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Write $c=⟨e_0|e_1⟩=v{e}^{i\alpha }$. The Cauchy-Schwarz inequality gives $0\le v\le 1$ because the two environment states are normalized. Immediately after the interaction, the joint quantum state isApplying the last Hadamard gate givesSince the environment is unobserved, the Born rule uses the squared norm of the environment-valued coefficient of $|0⟩$:Thus the Hadamard interference with environment overlap formula isThe plus sign follows from the explicitly chosen phase gate and the order $⟨e_0|e_1⟩$ of the inner product. Equivalently, before the last Hadamard gate the partial trace over the environment givesThe complex environment overlap in qubit dephasing both reduces the quantum coherence by $v$ and adds a relative quantum phase $\alpha$. Orthogonal environment states have $v=0$ and give $P_0=1/2$ for every $\phi$. If $v=1$, the quantum interference fringe has full contrast and is only shifted; the qubit has not become mixed. For $v=0$, $\alpha$ is immaterial.

Moving the interaction to between the first Hadamard gate and the phase gate givesthe identical joint quantum state. Therefore the expression is unchanged. This is the commutation of computational-basis dephasing and phase gates; it uses the specified population-preserving interaction with the same environment states, rather than an arbitrary model of quantum decoherence.

A good atomic qubit needs two reproducible, orthogonal internal states whose populations and relative quantum phase survive for many gate times. Both should be long lived, with weak spontaneous emission and little decoherence from fluctuating fields or atomic motion. Their transition must nevertheless permit fast, selective, coherent control with little leakage to other states. Reliable initialization and state-selective quantum measurement are also required; optical pumping and fluorescence readout can supply these functions.

For rubidium-87, use the two ground state hyperfine levels in the $5{}^2{S}_{1/2}$ electronic configuration. The electronic angular momentum is $J=1/2$, and addition of angular momentum to the given nuclear spin $I=3/2$ gives $F=|I-J|,\dots ,I+J=1,2$. A suitable Rubidium-87 clock qubit isThese are coupled electron-nucleus angular momentum eigenstates, rather than two bare nuclear-spin projections. Their hyperfine structure supplies a resolvable splitting ${\nu }_{01}\simeq 6.835\mathrm{GHz}$. Both belong to the electronic ground state, avoiding a short optical excited-state lifetime. Since the weak-field Zeeman shift is proportional to $g_F{m}_{F}B$, both have zero linear shift at $B=0$: their rubidium-87 clock transition suppresses first-order magnetic-field dephasing near zero field. A quadratic Zeeman shift of a clock transition and differential optical shifts remain. A small stable bias field can define the quantization axis and help resolve unwanted transitions, at the cost of some residual magnetic sensitivity.^[ref]

Drive the magnetic dipole transition with a resonant microwave field of suitable polarization, here $\Delta m_F=0$, or use two phase-coherent optical fields for a far-detuned stimulated Raman transition with their frequency difference resonant with ${\nu }_{01}$. The optical excited state is then only virtually populated, reducing spontaneous emission. In the rotating frame and rotating-wave approximation, define the drive phase $\varphi$ by the effective HamiltonianHere $X,Y,Z$ are the Pauli matrices on the chosen qubit, and $\Omega$ is the Rabi angular frequency. Pulse duration controls the rotation gate angle, while phase selects the axis in the equatorial plane of the Bloch sphere. A $\pi$ pulse exchanges the populations; a $\pi /2$ pulse makes an equal quantum superposition from either basis state. Controlled Rabi detuning, a differential energy shift, or compositions of these pulses give $z$ rotations. For example, $R_z(\alpha )R_y(\beta )R_z(\gamma )$ realizes any one-qubit unitary operator up to a global phase. Thus the atomic encoding admits arbitrary single-qubit quantum gates.

Use the ordered computational basis $(|00⟩,|01⟩,|10⟩,|11⟩)$, with qubit 1 written first. The half-time gate requires a branch convention. Even stationary Hamiltonian evolution with $W(2\tau )=\mathrm{SWAP}$ does not uniquely specify $W(\tau )$: different accumulated eigenstate phases can agree at $2\tau$ and differ at $\tau$. In particular, both $S$ and $S^{†}$ below square to SWAP gate and leave $|00⟩,|11⟩$ unchanged. Adopt the conventional continuous isotropic exchange interaction branch, with a relative singlet phase increasing from $0$ to $\pi$ during the full pulse. This is the exchange square-root branch convention.

The symmetric and antisymmetric eigenstateshave SWAP gate eigenvalues $+1,-1$. Let $P_{-}=|{\psi }_{-}⟩⟨{\psi }_{-}|$ and $P_{+}=I_4-P_{-}$ be their antisymmetric and symmetric orthogonal projectors. The adopted exchange evolution isFor effective spin-one-half qubits, $H_{\mathrm{ex}}=\pi \hslash (X_1{X}_2+Y_1{Y}_2+Z_1{Z}_2-I_4)/(8\tau )$, an isotropic exchange interaction plus an energy offset that fixes the common phase. At half time choose $+1$ on the symmetric sector and $+i$ on the antisymmetric sector, the conventional principal square-root value of $-1$. Hence the square root of the SWAP gate isThus $S|00⟩=|00⟩$, $S|11⟩=|11⟩$, $S|01⟩=a|01⟩+b|10⟩$ and $S|10⟩=b|01⟩+a|10⟩$, where $a=(1+i)/2$ and $b=(1-i)/2$. The identities $a^2+b^2=0$ and $2ab=1$ verify $S^2=\mathrm{SWAP}$, and the projector form proves unitarity. The endpoint truth table alone would allow other matrix square roots; the stated exchange branch supplies the additional physical assumption.

For the network, fix the Pauli Z gate, Hadamard gate and S gate conventionsThe letter $P$ distinguishes this local S gate from the two-qubit exchange gate $S$. In particular $P^2=Z$; replacing $P$ by $P^{†}$ changes the network. Subscripts specify the acted-on qubit: $Z_1=Z\otimes I_2$, $P_2=I_2\otimes P$ and $H_2=I_2\otimes H$. Operators act rightmost first, so the printed quantum circuit is

First multiply the central two exchange pulses. Since $a^2-b^2=i$, their off-diagonal terms cancel:This reduces the entire network to $U=H_{2}D{H}_2$. Its action on all four computational basis states, retaining every relative quantum phase, isEquivalently,It flips the second qubit when the first is zero and contributes $-i$ when the first is one. That conditional phase cannot be discarded as a global phase.

To obtain the standard CNOT gate controlled by qubit 1, append an $X$ gate to qubit 2 and $P$ to qubit 1. They commute, andFor control zero, $X_2$ undoes the original flip. For control one, it supplies the desired flip and $P_1$ cancels the phase because $i(-i)=1$. This phase-corrected exchange network for a CNOT gate realizes the standard gate exactly with the declared conventions.

Finally, $S|01⟩=(e^{i\pi /4}|01⟩+e^{-i\pi /4}|10⟩)/\sqrt{2}$ has equal Schmidt coefficients and is maximally entangled: a half exchange creates quantum entanglement although a full SWAP gate does not. More constructively, the network above produces a CNOT gate from two half exchanges and local gates. Since CNOT gates and arbitrary single-qubit unitary operators form a universal quantum gate set, the square root of the SWAP gate and arbitrary single-qubit gates are universal. The alternative inverse exchange branch remains entangling and universal, but its phases and local correction must be recomputed consistently.

Use the Hadamard basis states $|\pm ⟩=(|0⟩\pm |1⟩)/\sqrt{2}$. The Hadamard gate satisfies $H|0⟩=|+⟩$, $H|1⟩=|-⟩$, and $H^2=I$. After the first three Hadamard gates, the full quantum state isThe Pauli X gate obeys $X|-⟩=-|-⟩$. Consequently the Boolean quantum oracle produces phase kickback:Factoring out the ancilla qubit, the four states immediately after the function oracle areThe final Hadamard gates send $|+⟩$ to $|0⟩$ and $|-⟩$ to $|1⟩$. In particular, the printed quantum circuit includes a final Hadamard gate on the ancilla qubit too. Its full outputs, ordered as $x_1,x_2,y$, areThus one oracle call identifies $a$ with certainty: a computational-basis measurement of the first two qubits returns $a_1{a}_2$. The ancilla qubit returns to $|1⟩$ and carries no additional information about $a$. The Bernstein-Vazirani algorithm turns the pattern of relative quantum phases into a definite computational-basis state.

A full fringe scan is unnecessary. Use the same components to make three fixed projective measurements, on separate batches of identically prepared photons. Direct path detection after the polarizing beam splitter measures the $H/V$ populations. Recombine after matching the arm photon polarizations, once with $\theta =0$ and once with $\theta =\pi /2$, to measure the two quadratures of the quantum coherence.

In the $H/V$ basis, write the density matrix in terms of its Bloch vector as $\rho =(I+r_{x}X+r_{y}Y+r_{z}Z)/2$. The previous formula becomesThus three measured relative frequencies giveand determineThe two interferometric settings analyze the diagonal linear polarizations $(|H⟩\pm |V⟩)/\sqrt{2}$ and the circular polarizations $(|H⟩\pm i|V⟩)/\sqrt{2}$, with detector labels fixed by the calibrated quantum phase convention. This is informationally complete three-observable qubit tomography: direct measurements in three bases replace fitting an entire phase-dependent fringe. It also reconstructs a mixed input, for which $|r|\le 1$, although the source here is assumed pure.

Simply measuring transmission through one fixed polarizing beam splitter would determine only $r_z$. Rotating a linear analyzer through different angles can additionally determine $r_x$, but cannot distinguish the two signs of $r_y$ without a relative-phase measurement. For arbitrary, possibly elliptical photon polarization, all three settings above are needed; no extra analyzer element beyond the described path-phase control and photon polarization rotator is required.

Write the unknown pure photon polarization qubit asThe polarizing beam splitter sends the $H$ component down the lower path and the $V$ component up the upper path. After rotating the upper photon polarization to $H$, both paths have the same internal quantum state and can interfere at BS2. Absorb fixed apparatus phases into a calibrated zero for the tunable quantum phase. With the preceding reflection convention, the quantum state before BS2 is $(\alpha |\ell ⟩+i\beta e^{i\theta }|u⟩)|H⟩$, so the two output amplitudes areConsequentlyRepeated counts while scanning $\theta$ determine the complex quantum coherence $\gamma$. The interference visibility is $2|\gamma |$, and the fringe position determines the relative quantum phase $\arg \beta -\arg \alpha$ whenever both components are nonzero.

A fringe alone is not enough to determine an arbitrary photon polarization quantum state: it fixes $|\alpha \beta |$ but not which of $|\alpha {|}^2$ and $|\beta {|}^2$ is larger. Indeed, $|H⟩$ and $|V⟩$ both give the flat fringe $P_1=1/2$. For a general pure quantum state, the two possible population differences areResolve this ambiguity by temporarily detecting the two paths before BS2. Their relative counts give $h=|\alpha {|}^2$ and $1-h=|\beta {|}^2$ using the same polarizing beam splitter and detectors. Taking $\alpha =\sqrt{h}$ as a global-phase choice, the reconstructed quantum state for $h>0$ isIf $h=0$, it is $|V⟩$ up to a global phase. This completes interferometric polarization tomography. The counts are obtained from many identically prepared photons; they do not identify an arbitrary quantum state from one photon.

Use the symmetric lossless beam splitter convention with transmission amplitude $t=1/\sqrt{2}$ and reflection amplitude $r=i/\sqrt{2}$. The two mirrors contribute the same fixed quantum phase, which can be suppressed. At the rightward port $D_1$, the upper-arm route has one reflection at BS1 and one transmission at BS2, while the lower-arm route has one transmission at BS1 and one reflection at BS2. Their probability amplitudes therefore add asBy the Born rule,For the upward port, the two amplitudes are $r^2{e}^{i\theta }$ and $t^2$, giving $a_0=(1-e^{i\theta })/2$ and $P(D_0)={\sin }^2(\theta /2)$. They sum to one, as required for an ideal lossless Mach-Zehnder interferometer. Thus $D_1$ is bright at zero relative quantum phase and dark at quantum phase $\pi$. A fixed unequal reflection quantum phase or a swapped port labeling would shift the fringe; the stated convention fixes the port and quantum phase calibration used below.

Without the upper-arm rotation, the photon polarization still labels the path. Before BS2 the quantum state isAfter the balanced beam splitter, the polarization-valued output amplitudes areThe detectors do not resolve photon polarization. Since $⟨H|V⟩=0$, the cross terms vanish in the Born rule:These probabilities are independent of the tunable quantum phase and the input photon polarization, so there is no usable wave interference fringe. Equivalently, taking the partial trace over photon polarization removes the quantum coherence between the two path quantum states. The which-path information is retained in orthogonal photon polarizations even without anybody measuring it. Rotating the two occupied photon polarization quantum states to the same quantum state removes that distinction and allows the cross term in part (ii).

A central application is quantum order finding, which supplies the quantum step in the Shor algorithm for factoring an integer $M$. Choose an integer $a$ coprime to $M$, and let $r$ be its multiplicative order modulo $M$. On valid residue quantum states define the reversible modular-multiplication unitary operator $U_a|x⟩=|ax\mathrm{mod}M⟩$, extending it reversibly to any unused computational basis quantum states.

The orbit $|1⟩,|a⟩,\dots ,|a^{r-1}⟩$ has the orthonormal eigenstatesfor $s=0,\dots ,r-1$. One need not know $r$ to prepare an eigenstate: the easy input $|1⟩=r^{-1/2}{\sum }_s|u_s⟩$ makes quantum phase estimation sample one of these eigenphases. Its control register estimates $s/r$. With $N=2^t$ chosen at least of order $M^2$, the nearest quantum phase label has rational approximation error small enough for continued-fraction recovery in quantum order finding. For instance, $N>2M^2$ gives $|m/N-s/r|<1/(2r^2)$ on the nearest-label event. If $s$ and $r$ are coprime, the recovered denominator is $r$; otherwise one obtains a divisor and can repeat, combine candidates and verify them by modular exponentiation.

When the verified order is even and $a^{r/2}\not\equiv -1(\mathrm{mod}M)$, the greatest common divisorsgive nontrivial factors: their product is divisible by $M$, while neither factor inside the gcd is zero modulo $M$. Unsuccessful choices of $a$ or quantum phase samples are handled by repetition and classical verification. Controlled powers of $U_a$ are implemented by efficient reversible modular arithmetic, rather than by naively applying $U_a$ exponentially many times.

Quantum phase estimation turns modular-multiplication eigenphases into an efficiently usable period, enabling quantum factoring in time polynomial in the bit length of the integer. Factoring is important to public-key cryptography, and this is a major example of a quantum algorithm improving on known classical methods. Another use is extracting energy eigenvalues by applying quantum phase estimation to a controlled time-evolution operator, with care to avoid energy aliasing and to prepare an input having overlap with the desired eigenstate.

Write $\varphi =2\pi n/N+ϵ$, with $|ϵ|\le \delta \ll 2\pi /N$. Decode the measured bits as above. The probability amplitude of label $m$ isUsing the finite geometric series,Interpret removable singularities by continuity. In particular, near-grid quantum phase estimation returns the nearby exact label $n\mathrm{mod}N$ with probabilityThis is near one under the stated small-offset assumption, but not exactly one when $ϵ\ne 0$.

There is also a rigorous error bound. Taking the squared modulus of the sum giveswhere $1-\cos x\le x^2/2$ and the elementary sums of $k$ and $k^2$ evaluate the last expression. HenceOn that event the circular quantum phase error is at most $\delta$. For other labels, the exact probabilities areTheir total is $1-P(n)=O(N^2{\delta }^2)$, describing the leakage from the perfect solution. The physical output bit string remains reversed in the printed circuit; this changes its decoding, not these probabilities.

Put $N=2^t$ and write $U|u⟩=e^{i\varphi }|u⟩$, with $|u⟩$ normalized. The Hadamard gates prepare the control register in $N^{-1/2}{\sum }_{k=0}^{N-1}|k⟩$. Controlled powers of the unitary operator then give quantum phase kickback:Thus the register quantum state is $N^{-1/2}{\sum }_k{e}^{ik\varphi }|k⟩$, while the target remains $|u⟩$. Define the quantum Fourier transform by $F_N|m⟩=N^{-1/2}{\sum }_k{e}^{2\pi ikm/N}|k⟩$. Its inverse givesWhen $\varphi =2\pi n/N$, the inner finite geometric series equals $N$ if $m\equiv n(\mathrm{mod}N)$ and zero otherwise. Therefore exact quantum phase estimation givesThe quantum phase, rather than an unrestricted integer $n$, is recovered modulo $2\pi$.

The printed circuit has no bit-reversal swaps. Its top control uses $U^{2^{t-1}}$, and its measured top wire gives the least significant output bit. Write $n\mathrm{mod}N={\sum }_{j=0}^{t-1}{2}^j{b}_j$. Before the top wire's final Hadamard gate, its relative quantum phase is $2^{t-1}\varphi =\pi n$, so that gate produces $b_0$. The controlled inverse quantum phase gates then subtract its contribution from lower wires. More generally, before the final Hadamard gate on wire $j$, the relative quantum phase isso that wire gives $b_j$ deterministically. For the three-wire drawing, the correct interpretation isEquivalently, its Fourier-inversion circuit is $R{F}_N^{†}$, where $R$ reverses the register bits. Relabeling the measured bits or appending swaps gives the usual integer output. This bit convention matters, for example, when quantum phase $\pi$ produces top-to-bottom bits $001$, representing the label four rather than one.

Number the physical wires $1,\dots ,5$ from top to bottom and write a computational basis ket in that order. The first four are ancilla qubits and the fifth carries the input. It is enough to follow the basis input $|b⟩$, $b\in \{0,1\}$, because the encoding quantum circuit acts linearly.

The first fan-out consists of four controlled-NOT gates from wire 5 to wires 1–4. It sends $|0000b⟩$ to $|bbbbb⟩$. The three Hadamard gates then giveThe two Pauli X gates on wires 2 and 3, followed by a Controlled-Z gate and the same two X gates, restore the bits but contribute $(-1{)}^{(1-c)(1-d)}$: the sign is negative exactly when the original bits on both wires were zero. Finally, the right-hand controlled-NOT gates are $3\to 4$, $2\to 5$, $1\to 4$ and $1\to 5$. They leave the first three bits unchanged and turn the last two into $b\oplus a\oplus d$ and $b\oplus a\oplus c$. ThusExplicitly, the logical basis quantum states produced by the printed gates areEach has eight orthogonal terms of equal magnitude, and their supports are disjoint, so they are normalized and mutually orthogonal. The arbitrary input is therefore encoded asThis is a realization of the five-qubit error correcting code. The input amplitudes and their relative quantum phase are preserved in the logical subspace; the encoding is an entangled-state linear isometry, not five independent copies of the unknown input. The explicit signs and bit ordering above are fixed by this circuit, rather than by a convention for a different encoder.

At fixed particle number $N_p=\int |\psi {|}^{2}d\mathbf{x}$, an equilibrium wavefunction is a constrained critical point of the energy functional. Introduce the chemical potential $\mu$ as the Lagrange multiplier for $N_p$, and impose $\delta (E-\mu N_p)=0$. The resulting stationary cubic-quintic Gross–Pitaevskii equation isThe time-dependent equilibrium has $\psi (\mathbf{x},t)=e^{-i\mu t/\hslash }{\psi }_0(\mathbf{x})$, so its number density is stationary even though its quantum phase changes uniformly. The specified particle number fixes $\mu$ together with the boundary conditions. In an infinite uniform bulk, particle number is understood through a finite volume or a prescribed bulk density.

For the displayed scaling, assume $\mu >0$ and $n_0>0$. SetSubstituting in the stationary cubic-quintic Gross–Pitaevskii equation and dividing by $\mu \sqrt{n_0}$ givesThe uniform-state relation proves $\alpha +\beta =1$, so the real bulk amplitude is one; its constant quantum phase can be chosen to make $\tilde{\psi }\to 1$ in the zero-winding sector. The unit of distance $\ell$ is the kinetic-to-chemical-potential scale, reducing to the stated healing length convention for purely repulsive two-body coupling.

The positive-$\mu$ assumption is necessary for the printed sign convention. If $\mu <0$, the real length is $\hslash /\sqrt{2m|\mu |}$, and the dimensionless linear term is $-1$, with interaction coefficients divided by $|\mu |$; the displayed equation with linear term $+1$ does not follow. At $\mu =0$ this scaling is undefined. These are limitations of this particular nondimensionalization, not a claim that all cubic-quintic bulk states have positive chemical potential.

Define the physical velocity potential by $\psi =\sqrt{n}\exp (im\varphi /\hslash )$, so the superfluid velocity is $\mathbf{v}=\nabla \varphi$. The condensate number current isMultiplying the pumped Gross–Pitaevskii equation by ${\psi }^{\ast }$ and subtracting its complex conjugate gives the density equation. Equivalently, substituting the amplitude and phase and equating imaginary parts yieldsEquating real parts gives the other gain-loss Madelung equations relation,The last term is the quantum potential, retained wherever $n>0$. Taking a gradient gives the equivalent velocity equationin an irrotational region. In equilibrium $n_t=0$ and $\varphi (\mathbf{x},t)=\Phi (\mathbf{x})-\mu t/m$, soThis convention distinguishes the physical velocity potential from the dimensionless quantum phase $m\varphi /\hslash$; using the latter instead would change the divergence prefactor.

Here equilibrium means a time-independent number density and current, with a common temporal quantum phase. Write $\psi (\mathbf{x},t)=e^{-i\mu t/\hslash }\Psi (\mathbf{x})$ with real $\mu$, since an imaginary temporal frequency would change the density. The stationary pumped Gross–Pitaevskii equation isThe chemical potential is the common phase-rotation frequency times $\hslash$; this open system has no conserved particle-number constraint fixing it as in the conservative problem. Local gain and loss need not cancel separately in an inhomogeneous equilibrium, because a stationary condensate number current can transport particles between gain and loss regions.

Use the dimensionless quantum phase $S$, so that $\psi =R{e}^{iS}$ with $R\ge 0$. Work locally away from wavefunction nodes, where $R$ and $S$ are differentiable. Differentiating the wavefunction for substitution in the Time-dependent Schrödinger equation gives:Cancel $e^{iS}$ in the time-dependent equation and equate real and imaginary parts. The resulting Madelung equations areThe second equation is a Hamilton-Jacobi equation for the action $\hslash S$, with an additional quantum potential. To see the meaning of the first, multiply it by $2R$ and set $\rho =R^2$. It becomes the probability continuity equationThus the probability density is transported by the velocity field $\hslash \nabla S/m$. The Madelung equations are a local rewriting of the linear wave equation; their apparent nonlinearity comes from expressing a complex wavefunction in modulus and phase variables.

For the separated stationary state, restore its time factor $e^{-iEt/\hslash }$. Away from radial nodes, the quantum phase is $S=k\varphi -Et/\hslash$, up to a constant $0$ or $\pi$ where the real radial function has fixed sign. The guidance equation in plane polar coordinates therefore gives the Bohmian circulation of an angular-momentum eigenstateThe direction is $\hat{\mathbf{\varphi }}$ for $k>0$ and $-\hat{\mathbf{\varphi }}$ for $k<0$; the velocity is zero for $k=0$. Each admissible trajectory is a circle:The orbital angular momentum along the trajectory is $m{r}_0{v}_{\varphi }=\hslash k$. The origin or any zero-amplitude circle is excluded from this local formula. For the real degenerate superposition constructed above, the spatial quantum phase is constant on each nodal sector, so its Bohmian mechanics velocity is instead zero. Thus the circular motion is a conclusion about the separated angular-momentum eigenstate, not an arbitrary energy eigenstate.

The Born rule depends only on transition probabilitiesbetween rays. A physical transformation must preserve these probabilities. By Wigner theorem, such a ray transformation is induced by a unitary or antiunitary operator. For ordinary continuously connected transformations one chooses the unitary branch, so

The states $U(g_1)U(g_2)|\psi ⟩$ and $U(g_1{g}_2)|\psi ⟩$ represent the same ray. They may therefore differ by a state-independent quantum phase:Thus quantum transformations form a projective unitary representation.

If $g$ is a symmetry of the Hamiltonian operator $H$, it also obeysequivalently $[U(g),H]=0$. For a differentiable one-parameter subgroup, writewith a self-adjoint generator $Q$. Differentiating the symmetry equation at $s=0$ yieldsFor a time-independent $Q$, the Heisenberg picture equation isHence the generator is a conserved observable. Conversely, a self-adjoint conserved $Q$ commutes with $H$ and its unitary group generates symmetries. This is the conserved generator of a continuous quantum symmetry.

Forthe phase of the $n$th stationary state isReflection in the midpoint givessoThe minus sign is a global quantum phase and therefore does not change the physical state. The relative even--odd phases first acquire this reflection pattern at $T$. Applying it twice givesIf $f$ vanishes on the right half of the well, then $\psi (x,T)$ vanishes on the left half. The Born rule consequently gives zero probability of finding the particle in $0\le x\le a/2$ at time $T$.