1). In the standard (energy) basis, the two clocks are modeled per Figure 4, with the measurement of either 0Hz or 1Hz being equally likely at any point in time. In the Fourier (time) basis, the relative phase between the two clocks are modeled per Figure 5, with the probability of measuring either T0 or T1 (which is + or -, in-phase or out-of-phase, 0 or 1) being dependent on the time of measurement.
Relative phase intuition: In Figure 4, the straight bright purple line models the 0Hz stopped clock hand whereas the curved dim purple line models the 1Hz moving clock hand. In Figure 5, the bright green line models how correlated the two clock hands are (how in-phase they are, i.e., the phase difference) while the dim green line models how anti-correlated the two clock hands are (how out-of-phase they are). So at t=0, the two hands point in the same direction, while at t=0.5, the two hands point in the opposite direction. Note that the purple lines don't exhibit interference since just the stopped clock or the moving clock state is measured (but, to reiterate, not its phase). This is analogous to measuring a particle at just one of the slits in the double-slit experiment. However the green lines are each interference patterns since they represent the phase between both clock hands. Remove one of the clocks and the interference pattern disappears.
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 of 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
The Hamiltonian \(H\) is a diagonal matrix that encodes the system's energy levels \(E_k\) along the diagonal. For a one-state system, it reduces to a scalar.
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
\[|T_0\rangle=\frac{1}{\sqrt2}(|E_0\rangle+|E_1\rangle)=\frac{1}{\sqrt2}\begin{bmatrix}1\\[2pt]1\end{bmatrix},\qquad |T_1\rangle=\frac{1}{\sqrt2}(|E_0\rangle-|E_1\rangle)=\frac{1}{\sqrt2}\begin{bmatrix}1\\[2pt]-1\end{bmatrix}.\]
[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:
\[\begin{align}\psi_{time}(t) &= \mathcal{F}^{-1} \psi_{energy}(t) \\ &= \frac{1}{\sqrt2} \begin{bmatrix}1 & 1\\1 & -1\end{bmatrix} \frac{1}{\sqrt2}\begin{bmatrix} 1 \\ e^{−i \tau t} \end{bmatrix} \\ &= \frac{1}{2}\begin{bmatrix}1 + e^{−i \tau t} \\ 1 - e^{−i \tau t}\end{bmatrix} \end{align} \]
