QuantumCore

Lesson 10 · 55 min · core

Postulate 2: Unitary Evolution & the Schrödinger Equation

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The second postulate says how the state of a closed system changes in time: by a unitary operator in discrete time, generated by the Schrödinger equation iℏ ddt∣ψ⟩=H∣ψ⟩i\hbar\,\tfrac{d}{dt}\lvert\psi\rangle=H\lvert\psi\rangle in continuous time. This lesson solves the time-independent Schrödinger equation in the energy eigenbasis to obtain U(t)=e−iHt/ℏU(t)=e^{-iHt/\hbar}, proves unitarity and norm conservation from the Hermiticity of HH, proves the converse that every unitary is e−iKe^{-iK} for a Hermitian KK, derives Rabi oscillations P1(t)=sin⁡2(ωt/2)P_1(t)=\sin^2(\omega t/2) for H=ℏω2XH=\tfrac{\hbar\omega}{2}X and identifies U(t)=RX(ωt)U(t)=R_X(\omega t), and shows that energy eigenstates are stationary.

Prerequisites: Postulate 1: State Space & Global Phase, Spectral Theorem & Functions of Operators

Objectives and key results

  • State Postulate 2 in its discrete form ∣ψ′⟩=U∣ψ⟩\lvert\psi'\rangle=U\lvert\psi\rangle and its continuous form iℏ ddt∣ψ(t)⟩=H∣ψ(t)⟩i\hbar\,\tfrac{d}{dt}\lvert\psi(t)\rangle=H\lvert\psi(t)\rangle.

  • Solve the Schrödinger equation for a time-independent Hamiltonian H=∑kEk∣k⟩⟨k∣H=\sum_kE_k\lvert k\rangle\langle k\rvert and obtain U(t)=e−iHt/ℏ=∑ke−iEkt/ℏ∣k⟩⟨k∣U(t)=e^{-iHt/\hbar}=\sum_ke^{-iE_kt/\hbar}\lvert k\rangle\langle k\rvert.

  • Prove U(t)†U(t)=IU(t)^\dagger U(t)=I and ddt⟨ψ∣ψ⟩=0\tfrac{d}{dt}\langle\psi\vert\psi\rangle=0, and prove that every unitary UU equals e−iKe^{-iK} with K=K†K=K^\dagger.

  • Compute Rabi oscillations for H=ℏω2XH=\tfrac{\hbar\omega}{2}X: U(t)=RX(ωt)U(t)=R_X(\omega t) and P1(t)=sin⁡2(ωt/2)P_1(t)=\sin^2(\omega t/2), and explain why energy eigenstates only acquire a phase.

definition · Schrödinger equation
iℏ ddt∣ψ(t)⟩=H∣ψ(t)⟩,H=H†i\hbar\,\frac{d}{dt}\lvert\psi(t)\rangle=H\lvert\psi(t)\rangle,\qquad H=H^\dagger
formula · Time-independent propagator
U(t)=e−iHt/ℏ=∑ke−iEkt/ℏ∣k⟩⟨k∣U(t)=e^{-iHt/\hbar}=\sum_ke^{-iE_kt/\hbar}\lvert k\rangle\langle k\rvert
theorem · Unitarity and norm conservation
U(t)†U(t)=I,ddt⟨ψ(t)∣ψ(t)⟩=0U(t)^\dagger U(t)=I,\qquad \frac{d}{dt}\langle\psi(t)\vert\psi(t)\rangle=0
theorem · Every unitary is generated by a Hermitian operator
U=∑keiϕkPk=e−iK,K=−∑kϕkPk=K†U=\sum_ke^{i\phi_k}P_k=e^{-iK},\qquad K=-\sum_k\phi_kP_k=K^\dagger
formula · Rabi oscillation
H=ℏω2X:U(t)=RX(ωt)=cos⁡ωt2 I−isin⁡ωt2 X,P1(t)=sin⁡2ωt2H=\tfrac{\hbar\omega}{2}X:\quad U(t)=R_X(\omega t)=\cos\tfrac{\omega t}{2}\,I-i\sin\tfrac{\omega t}{2}\,X,\qquad P_1(t)=\sin^2\tfrac{\omega t}{2}

The evolution postulate#

Postulate 1 told us what a state is: a unit vector of a complex Hilbert space, defined up to a global phase. Postulate 2 tells us how it changes while the system is left alone — closed, not measured, not coupled to an environment. In a quantum computer this is the regime of gates: a pulse of microwaves or a laser drives the qubit for a fixed duration, and the net effect is a fixed unitary matrix. The postulate comes in two equivalent forms. The discrete form is the one circuit diagrams use; the continuous form is the one hardware engineers use, because the physical control knob is a Hamiltonian applied for a time. The main result of this lesson is that the two forms are the same thing: every Hamiltonian applied for a time produces a unitary, and every unitary is produced by some Hamiltonian applied for unit time.

Definition

Postulate 2 (discrete form)

The evolution of a closed quantum system from time t1t_1 to time t2t_2 is described by a unitary operator U=U(t2,t1)U=U(t_2,t_1) that depends only on t1t_1 and t2t_2: ∣ψ(t2)⟩=U∣ψ(t1)⟩\lvert\psi(t_2)\rangle=U\lvert\psi(t_1)\rangle, with U†U=UU†=IU^\dagger U=UU^\dagger=I. Unitarity guarantees that unit vectors stay unit vectors and that orthogonal (perfectly distinguishable) states stay orthogonal, because ⟨Uϕ∣Uψ⟩=⟨ϕ∣U†U∣ψ⟩=⟨ϕ∣ψ⟩\langle U\phi\vert U\psi\rangle=\langle\phi\vert U^\dagger U\vert\psi\rangle=\langle\phi\vert\psi\rangle.

Definition

Postulate 2 (continuous form, Schrödinger equation)

The time evolution of the state of a closed system is governed by iℏ ddt∣ψ(t)⟩=H∣ψ(t)⟩,i\hbar\,\frac{d}{dt}\lvert\psi(t)\rangle=H\lvert\psi(t)\rangle, where ℏ\hbar is the reduced Planck constant and HH is a fixed Hermitian operator, the Hamiltonian of the system. Its eigenvalues EkE_k are the possible energies and its eigenvectors ∣k⟩\lvert k\rangle the energy eigenstates. By the spectral theorem H=∑kEk∣k⟩⟨k∣H=\sum_kE_k\lvert k\rangle\langle k\rvert with real EkE_k and orthonormal {∣k⟩}\{\lvert k\rangle\}.

Solving the Schrödinger equation#

Derivation

Solution for a time-independent Hamiltonian

  1. Step 1:
    H=∑kEk∣k⟩⟨k∣,⟨k∣l⟩=δkl,Ek∈RH=\sum_kE_k\lvert k\rangle\langle k\rvert,\qquad \langle k\vert l\rangle=\delta_{kl},\qquad E_k\in\mathbb R
    Spectral theorem for the Hermitian operator HH (previous module): an orthonormal eigenbasis with real eigenvalues.
  2. Step 2:
    ∣ψ(t)⟩=∑kck(t)∣k⟩,ck(t)=⟨k∣ψ(t)⟩\lvert\psi(t)\rangle=\sum_kc_k(t)\lvert k\rangle,\qquad c_k(t)=\langle k\vert\psi(t)\rangle
    Expand the unknown state in the fixed eigenbasis; all time dependence is in the coefficients.
  3. Step 3:
    iℏ∑kc˙k(t)∣k⟩=∑kck(t) H∣k⟩=∑kEk ck(t)∣k⟩i\hbar\sum_k\dot c_k(t)\lvert k\rangle=\sum_kc_k(t)\,H\lvert k\rangle=\sum_kE_k\,c_k(t)\lvert k\rangle
    Substitute into the Schrödinger equation; the basis vectors do not depend on tt, and H∣k⟩=Ek∣k⟩H\lvert k\rangle=E_k\lvert k\rangle.
  4. Step 4:
    iℏ c˙k(t)=Ek ck(t)for each ki\hbar\,\dot c_k(t)=E_k\,c_k(t)\quad\text{for each }k
    Take the inner product with ⟨k∣\langle k\rvert on both sides and use orthonormality: the vector equation decouples into independent scalar ODEs.
  5. Step 5:
    ck(t)=e−iEkt/ℏ ck(0)c_k(t)=e^{-iE_kt/\hbar}\,c_k(0)
    The unique solution of the linear ODE c˙=−iEkℏc\dot c=-\tfrac{iE_k}{\hbar}c with initial value ck(0)c_k(0) (check: ddte−iEkt/ℏ=−iEkℏe−iEkt/ℏ\tfrac{d}{dt}e^{-iE_kt/\hbar}=-\tfrac{iE_k}{\hbar}e^{-iE_kt/\hbar}).
  6. Step 6:
    ∣ψ(t)⟩=∑ke−iEkt/ℏ∣k⟩⟨k∣ψ(0)⟩=(∑ke−iEkt/ℏ∣k⟩⟨k∣)∣ψ(0)⟩\lvert\psi(t)\rangle=\sum_ke^{-iE_kt/\hbar}\lvert k\rangle\langle k\vert\psi(0)\rangle=\Big(\sum_ke^{-iE_kt/\hbar}\lvert k\rangle\langle k\rvert\Big)\lvert\psi(0)\rangle
    Insert ck(0)=⟨k∣ψ(0)⟩c_k(0)=\langle k\vert\psi(0)\rangle and pull the initial state out of the sum.
  7. Step 7:
    U(t)=∑ke−iEkt/ℏ∣k⟩⟨k∣=f(H),f(E)=e−iEt/ℏ⟹U(t)=e−iHt/ℏU(t)=\sum_ke^{-iE_kt/\hbar}\lvert k\rangle\langle k\rvert=f(H),\quad f(E)=e^{-iEt/\hbar}\quad\Longrightarrow\quad U(t)=e^{-iHt/\hbar}
    The operator in brackets is the function ff applied to HH through its spectral decomposition, which is how e−iHt/ℏe^{-iHt/\hbar} is defined (and agrees with the power series ∑n(−iHt/ℏ)n/n!\sum_n(-iHt/\hbar)^n/n!).

Result

∣ψ(t)⟩=e−iHt/ℏ∣ψ(0)⟩\boxed{\lvert\psi(t)\rangle=e^{-iHt/\hbar}\lvert\psi(0)\rangle}
Theorem

Schrödinger evolution is unitary and conserves the norm

If H=H†H=H^\dagger is time-independent, then U(t)=e−iHt/ℏU(t)=e^{-iHt/\hbar} satisfies U(t)†U(t)=U(t)U(t)†=IU(t)^\dagger U(t)=U(t)U(t)^\dagger=I for all real tt, and U(t1)U(t2)=U(t1+t2)U(t_1)U(t_2)=U(t_1+t_2). Independently of the explicit solution, any solution of the Schrödinger equation with Hermitian (possibly time-dependent) HH satisfies ddt⟨ψ(t)∣ψ(t)⟩=0\tfrac{d}{dt}\langle\psi(t)\vert\psi(t)\rangle=0.

Proof (6 steps)
  1. Step 1:
    U(t)†=(∑ke−iEkt/ℏ∣k⟩⟨k∣)†=∑ke−iEkt/ℏ‾ ∣k⟩⟨k∣=∑ke+iEkt/ℏ∣k⟩⟨k∣U(t)^\dagger=\Big(\sum_ke^{-iE_kt/\hbar}\lvert k\rangle\langle k\rvert\Big)^\dagger=\sum_k\overline{e^{-iE_kt/\hbar}}\,\lvert k\rangle\langle k\rvert=\sum_ke^{+iE_kt/\hbar}\lvert k\rangle\langle k\rvert
    (λ∣k⟩⟨k∣)†=λˉ∣k⟩⟨k∣(\lambda\lvert k\rangle\langle k\rvert)^\dagger=\bar\lambda\lvert k\rangle\langle k\rvert; the conjugate of e−iEkt/ℏe^{-iE_kt/\hbar} is e+iEkt/ℏe^{+iE_kt/\hbar} because EkE_k and tt are real — this is where Hermiticity of HH enters.
  2. Step 2:
    U(t)†U(t)=∑k,leiEkt/ℏe−iElt/ℏ∣k⟩⟨k∣l⟩⟨l∣=∑keiEkt/ℏe−iEkt/ℏ∣k⟩⟨k∣U(t)^\dagger U(t)=\sum_{k,l}e^{iE_kt/\hbar}e^{-iE_lt/\hbar}\lvert k\rangle\langle k\vert l\rangle\langle l\rvert=\sum_ke^{iE_kt/\hbar}e^{-iE_kt/\hbar}\lvert k\rangle\langle k\rvert
    Multiply the two sums; ⟨k∣l⟩=δkl\langle k\vert l\rangle=\delta_{kl} kills the cross terms.
  3. Step 3:
    =∑k∣k⟩⟨k∣=I,and likewise U(t)U(t)†=I=\sum_k\lvert k\rangle\langle k\rvert=I,\qquad\text{and likewise}\ U(t)U(t)^\dagger=I
    eixe−ix=1e^{ix}e^{-ix}=1 and completeness of the orthonormal eigenbasis; the product in the other order is identical term by term. Equivalently eiHt/ℏe−iHt/ℏ=Ie^{iHt/\hbar}e^{-iHt/\hbar}=I because the exponents commute.
  4. Step 4:
    U(t1)U(t2)=∑ke−iEk(t1+t2)/ℏ∣k⟩⟨k∣=U(t1+t2)U(t_1)U(t_2)=\sum_ke^{-iE_k(t_1+t_2)/\hbar}\lvert k\rangle\langle k\rvert=U(t_1+t_2)
    Same orthonormality computation with e−iEkt1/ℏe−iEkt2/ℏ=e−iEk(t1+t2)/ℏe^{-iE_kt_1/\hbar}e^{-iE_kt_2/\hbar}=e^{-iE_k(t_1+t_2)/\hbar}: evolving for t2t_2 then t1t_1 equals evolving for t1+t2t_1+t_2, and U(−t)=U(t)†U(-t)=U(t)^\dagger undoes U(t)U(t).
  5. Step 5:
    ddt∣ψ⟩=−iℏH∣ψ⟩,ddt⟨ψ∣=(−iℏH∣ψ⟩)†=iℏ⟨ψ∣H†=iℏ⟨ψ∣H\frac{d}{dt}\lvert\psi\rangle=-\frac{i}{\hbar}H\lvert\psi\rangle,\qquad \frac{d}{dt}\langle\psi\rvert=\Big(-\frac{i}{\hbar}H\lvert\psi\rangle\Big)^\dagger=\frac{i}{\hbar}\langle\psi\vert H^\dagger=\frac{i}{\hbar}\langle\psi\vert H
    Divide the Schrödinger equation by iℏi\hbar (using 1/i=−i1/i=-i), then take the adjoint: −i‾=i\overline{-i}=i and H†=HH^\dagger=H.
  6. Step 6:
    ddt⟨ψ∣ψ⟩=(ddt⟨ψ∣)∣ψ⟩+⟨ψ∣(ddt∣ψ⟩)=iℏ⟨ψ∣H∣ψ⟩−iℏ⟨ψ∣H∣ψ⟩=0\frac{d}{dt}\langle\psi\vert\psi\rangle=\Big(\frac{d}{dt}\langle\psi\rvert\Big)\lvert\psi\rangle+\langle\psi\rvert\Big(\frac{d}{dt}\lvert\psi\rangle\Big)=\frac{i}{\hbar}\langle\psi\vert H\vert\psi\rangle-\frac{i}{\hbar}\langle\psi\vert H\vert\psi\rangle=0
    Product rule for the inner product, then substitute both derivatives: the two terms are equal and opposite, so the norm ⟨ψ∣ψ⟩=1\langle\psi\vert\psi\rangle=1 is conserved for all tt.

∎

Why Hermiticity is essential. If some EkE_k had an imaginary part, say Ek=a−ibE_k=a-ib with b>0b>0, then ∣e−iEkt/ℏ∣=e−bt/ℏ\lvert e^{-iE_kt/\hbar}\rvert=e^{-bt/\hbar} and the amplitude on ∣k⟩\lvert k\rangle would decay: probability would leak out and the total would drop below one. Effective non-Hermitian Hamiltonians are indeed used to model decay into an environment, but that is exactly the situation in which the system is not closed; Postulate 2 applies to closed systems only, and open-system dynamics is treated with channels on density operators later in the course.

Numerical view. The platform's simulator computes e−iHte^{-iHt} exactly this way: its expmIHermitian function diagonalises HH with a Hermitian eigensolver and returns ∑ke−iEkt∣k⟩⟨k∣\sum_ke^{-iE_kt}\lvert k\rangle\langle k\rvert, so the result is unitary to machine precision for every tt — unlike a truncated Taylor series, which is only approximately unitary.

Quick check

A qubit has Hamiltonian H=diag(0,E)H=\mathrm{diag}(0,E) with ℏ=1\hbar=1 and starts in 12(∣0⟩+∣1⟩)\tfrac1{\sqrt2}(\lvert0\rangle+\lvert1\rangle). Which vector is ∣ψ(t)⟩\lvert\psi(t)\rangle?

Select one answer

Every unitary is a Hamiltonian evolution#

Theorem

Every unitary is the exponential of a Hermitian operator

For every unitary UU on a finite-dimensional Hilbert space there is a Hermitian KK with U=e−iKU=e^{-iK}. Explicitly, if U=∑keiϕkPkU=\sum_ke^{i\phi_k}P_k is its spectral decomposition with distinct eigenvalues eiϕke^{i\phi_k}, ϕk∈(−π,π]\phi_k\in(-\pi,\pi], and orthogonal eigenprojectors PkP_k, then K=−∑kϕkPkK=-\sum_k\phi_kP_k. Physically, the Hamiltonian H=ℏK/t0H=\hbar K/t_0 applied for time t0t_0 implements UU.

Proof (6 steps)
  1. Step 1:
    U†U=UU†=I ⟹ U is normal ⟹ U=∑kλkPk,PkPl=δklPk, Pk†=Pk, ∑kPk=IU^\dagger U=UU^\dagger=I\ \Longrightarrow\ U\ \text{is normal}\ \Longrightarrow\ U=\sum_k\lambda_kP_k,\quad P_kP_l=\delta_{kl}P_k,\ P_k^\dagger=P_k,\ \sum_kP_k=I
    A unitary commutes with its adjoint, and the spectral theorem for normal operators gives an orthogonal decomposition into eigenprojectors.
  2. Step 2:
    U∣v⟩=λ∣v⟩ ⇒ ⟨v∣v⟩=⟨v∣U†U∣v⟩=λ‾λ⟨v∣v⟩=∣λ∣2⟨v∣v⟩U\lvert v\rangle=\lambda\lvert v\rangle\ \Rightarrow\ \langle v\vert v\rangle=\langle v\vert U^\dagger U\vert v\rangle=\overline{\lambda}\lambda\langle v\vert v\rangle=\lvert\lambda\rvert^2\langle v\vert v\rangle
    For an eigenvector ∣v⟩≠0\lvert v\rangle\neq0: insert U†U=IU^\dagger U=I and use ⟨v∣U†=λˉ⟨v∣\langle v\vert U^\dagger=\bar\lambda\langle v\rvert.
  3. Step 3:
    ∣λk∣=1 ⟹ λk=eiϕk,ϕk=arg⁡λk∈(−π,π]\lvert\lambda_k\rvert=1\ \Longrightarrow\ \lambda_k=e^{i\phi_k},\quad \phi_k=\arg\lambda_k\in(-\pi,\pi]
    Divide by ⟨v∣v⟩>0\langle v\vert v\rangle>0; every unimodular complex number has a polar form with a real angle.
  4. Step 4:
    K:=−∑kϕkPk,K†=−∑kϕk‾ Pk†=−∑kϕkPk=KK:=-\sum_k\phi_kP_k,\qquad K^\dagger=-\sum_k\overline{\phi_k}\,P_k^\dagger=-\sum_k\phi_kP_k=K
    Define KK by its spectral decomposition; it is Hermitian because the ϕk\phi_k are real and the PkP_k are self-adjoint.
  5. Step 5:
    e−iK=∑ke−i(−ϕk)Pk=∑keiϕkPk=Ue^{-iK}=\sum_ke^{-i(-\phi_k)}P_k=\sum_ke^{i\phi_k}P_k=U
    A function of a Hermitian operator acts on its eigenvalues: KK has eigenvalue −ϕk-\phi_k on ran⁡Pk\operatorname{ran}P_k, and e−i(−ϕk)=eiϕke^{-i(-\phi_k)}=e^{i\phi_k}.
  6. Step 6:
    e−iHt0/ℏ=e−i(ℏK/t0)t0/ℏ=e−iK=Ue^{-iHt_0/\hbar}=e^{-i(\hbar K/t_0)t_0/\hbar}=e^{-iK}=U
    Choosing H=ℏK/t0H=\hbar K/t_0, which is Hermitian, reproduces UU at time t0t_0 via the solution derived above.

∎

Derivation

Example: the Hadamard gate as e^{−iK}

  1. Step 1:
    H2=I, H=H† ⇒ spec⁡(H)={+1,−1},P±=12(I±H)H^2=I,\ H=H^\dagger\ \Rightarrow\ \operatorname{spec}(H)=\{+1,-1\},\qquad P_\pm=\tfrac12(I\pm H)
    In this example HH denotes the Hadamard gate, not a Hamiltonian. An involution has eigenvalues ±1\pm1; P±P_\pm are the eigenprojectors because HP±=12(H±I)=±P±HP_\pm=\tfrac12(H\pm I)=\pm P_\pm.
  2. Step 2:
    H=(+1)P++(−1)P−=ei0P++eiπP−H=(+1)P_++(-1)P_-=e^{i0}P_++e^{i\pi}P_-
    Spectral decomposition with phases ϕ+=0\phi_+=0 and ϕ−=π\phi_-=\pi.
  3. Step 3:
    K=−(0⋅P++πP−)=−πP−=−π2(I−H)=π2(H−I)K=-(0\cdot P_++\pi P_-)=-\pi P_-=-\frac{\pi}{2}(I-H)=\frac{\pi}{2}(H-I)
    The recipe of the theorem.
  4. Step 4:
    e−iK=eiπ/2 e−iπ2H=i(cos⁡π2I−isin⁡π2H)=i(−iH)=He^{-iK}=e^{i\pi/2}\,e^{-i\frac{\pi}{2}H}=i\big(\cos\tfrac{\pi}{2}I-i\sin\tfrac{\pi}{2}H\big)=i(-iH)=H
    II commutes with HH, so the exponential splits; then e−iαH=cos⁡α I−isin⁡α He^{-i\alpha H}=\cos\alpha\,I-i\sin\alpha\,H for an involution (shown for XX below).

Result

H=e−iπ2(H−I)\boxed{H=e^{-i\frac{\pi}{2}(H-I)}}

The generator is not unique: adding 2πmk2\pi m_k to any ϕk\phi_k gives another valid KK, and adding a multiple of the identity to KK only changes the global phase of UU, which by Postulate 1 is unobservable. The theorem justifies the circuit model: any gate a circuit diagram uses can, in principle, be realised by switching on a suitable Hamiltonian for a suitable time. In practice hardware offers only a few Hamiltonians (for example XX, YY drives and a ZZZZ coupling), and compilation decomposes an arbitrary unitary into gates generated by those.

Rabi oscillations#

The simplest non-trivial dynamics: a qubit initially in ∣0⟩\lvert0\rangle, driven by a resonant field. In the rotating frame the Hamiltonian is H=ℏω2XH=\tfrac{\hbar\omega}{2}X, where ω\omega is the Rabi frequency, set by the drive amplitude. Since XX does not commute with the projector ∣0⟩⟨0∣\lvert0\rangle\langle0\rvert, population is transferred back and forth between ∣0⟩\lvert0\rangle and ∣1⟩\lvert1\rangle — the Rabi oscillation that every superconducting and trapped-ion lab uses to calibrate its π\pi pulse.

Derivation

Rabi problem: U(t) = R_X(ωt) and P₁(t) = sin²(ωt/2)

  1. Step 1:
    U(t)=e−iHt/ℏ=e−iωt2X=e−iαX,α=ωt2U(t)=e^{-iHt/\hbar}=e^{-i\frac{\omega t}{2}X}=e^{-i\alpha X},\qquad \alpha=\frac{\omega t}{2}
    Time-independent solution with H=ℏω2XH=\tfrac{\hbar\omega}{2}X; the ℏ\hbar cancels.
  2. Step 2:
    X2=(0110)(0110)=(1001)=I ⇒ X2m=I, X2m+1=XX^2=\begin{pmatrix}0&1\\1&0\end{pmatrix}\begin{pmatrix}0&1\\1&0\end{pmatrix}=\begin{pmatrix}1&0\\0&1\end{pmatrix}=I\ \Rightarrow\ X^{2m}=I,\ X^{2m+1}=X
    Direct multiplication, then induction on mm.
  3. Step 3:
    e−iαX=∑n=0∞(−iα)nn!Xn=∑m=0∞(−iα)2m(2m)! I+∑m=0∞(−iα)2m+1(2m+1)! Xe^{-i\alpha X}=\sum_{n=0}^{\infty}\frac{(-i\alpha)^n}{n!}X^n=\sum_{m=0}^{\infty}\frac{(-i\alpha)^{2m}}{(2m)!}\,I+\sum_{m=0}^{\infty}\frac{(-i\alpha)^{2m+1}}{(2m+1)!}\,X
    Power series of the exponential (absolutely convergent), split into even and odd nn and use the powers of XX.
  4. Step 4:
    (−i)2m=(−1)m,(−i)2m+1=−i (−1)m(-i)^{2m}=(-1)^m,\qquad (-i)^{2m+1}=-i\,(-1)^m
    (−i)2=−1(-i)^2=-1.
  5. Step 5:
    e−iαX=(∑m(−1)mα2m(2m)!)I−i(∑m(−1)mα2m+1(2m+1)!)X=cos⁡α I−isin⁡α Xe^{-i\alpha X}=\Big(\sum_{m}\frac{(-1)^m\alpha^{2m}}{(2m)!}\Big)I-i\Big(\sum_m\frac{(-1)^m\alpha^{2m+1}}{(2m+1)!}\Big)X=\cos\alpha\,I-i\sin\alpha\,X
    Recognise the Taylor series of cos⁡\cos and sin⁡\sin.
  6. Step 6:
    U(t)=cos⁡ωt2 I−isin⁡ωt2 X=(cos⁡ωt2−isin⁡ωt2−isin⁡ωt2cos⁡ωt2)=RX(ωt)U(t)=\cos\tfrac{\omega t}{2}\,I-i\sin\tfrac{\omega t}{2}\,X=\begin{pmatrix}\cos\frac{\omega t}{2}&-i\sin\frac{\omega t}{2}\\-i\sin\frac{\omega t}{2}&\cos\frac{\omega t}{2}\end{pmatrix}=R_X(\omega t)
    Substitute α=ωt/2\alpha=\omega t/2 and compare with RX(θ)=e−iθX/2R_X(\theta)=e^{-i\theta X/2} at θ=ωt\theta=\omega t.
  7. Step 7:
    ∣ψ(t)⟩=U(t)∣0⟩=cos⁡ωt2∣0⟩−isin⁡ωt2∣1⟩\lvert\psi(t)\rangle=U(t)\lvert0\rangle=\cos\tfrac{\omega t}{2}\lvert0\rangle-i\sin\tfrac{\omega t}{2}\lvert1\rangle
    U(t)∣0⟩U(t)\lvert0\rangle is the first column of the matrix.
  8. Step 8:
    P1(t)=∣−isin⁡ωt2∣2=sin⁡2ωt2=1−cos⁡ωt2,P0(t)=cos⁡2ωt2P_1(t)=\Big\lvert-i\sin\tfrac{\omega t}{2}\Big\rvert^2=\sin^2\tfrac{\omega t}{2}=\frac{1-\cos\omega t}{2},\qquad P_0(t)=\cos^2\tfrac{\omega t}{2}
    Born rule; the population oscillates with angular frequency ω\omega and period 2π/ω2\pi/\omega.
  9. Step 9:
    ∣0⟩=12(∣+⟩+∣−⟩) ⇒ ∣ψ(t)⟩=12(e−iωt/2∣+⟩+eiωt/2∣−⟩)=cos⁡ωt2∣0⟩−isin⁡ωt2∣1⟩\lvert0\rangle=\tfrac{1}{\sqrt2}(\lvert+\rangle+\lvert-\rangle)\ \Rightarrow\ \lvert\psi(t)\rangle=\tfrac{1}{\sqrt2}\big(e^{-i\omega t/2}\lvert+\rangle+e^{i\omega t/2}\lvert-\rangle\big)=\cos\tfrac{\omega t}{2}\lvert0\rangle-i\sin\tfrac{\omega t}{2}\lvert1\rangle
    Cross-check with the eigenbasis method: X∣±⟩=±∣±⟩X\lvert\pm\rangle=\pm\lvert\pm\rangle, so E±=±ℏω/2E_\pm=\pm\hbar\omega/2; expand ∣±⟩\lvert\pm\rangle back in the computational basis and use Euler's formulas.

Result

P1(t)=sin⁡2ωt2\boxed{P_1(t)=\sin^2\tfrac{\omega t}{2}}
RX(π2)=12(1−i−i1),RX(π2)†RX(π2)=12(1ii1)(1−i−i1)=12(2002)=IR_X(\tfrac{\pi}{2})=\frac{1}{\sqrt2}\begin{pmatrix}1&-i\\-i&1\end{pmatrix},\qquad R_X(\tfrac{\pi}{2})^\dagger R_X(\tfrac{\pi}{2})=\frac12\begin{pmatrix}1&i\\i&1\end{pmatrix}\begin{pmatrix}1&-i\\-i&1\end{pmatrix}=\frac12\begin{pmatrix}2&0\\0&2\end{pmatrix}=I
R_X(π/2) = U(t) at ωt = π/22×2

Unitary verified: U†U=IU^\dagger U = I (max deviation 2.2e-16)

Circuit · 1 qubit

Quantum circuit on 1 qubitRX(π/2) on q0q0RX(π/2)

A single RX(π/2)R_X(\pi/2) is the evolution under H=ω2XH=\tfrac\omega2X for ωt=π/2\omega t=\pi/2 (a "π/2\pi/2 pulse"). The statevector is exactly (12, −i2)\big(\tfrac1{\sqrt2},\,-\tfrac{i}{\sqrt2}\big), so P1=sin⁡2π4=12P_1=\sin^2\tfrac\pi4=\tfrac12.

∣ψ⟩=12∣0⟩−12 i∣1⟩\lvert\psi\rangle = \tfrac{1}{\sqrt{2}}\lvert 0\rangle - \tfrac{1}{\sqrt{2}}\,i\lvert 1\rangle

In the SchrodingerPlayer the Bloch vector starts at the north pole and precesses about the xx axis at angular frequency ω\omega: at ωt=π/2\omega t=\pi/2 it reaches −y^-\hat y (the state ∣−i⟩\lvert{-i}\rangle), at ωt=π\omega t=\pi the south pole (−i∣1⟩-i\lvert1\rangle, a complete population inversion up to global phase), and at ωt=2π\omega t=2\pi it is back at ∣0⟩\lvert0\rangle — with the global phase −1-1, since RX(2π)=−IR_X(2\pi)=-I. Change the axis to (0,0,1)(0,0,1) and the vector stays at the north pole: ∣0⟩\lvert0\rangle is an eigenstate of ZZ, which leads to the last section. The snippet below computes e−iHte^{-iHt} with the engine's spectral exponential and confirms U(t)=RX(ωt)U(t)=R_X(\omega t) and P1=sin⁡2(ωt/2)P_1=\sin^2(\omega t/2).

TypeScriptrunnable
import { expmIHermitian, matrix, PAULI_X, rxMatrix } from '@/lib/quantum-simulator';

// Rabi problem with hbar = 1: H = (omega / 2) X and |psi(0)> = |0>.
const omega = 2;
const H = matrix.scale(PAULI_X, omega / 2);
for (const wt of [Math.PI / 3, Math.PI / 2, Math.PI]) {
  const t = wt / omega;
  const U = expmIHermitian(H, t); // U(t) = exp(-iHt), built from the spectral decomposition of H
  const a1 = matrix.get(U, 1, 0); // <1|U(t)|0>: second entry of the first column
  const p1 = a1.re * a1.re + a1.im * a1.im;
  const sameAsRx = matrix.maxAbsDiff(U, rxMatrix(omega * t)) < 1e-12;
  console.log('wt=' + wt.toFixed(4) + ' P1=' + p1.toFixed(4) + ' sin^2(wt/2)=' + (Math.sin(wt / 2) ** 2).toFixed(4)
    + ' U==RX(wt): ' + sameAsRx + ' unitary: ' + matrix.isUnitary(U));
}

Expected output

wt=1.0472 P1=0.2500 sin^2(wt/2)=0.2500 U==RX(wt): true unitary: true
wt=1.5708 P1=0.5000 sin^2(wt/2)=0.5000 U==RX(wt): true unitary: true
wt=3.1416 P1=1.0000 sin^2(wt/2)=1.0000 U==RX(wt): true unitary: true

Quick check

A qubit starts in ∣0⟩\lvert0\rangle and evolves under H=ℏω2XH=\tfrac{\hbar\omega}{2}X. What is P1P_1 when ωt=π/3\omega t=\pi/3?

tolerance ±0.000001

Stationary states#

Proposition

Energy eigenstates only acquire a global phase

If H∣E⟩=E∣E⟩H\lvert E\rangle=E\lvert E\rangle, then ∣ψ(t)⟩=e−iHt/ℏ∣E⟩=e−iEt/ℏ∣E⟩\lvert\psi(t)\rangle=e^{-iHt/\hbar}\lvert E\rangle=e^{-iEt/\hbar}\lvert E\rangle. Consequently every outcome probability and every expectation value ⟨ψ(t)∣A∣ψ(t)⟩\langle\psi(t)\vert A\vert\psi(t)\rangle is independent of tt: energy eigenstates are stationary.

Proof (3 steps)
  1. Step 1:
    Hn∣E⟩=En∣E⟩(n≥0)H^n\lvert E\rangle=E^n\lvert E\rangle\quad(n\ge0)
    Induction: Hn+1∣E⟩=H(En∣E⟩)=EnH∣E⟩=En+1∣E⟩H^{n+1}\lvert E\rangle=H(E^n\lvert E\rangle)=E^nH\lvert E\rangle=E^{n+1}\lvert E\rangle.
  2. Step 2:
    e−iHt/ℏ∣E⟩=∑n(−it/ℏ)nn!Hn∣E⟩=∑n(−iEt/ℏ)nn!∣E⟩=e−iEt/ℏ∣E⟩e^{-iHt/\hbar}\lvert E\rangle=\sum_n\frac{(-it/\hbar)^n}{n!}H^n\lvert E\rangle=\sum_n\frac{(-iEt/\hbar)^n}{n!}\lvert E\rangle=e^{-iEt/\hbar}\lvert E\rangle
    Apply the power series term by term and resum the scalar exponential.
  3. Step 3:
    ⟨ψ(t)∣A∣ψ(t)⟩=eiEt/ℏe−iEt/ℏ⟨E∣A∣E⟩=⟨E∣A∣E⟩\langle\psi(t)\vert A\vert\psi(t)\rangle=e^{iEt/\hbar}e^{-iEt/\hbar}\langle E\vert A\vert E\rangle=\langle E\vert A\vert E\rangle
    The time-dependent factor is a global phase; by the corollary of the state-space lesson it cancels in every expectation value (and likewise in every probability ∥Pψ(t)∥2\lVert P\psi(t)\rVert^2).

∎

Derivation

Superpositions of energies: the relative phase evolves

  1. Step 1:
    H=ω2Z,∣ψ(0)⟩=∣+⟩ ⇒ ∣ψ(t)⟩=12(e−iωt/2∣0⟩+eiωt/2∣1⟩)H=\tfrac{\omega}{2}Z,\quad \lvert\psi(0)\rangle=\lvert+\rangle\ \Rightarrow\ \lvert\psi(t)\rangle=\tfrac{1}{\sqrt2}\big(e^{-i\omega t/2}\lvert0\rangle+e^{i\omega t/2}\lvert1\rangle\big)
    ℏ=1\hbar=1; ∣0⟩\lvert0\rangle and ∣1⟩\lvert1\rangle are eigenstates with E0=ω/2E_0=\omega/2, E1=−ω/2E_1=-\omega/2, each acquiring its own phase.
  2. Step 2:
    =e−iωt/2 12(∣0⟩+eiωt∣1⟩)=e^{-i\omega t/2}\,\tfrac{1}{\sqrt2}\big(\lvert0\rangle+e^{i\omega t}\lvert1\rangle\big)
    Factor out a global phase: what remains is a relative phase φ=ωt=(E0−E1)t\varphi=\omega t=(E_0-E_1)t, the Bohr frequency times tt.
  3. Step 3:
    ⟨X⟩t=cos⁡ωt,PH then 0(t)=cos⁡2ωt2\langle X\rangle_t=\cos\omega t,\qquad P_{H\text{ then }0}(t)=\cos^2\tfrac{\omega t}{2}
    ⟨X⟩=2Re⁡(c0‾c1)=Re⁡eiωt\langle X\rangle=2\operatorname{Re}(\overline{c_0}c_1)=\operatorname{Re}e^{i\omega t}; reading out with HH gives cos⁡2(φ/2)\cos^2(\varphi/2) from the state-space lesson. The ZZ probabilities stay 12,12\tfrac12,\tfrac12 for all tt.

Result

only energy differences Ej−Ek are observable\boxed{\text{only energy differences }E_j-E_k\text{ are observable}}

Quick check

Under H=ω2ZH=\tfrac\omega2Z, which initial state gives time-independent measurement statistics in every basis?

Select one answer

References

  1. M. A. Nielsen and I. L. Chuang, Quantum Computation and Quantum Information, 10th anniversary ed., Cambridge University Press (2010), §2.2.2 (evolution, Postulate 2 and the Schrödinger equation) and §4.2 (single-qubit rotations and Exercise 4.2).
  2. J. J. Sakurai and J. Napolitano, Modern Quantum Mechanics, 3rd ed., Cambridge University Press (2020), §2.1 (the time-evolution operator, the Schrödinger equation, energy eigenkets and the time dependence of expectation values).
  3. P. Krantz, M. Kjaergaard, F. Yan, T. P. Orlando, S. Gustavsson and W. D. Oliver, A quantum engineer's guide to superconducting qubits, Applied Physics Reviews 6, 021318 (2019), §IV (qubit control and Rabi oscillations).doi:10.1063/1.5089550

Hands-on labs

  • Rabi evolution as a rotation

    1 step · auto-graded

    Solve the Schrödinger equation for H=ω2XH=\tfrac{\omega}{2}X (with ℏ=1\hbar=1): U(t)=e−iHt=e−i(ωt/2)X=RX(ωt)U(t)=e^{-iHt}=e^{-i(\omega t/2)X}=R_X(\omega t). Set the rotation angle for ωt=π/2\omega t=\pi/2 and reproduce the exact state, phases included.

    Open lab