A Qubit as a Two-Dimensional Subspace

A qubit has Hilbert space $\mathbb C^2$, but identifying two levels inside a larger system does not automatically produce a closed qubit model. An isometric encoding specifies the states and observables; a reducing-subspace condition specifies whether the dynamics remains inside that encoding. When it does not, leakage must be estimated separately from errors within the two-level Hamiltonian.

Helpful background. Spin-1/2 Hilbert Space supplies the elementary two-component model, and Bloch Sphere introduces its state coordinates. Finite-Dimensional Systems supplies the perturbation estimate used below.

An isometry identifies the encoded Hilbert space

Choose orthonormal physical vectors $u_0,u_1\in\mathcal H$ and define

$$V:\mathbb C^2\to\mathcal H,\qquad V(a,b)=a u_0+b u_1.$$

Then $V^*V=I_2$, while $P=VV^*$ is the orthogonal projection onto the encoded subspace. A qubit density operator $\rho$ becomes $V\rho V^*$ on $\mathcal H$.

A bounded physical observable $A$ has compressed qubit observable $V^*AV$, since

$$\operatorname{Tr}(V\rho V^*A) =\operatorname{Tr}(\rho V^*AV).$$

Compression preserves adjoints and positivity, but not arbitrary products. In fact,

$$V^*ABV-(V^*AV)(V^*BV) =V^*A(I-P)BV.$$

The missing term records excursions through excluded levels. Consequently one cannot compress each factor of a physical calculation independently and assume the original algebraic relations survive.

Exact two-level dynamics requires a reducing subspace

First take a bounded self-adjoint physical Hamiltonian $H$. The candidate qubit Hamiltonian is $h=V^*HV$. It generates the exact encoded evolution for every encoded initial vector precisely when the subspace is invariant under the physical unitary group.

For a self-adjoint Hamiltonian this is equivalent to reduction,

$$[H,P]=0.$$

Under this condition, $HV=Vh$ and the power series or functional calculus gives

$$e^{-itH/\hbar}V=Ve^{-ith/\hbar}.$$

For an unbounded self-adjoint $H$, the appropriate statement uses reduction by its spectral projections, equivalently commutation of $P$ with its unitary group. A formal matrix of selected entries does not establish domain invariance. This page’s quantitative estimate below assumes bounded $H$ to keep that issue separate.

Choosing two energy eigenvectors of a fixed Hamiltonian gives a reducing subspace. Adding a drive may break the reduction by coupling them to other levels. Exact closure must be checked for the Hamiltonian actually used during the operation.

A bounded off-diagonal block controls leakage

Decompose the physical Hilbert space into $P\mathcal H\oplus(I-P)\mathcal H$ and write

$$H=H_{\rm d}+C,\qquad H_{\rm d}=PHP+(I-P)H(I-P).$$

The coupling is $C=PH(I-P)+(I-P)HP$. If $B=(I-P)HP$, its block structure gives $\|C\|=\|B\|$.

The Duhamel bound implies

$$\|e^{-itH/\hbar}V-Ve^{-ith/\hbar}\| \leq\frac{|t|}{\hbar}\|B\|.$$

The diagonal Hamiltonian has no leakage, so every normalized encoded state satisfies

$$\|(I-P)e^{-itH/\hbar}V\psi\|^2 \leq\min\!\left(1,\frac{t^2\|B\|^2}{\hbar^2}\right).$$

This bound is independent of a spectral gap and is therefore useful for short-time control. It can be crude for detuned couplings, where oscillatory cancellation improves the estimate. A gap-dependent effective Hamiltonian needs a separate perturbative argument.

A three-level model exposes the missing transition

Let $u_0,u_1,u_2$ be orthonormal and take real $g$ with

$$H=g\bigl(|u_1\rangle\langle u_2| +|u_2\rangle\langle u_1|\bigr).$$

Encode the qubit in $u_0,u_1$. Its compressed Hamiltonian is zero, yet an initial $u_1$ evolves into

$$\cos(gt/\hbar)u_1-i\sin(gt/\hbar)u_2.$$

The exact leakage probability is $\sin^2(gt/\hbar)$, consistent with the short-time bound. A vanishing compressed Hamiltonian therefore does not imply stationary physical evolution.

Projecting the evolved state back into the encoding gives a subnormalized state whose trace is the survival probability. Renormalizing it describes a conditional preparation after a successful projection. That conditioning is generally nonlinear in the initial density operator and must not be confused with an unconditional trace-preserving qubit channel. State-Update Maps distinguishes these operations, including the special case of a state-independent success probability.

Exercise: compression can destroy a square identity

For the three-level example, compare $V^*H^2V$ with $(V^*HV)^2$. What physical quantity does the discrepancy detect?

Solution

The compressed Hamiltonian vanishes, so its square is zero. But $H^2=g^2(|u_1\rangle\langle u_1|+|u_2\rangle\langle u_2|)$, giving $V^*H^2V=g^2|1\rangle\langle1|$. The discrepancy is the squared coupling through the excluded level. It determines the initial quadratic leakage from $u_1$ and shows why matching selected matrix elements of $H$ does not control all dynamical predictions.

References

• Hall, Brian C., Quantum Theory for Mathematicians, Graduate Texts in Mathematics 267, Springer, 2013, doi:10.1007/978-1-4614-7116-5 — invariant subspaces and finite-dimensional quantum models.
• Teschl, Gerald, Mathematical Methods in Quantum Mechanics, 2nd ed., Graduate Studies in Mathematics 157, American Mathematical Society, 2014, author’s text — spectral reduction and perturbation estimates for evolution.