Starting point: Probabilities, conditioning and entropy; complex amplitudes and vector notation are developed within the examples.
Keep in view: A two-alternative interference experiment with a controllable relative phase.
So far every uncertainty has been classical. Bob did not know which card or which bit Alice sent, and we described his ignorance with probabilities. Probabilities of alternatives that exclude one another simply add. On that basis we measured uncertainty by entropy and found how many bits a noisy channel can carry.
A simple experiment with light breaks the rule that probabilities of alternatives add. A photon can reach a detector by two routes, and opening the second route can stop it from arriving at all. If no probability can describe this, what can? We will need a quantity attached to each route that adds before any probability is formed: an amplitude. One experiment then forces amplitudes to be complex numbers. Comparing two states turns amplitudes into vectors and gives the inner product. Requiring that probabilities always total one restricts reversible operations to unitary matrices. Finally, an observation needs its own rule, because it gives one outcome and leaves a new state behind.
3.1 Amplitudes
Can a probability attached to each route describe a photon that can travel two routes? We test the idea on a two-mirror experiment and watch it fail. The failure points to amplitudes, which add before they are squared. A second experiment then asks for a preparation that real amplitudes cannot describe, and complex numbers supply it.
A half-silvered mirror, or beam splitter, lets half of the light through and reflects the other half. Send single photons at one. Put a detector in each outgoing path, and call the paths 0 and 1. Each detector fires on half of the photons. That much is just a fair coin.
Now let the two paths meet again at a second beam splitter, with the detectors behind it. What should happen? Reason with probabilities. After the first mirror the photon is in path 0 or path 1, each with probability \(1/2\). Whichever path it took, the second mirror sends it to either detector with probability \(1/2\). Adding the two exclusive routes predicts each detector fires with probability \(\tfrac12\cdot\tfrac12+\tfrac12\cdot\tfrac12=\tfrac12\).
That is not what happens. With the paths carefully aligned, every photon reaches detector 0. Detector 1 never fires. Yet if we block path 1, the photons that get through fire both detectors equally often. Opening a second route has removed arrivals at detector 1. No probabilities, being nonnegative, can cancel like that.
So we try a different quantity. Attach a number to each path, called an amplitude, which may be negative. A photon definitely in path 0 has amplitudes \((1,0)\). We model a beam splitter by the operation \(H\):
The minus sign belongs to one of the reflections. The amplitude for a detector to fire is the sum of the contributions from both paths. The probability is the square of that sum.
Check this against the experiment. After one \(H\), path 0 gives \((1/\sqrt2,1/\sqrt2)\): probabilities \(1/2\) and \(1/2\), as observed. After the second \(H\), the contributions to output 0 are \(1/2\) and \(1/2\), which add to 1. The contributions to output 1 are \(1/2\) and \(-1/2\), which cancel. The amplitudes are \((1,0)\), so detector 0 always fires. Squaring each route first would have given \((1/2,1/2)\) again. Adding contributions before squaring is called interference.
The rule that turns amplitudes into probabilities is a physical postulate, chosen because it matches experiments like this one. For real amplitudes \((a,b)\) the probabilities are \(a^2\) and \(b^2\), so we require \(a^2+b^2=1\). Negative amplitudes are allowed; negative probabilities are not.
A system with two configurations, described by a pair of amplitudes in this way, is called a qubit. The photon's two paths form one qubit, just as a card with two labels carried one bit. The configurations 0 and 1 could equally be two energy levels of an atom or two polarizations of light.
One more observation will matter later. If a device beside the paths records which path each photon took, the interference disappears. The detectors behind the second mirror then fire equally often. Accounting for this needs a description of the photon and the recording device together, which we do not yet have.
We now have two ways to read a qubit. Detectors placed directly after the first mirror read the path. A second beam splitter before the detectors applies \(H\) and then reads. Is there a preparation that gives fair results in both readings?
Try real amplitudes. A fair direct reading needs \(a^2=b^2=1/2\), so \(b=a\) or \(b=-a\). But \((1,1)/\sqrt2\) gives detector 0 with certainty after \(H\), and \((1,-1)/\sqrt2\) gives detector 1 with certainty. No real pair is fair in both readings.
Experiment says such preparations exist. Place a thin glass plate in path 1, so that its light is delayed by a quarter of a wavelength. The direct reading is still fair, and so is the reading through \(H\). Real amplitudes cannot describe this delay. We need numbers with more room in them.
A complex number \(z=u+iv\) has two real components \(u\) and \(v\), and the new number \(i\) satisfies \(i^2=-1\). Its conjugate is \(z^*=u-iv\). Multiplying gives \(zz^*=u^2+v^2\), a nonnegative number written \(|z|^2\), the squared magnitude. The probability rule becomes: an amplitude \(z\) gives probability \(|z|^2\). For a real amplitude this is the old square. The square itself would not do: \(((1+i)/2)^2=i/2\), which is not a probability, while \(|(1+i)/2|^2=1/2\).
A complex number is a point in a plane, so it has a direction. An angle \(\phi\) picks the point \((\cos\phi,\sin\phi)\) on the unit circle, and we write
\[e^{i\phi}=\cos\phi+i\sin\phi.\]
Angles are measured in radians, so a full turn is \(2\pi\). At \(\phi=0,\pi/2,\pi\) the number \(e^{i\phi}\) is \(1,i,-1\). Its squared magnitude is \(\cos^2\phi+\sin^2\phi=1\), so it is called a phase: multiplying by it turns an amplitude without changing its size.
The glass plate multiplies the path-1 amplitude by a phase. The quarter-wavelength plate gives \(e^{i\pi/2}=i\), turning \((1,1)/\sqrt2\) into \((1,i)/\sqrt2\). Now apply \(H\). The output-0 amplitude is \((1+i)/2\), with probability \(1/2\). The direct reading gives \(|1/\sqrt2|^2=|i/\sqrt2|^2=1/2\). Both readings are fair, as observed.
A general plate gives \((1,e^{i\phi})/\sqrt2\). After \(H\), the output-0 amplitude is \((1+e^{i\phi})/2\), and
The phase between the two amplitudes, the relative phase, changes no direct reading, yet it controls the reading through \(H\). What if both amplitudes are multiplied by the same phase? Every contribution to every output picks up the same factor, so no probability changes, now or after any later step. Such a global phase is unobservable, and two amplitude pairs differing only by one describe the same preparation.
Problems
Problem 3.1. Predict H applied to the amplitudes (1,−1)/√2.
Worked solution
This input is called minus and has amplitudes (1,−1)/√2. The output amplitudes are (0,1), so the result is one with certainty.
3.2 State Representations
We have several named preparations and two ways of reading them. For each new pair we have been recomputing amplitudes by hand. Is there one rule that predicts how any preparation responds to a test for any other? Writing amplitudes as vectors gives that rule, the inner product. Its size then tells us when two preparations can be told apart perfectly.
so that \(a\ket0+b\ket1\) is the column with entries \(a\) and \(b\). The angle bracket \(\ket{\cdot}\), called a ket, only labels a vector; it adds no physics. A general qubit state is \(\ket\psi=a\ket0+b\ket1\) with \(|a|^2+|b|^2=1\). Two states recur often enough to name:
Since \(H\ket+=\ket0\), the reading "apply \(H\), then see 0" is a test for plus. What is the amplitude for an arbitrary \(\ket\psi=a\ket0+b\ket1\) to pass it? The output-0 amplitude of \(H\) is \((a+b)/\sqrt2\). That is plus's amplitudes, \(1/\sqrt2\) and \(1/\sqrt2\), multiplied by \(\psi\)'s and summed.
Does the same pattern hold for a complex state? Take \(\ket\phi=(\ket0+i\ket1)/\sqrt2\). To test for it we turn it into \(\ket0\). A plate with phase \(-i\) on path 1 turns \(\phi\) into \(\ket+\), and \(H\) then gives \(\ket0\). Apply the same two steps to \(\ket\psi\): the plate gives \((a,-ib)\), and the output-0 amplitude of \(H\) is \((a-ib)/\sqrt2\). This is not \(\phi\)'s amplitudes times \(\psi\)'s. It is their conjugates, \(1/\sqrt2\) and \(-i/\sqrt2\), times \(\psi\)'s.
So we define, for \(\ket\phi=c\ket0+d\ket1\),
\[\braket\phi\psi=c^*a+d^*b.\]
This number is the inner product of \(\phi\) and \(\psi\). We can write it as a row times a column. The row of conjugated amplitudes, \(\bra\phi=(c^*,d^*)\), is called a bra. In both tests above, the amplitude for \(\psi\) to pass the test for \(\phi\) was \(\braket\phi\psi\), and the probability was \(|\braket\phi\psi|^2\). We adopt this as the rule for every test of a normalized state \(\phi\). Once we know which operations are allowed, we will see that it follows for every test built from them.
Why the conjugate? Without it, \(\phi\) compared with itself would give \(\tfrac12+\tfrac{i^2}2=0\), as if it had no length. With it, \(\braket\phi\phi=\tfrac12+\tfrac{(-i)i}2=1\). And the test for plus on \(\phi\) has amplitude \(\braket+\phi=(1+i)/2\) and probability \(1/2\), matching the fair reading through \(H\).
A set of unit vectors, at right angles to one another, from which every vector can be built is a basis. The pair \(\ket0,\ket1\) is the computational basis. The pair \(\ket+,\ket-\) is another basis, since \(\ket0=(\ket++\ket-)/\sqrt2\) and \(\ket1=(\ket+-\ket-)/\sqrt2\). Because \(H\ket+=\ket0\) and \(H\ket-=\ket1\), applying \(H\) and then reading the path tells plus from minus. That is a measurement in the plus/minus basis.
The inner product of a state with itself, \(\braket\psi\psi=|a|^2+|b|^2\), is the total probability of the two configurations. A state with \(\braket\psi\psi=1\) is normalized, and every state vector must be. When can a test tell two normalized states apart without error?
Suppose \(\braket\phi\psi=0\). Then \(\psi\) never passes the test for \(\phi\), while \(\phi\) always does. One test separates them perfectly. Two vectors with inner product zero are orthogonal. The pairs \(\ket0,\ket1\) and \(\ket+,\ket-\) are orthogonal, which is why each is read without error by its own measurement.
Now take \(\ket0\) and \(\ket+\). Their overlap is \(\braket0+=1/\sqrt2\). The test for zero passes on plus with probability \(1/2\), so a pass does not prove that zero was sent. Every test we have built makes errors on this pair. Whether some cleverer test could do better is a question we return to when we study discrimination.
What does a state vector tell us about a preparation? Plus gives a fair result in the computational basis and a certain result in the plus/minus basis. So a vector is not a list of answers fixed in advance for every question. It is the rule that predicts the probabilities of any test. A preparation that a single vector describes completely is called pure. We will meet less complete descriptions once preparations are mixed at random.
Problems
Problem 3.2. For (|0〉+i|1〉)/√2, what is the H readout probability of zero?
Worked solution
It is |(1+i)/2|²=1/2. The phase π/2 gives a fair readout, unlike phase zero or π.
3.3 Evolution and Readout
We have used two operations, the beam splitter and the phase plate, and one kind of reading. Which other operations could nature allow, and what is left after a reading? Writing an operation as a matrix lets us ask which matrices keep every total probability equal to one; the answer is the unitary matrices. A reading is not of this kind. It gives one outcome, and an immediate repeat must agree with it, which fixes the state it leaves.
The beam splitter's rule had a special shape. Each input amplitude contributed to each output, and the contributions added. An operation of this shape is linear: the output of a weighted sum of inputs is the same weighted sum of the separate outputs. A linear operation is fixed by what it does to \(\ket0\) and \(\ket1\). We record those two images as the columns of a matrix:
Applying the matrix to a column means multiplying each input amplitude by its column and adding. For example, \(H(1,i)^T/\sqrt2=((1+i)/2,(1-i)/2)^T\), where the superscript \(T\) turns a row into a column. Both outputs have probability one half. Applying \(H\) twice returns every input, as the two-mirror experiment showed for \(\ket0\).
Which linear operations are physically possible for an isolated qubit? At least, every normalized input must give a normalized output. Call the columns \(\ket{u_0}=U\ket0\) and \(\ket{u_1}=U\ket1\). What does normalization demand of them?
Inputs \(\ket0\) and \(\ket1\) force \(\braket{u_0}{u_0}=\braket{u_1}{u_1}=1\). Next take the unnormalized input \(\ket0+\ket1\), whose squared length is 2. Its output \(\ket{u_0}+\ket{u_1}\) has squared length \(2+2\,\mathrm{Re}\braket{u_0}{u_1}\), where Re takes the real part. Keeping the length forces \(\mathrm{Re}\braket{u_0}{u_1}=0\). The input \(\ket0+i\ket1\) similarly gives \(2-2\,\mathrm{Im}\braket{u_0}{u_1}\), where Im takes the imaginary part, so \(\mathrm{Im}\braket{u_0}{u_1}=0\). The columns must be orthogonal unit vectors.
A compact way to say this needs one new operation. The adjoint \(U^\dagger\), read "U dagger", is the matrix obtained by transposing \(U\) and conjugating every entry. Its rows are the bras \(\bra{u_0}\) and \(\bra{u_1}\). So the entries of \(U^\dagger U\) are the inner products of the columns, and our two conditions say
\[U^\dagger U=I,\]
where the identity \(I\) leaves every vector unchanged. A matrix with this property is unitary. Conversely, any unitary keeps every length, because \(\braket{U\psi}{U\psi}=\bra\psi U^\dagger U\ket\psi=\braket\psi\psi\). By the same step it keeps every inner product. It is reversible, with inverse \(U^\dagger\). The matrix \(H\) is unitary and is its own inverse. It is called the Hadamard operation.
Normalization only tells us which operations are consistent. That the isolated evolution of a qubit is always unitary is a physical postulate, which experiments support. Measurements, and systems that are partly discarded, will need a broader description.
Now we can return to the rule for testing a state. Suppose a unitary \(U\) turns \(\ket\phi\) into \(\ket0\), and we then read the path. The amplitude to see 0 is \(\bra0U\ket\psi\). Since \(U\) keeps inner products and \(U\ket\phi=\ket0\), this is \(\braket{U\phi}{U\psi}=\braket\phi\psi\). So every test built from a unitary and a reading obeys the inner-product rule.
3.3.2 Projective measurements and conditional states§
A reading of \(a\ket0+b\ket1\) gives 0 with probability \(|a|^2\). Suppose it gave 0. What state should we use to predict the next reading? Experiment answers: an immediate repeat of the same reading gives 0 again. So after outcome 0 the state must be \(\ket0\).
We can write this step as an operation. A column times a row is a matrix, so \(\ket0\bra0\) is the matrix with a single 1 in the top-left corner. Applied to \(a\ket0+b\ket1\) it keeps \(a\ket0\) and removes the rest. An operation that keeps the component along some directions and removes the perpendicular part is a projector. The squared length of what it keeps, \(|a|^2\), is the outcome's probability. Dividing by \(|a|\) restores unit length. Any leftover phase is global and has no effect.
Measurement postulate. A projective measurement is described by projectors \(P_j\) onto mutually orthogonal directions, with \(\sum_jP_j=I\). Outcome \(j\) occurs with probability \(p_j=\|P_j\ket\psi\|^2\). Given that outcome, the state becomes \(P_j\ket\psi/\sqrt{p_j}\) (for \(p_j>0\)).
This update is a separate postulate. It is not unitary evolution: it depends on the outcome, and it cannot be undone. The plus/minus measurement uses \(P_\pm=\ket\pm\bra\pm\). On a single qubit each projector keeps one direction. On larger systems a projector may keep a whole set of directions and leave superpositions within that set intact.
Diagrams of these steps are called circuits. A horizontal line, or wire, is a system as time runs from left to right. A labelled box is a unitary. A box marked M is a computational-basis measurement, and a double line after it carries its classical record.
Two more unitaries will recur. The operation \(X\) exchanges the labels, \(X\ket0=\ket1\) and \(X\ket1=\ket0\), as a bit flip does. The operation \(Z\) reverses the sign of the second amplitude, \(Z\ket0=\ket0\) and \(Z\ket1=-\ket1\). A product acts from right to left: \(XZ\ket\psi\) means apply \(Z\) first, then \(X\).
Which states does each of these leave alone? \(Z\) leaves \(\ket0\) unchanged and only multiplies \(\ket1\) by \(-1\). \(X\) leaves \(\ket+\) unchanged and multiplies \(\ket-\) by \(-1\), since \(X\ket-=(\ket1-\ket0)/\sqrt2\). A nonzero vector that an operation only multiplies by a number is an eigenvector of it, and the number is its eigenvalue. Both \(X\) and \(Z\) have eigenvalues \(+1\) and \(-1\). So the computational-basis measurement reads the eigenvectors of \(Z\), and we call it a Z measurement. The plus/minus measurement reads those of \(X\), and we call it an X measurement.
Read a relative phase
Predict first. Predict the final zero probability at phases 0°, 90°, and 180°, then move the phase control.
The circuit prepares plus, applies phase to the one component, and reads out through H. The final zero probability is \((1+\cos\phi)/2\): one at 0°, one half at 90°, and zero at 180°.
One observation is still unexplained. A device that records the path destroys the interference. The record is a second physical system, and we have no way yet to write a state for two systems together. What does such a state look like?
Problems
Problem 3.3. Can a unitary map both zero and one to plus?
Worked solution
No. Their original inner product is zero, while identical output vectors have inner product one. Unitary evolution preserves this distinction.
Transfer problem. A source prepares either \(\ket+\) or \(\ket-\) with equal probability and hides the label. Does an H readout distinguish this source from a source that always prepares \((\ket0+i\ket1)/\sqrt2\)?
Worked solution
No. H gives fair outcomes for either source, so this one readout does not distinguish them. They are still different preparations: projection onto the second source’s pure state succeeds certainly on it and only half the time on the hidden plus/minus ensemble. Agreement in one experiment is insufficient.
Summary
Pure quantum states are unit vectors \(|\psi\rangle\) in a complex Hilbert space up to global phase, where coherent superpositions exhibit interference that classical probability mixtures cannot reproduce.
Projective measurements in an orthonormal basis \(\{|b_k\rangle\}\) yield outcome \(k\) with Born probability \(p(k) = |\langle b_k|\psi\rangle|^2\) and project the system onto \(|b_k\rangle\).
Unitary operators \(U^\dagger U = I\) preserve inner products \(\langle\phi|\psi\rangle\), so no unitary transformation can change the overlap or distinguishability of two states.
Questions and worked problems
These exercises draw on the whole chapter. Attempt each question before opening a hint, explain your reasoning, then compare it with the worked solution.
Let \(\ket{y_\pm}=(\ket0\pm i\ket1)/\sqrt2\). Compute \(\braket{-}{y_+}\) and \(\braket{y_+}{y_-}\), and give the probability of each projection test. Which pair is orthogonal?
Hint 1
Conjugate the coefficients of the left vector.
Hint 2
\(\bra{y_+}=(1,-i)/\sqrt2\).
Worked solution
\(\braket{-}{y_+}=\frac12(1\cdot1+(-1)\cdot i)=\frac{1-i}{2}\), with probability \(|1-i|^2/4=\frac12\). \(\braket{y_+}{y_-}=\frac12\big(1+(-i)(-i)\big)=\frac12(1-1)=0\), so \(\ket{y_+}\) and \(\ket{y_-}\) are orthogonal and a suitable test distinguishes them perfectly, even though both give fair results in the computational basis and in the \(\pm\) basis.
Normalize \(3\ket0+4i\ket1\), and give its computational-basis probabilities. Then decide whether \((\ket0+\ket1)/2\) is a valid state as written.
Hint 1
The squared length is \(|3|^2+|4i|^2\).
Hint 2
A state must satisfy \(\braket\psi\psi=1\).
Worked solution
The squared length is \(9+16=25\), so the normalized state is \(\frac35\ket0+\frac{4i}{5}\ket1\), with probabilities \(\frac{9}{25}\) and \(\frac{16}{25}\). The vector \((\ket0+\ket1)/2\) has squared length \(\frac14+\frac14=\frac12\), so its “probabilities” would add to \(\frac12\); dividing by \(1/\sqrt2\) gives \(\ket+\). Only the direction of a vector describes the state, but the probability rule requires the normalized representative.
Let \(S=\mathrm{diag}(1,i)\). Check that \(S\) is unitary, compute \(S\ket+\), and find the probability of outcome zero when \(S\ket+\) is read out through \(H\).
Hint 1
\(S^\dagger=\mathrm{diag}(1,-i)\).
Hint 2
Use \(p(0)=(1+\cos\phi)/2\) for amplitudes \((1,e^{i\phi})/\sqrt2\).
Worked solution
\(S^\dagger S=\mathrm{diag}(1\cdot1,(-i)i)=I\), so \(S\) is unitary. \(S\ket+=(\ket0+i\ket1)/\sqrt2\), a relative phase \(\phi=\pi/2\). The \(H\) readout gives \(p(0)=(1+\cos\frac\pi2)/2=\frac12\). \(S\) changes no computational-basis probability, yet it moves \(\ket+\) to a state that the \(\pm\) test no longer identifies with certainty.
A qubit is prepared in \(\ket\psi=\cos\theta\ket0+\sin\theta\ket1\) with real \(\theta\). (a) Find the probability of outcome \(+\) in a projective test onto \(\{\ket+,\ket-\}\), and the conditional state. (b) Evaluate at \(\theta=\pi/8\) and \(\theta=-\pi/4\). (c) For which \(\theta\) is the outcome certain?
Hint 1
\(\braket+\psi=(\cos\theta+\sin\theta)/\sqrt2\).
Hint 2
\((\cos\theta+\sin\theta)^2=1+\sin2\theta\).
Worked solution
(a) \(p(+)=\frac12(\cos\theta+\sin\theta)^2=\frac12(1+\sin2\theta)\), and \(p(-)=\frac12(1-\sin2\theta)\). After outcome \(+\) the state is \(\ket+\) (up to phase), whatever \(\theta\) was. (b) At \(\theta=\pi/8\), \(p(+)=\frac12(1+\frac{1}{\sqrt2})\approx0.854\). At \(\theta=-\pi/4\), \(\sin2\theta=-1\), so \(p(+)=0\): the state is \(-\ket-\). (c) The outcome is certain when \(\sin2\theta=\pm1\), that is, when \(\ket\psi\) is \(\pm\ket+\) or \(\pm\ket-\). A projective test gives certainty only for states lying along one of its directions.
Start with \(\ket0\). (a) Measure in the computational basis twice in a row: what are the probabilities of the outcome pairs? (b) Instead, first measure in the \(\pm\) basis, then in the computational basis: what is the probability of \(0\) at the end? (c) What does the comparison show?
Hint 1
After an outcome, the state is the corresponding basis vector.
Hint 2
\(\ket\pm\) each give \(0\) with probability \(\frac12\).
Worked solution
(a) Both measurements give \(0\) with certainty: the conditional state after the first is \(\ket0\). (b) The \(\pm\) measurement gives \(+\) or \(-\) with probability \(\frac12\) each, leaving \(\ket+\) or \(\ket-\). Each then gives \(0\) with probability \(\frac12\), so the final probability of \(0\) is \(\frac12\). (c) An intermediate measurement in a different basis changes the later statistics: a state that would have given \(0\) with certainty gives \(0\) only half the time. Measurement in one basis disturbs the state as seen in another, and that disturbance limits how much information can be gained without it.
A three-level system is prepared in \(\ket\psi=(\ket0+\ket1+\ket2)/\sqrt3\). A projective test has two outcomes: \(P_a=\ket0\bra0+\ket1\bra1\) and \(P_b=\ket2\bra2\). (a) Find the outcome probabilities and conditional states. (b) After outcome \(a\), apply \(H\) to the levels \(0,1\) (leaving \(\ket2\) alone) and measure all three levels. What results? (c) What would a test with three one-dimensional outcomes have done instead?
Hint 1
\(P_a\ket\psi=(\ket0+\ket1)/\sqrt3\).
Hint 2
The conditional state after \(a\) is \(\ket+\) on the levels \(0,1\).
Worked solution
(a) \(p(a)=\|(\ket0+\ket1)/\sqrt3\|^2=\frac23\), leaving \((\ket0+\ket1)/\sqrt2\); \(p(b)=\frac13\), leaving \(\ket2\). (b) After outcome \(a\), \(H\) maps \((\ket0+\ket1)/\sqrt2\) to \(\ket0\), so the final measurement gives \(0\) with certainty. (c) A test with three separate outcomes would leave \(\ket0\) or \(\ket1\) (each with probability \(\frac13\)), and \(H\) would then give \(0\) or \(1\) at random. The rank-two test answered only “is it level \(2\)?” and kept the superposition within the other two levels intact. A measurement disturbs only what it distinguishes.
Find every unitary \(U\) with \(U\ket0=\ket+\). Show that the second column must be \(e^{i\alpha}\ket-\) for some phase \(\alpha\), and use this to explain why no unitary maps both \(\ket0\) and \(\ket1\) to \(\ket+\).
Hint 1
The columns of a unitary are orthonormal.
Hint 2
In two dimensions, the unit vectors orthogonal to \(\ket+\) are the multiples \(e^{i\alpha}\ket-\).
Worked solution
The first column of \(U\) is \(U\ket0=\ket+\). The second column \(U\ket1=c\ket0+d\ket1\) must be a unit vector orthogonal to \(\ket+\): \(\frac{1}{\sqrt2}(c+d)=0\) gives \(d=-c\), and normalization gives \(|c|=1/\sqrt2\). So \(U\ket1=e^{i\alpha}\ket-\), and \(U=\ket+\bra0+e^{i\alpha}\ket-\bra1\); \(\alpha=0\) gives \(H\). A unitary preserves inner products, and \(\braket01=0\), so the images of \(\ket0\) and \(\ket1\) must be orthogonal. They cannot both be \(\ket+\).
A student computes probabilities by squaring amplitudes without conjugating: \(p_j=v_j^2\). Apply this rule to \(\ket\psi=\frac{1+i}{2}\ket0+\frac{1-i}{2}\ket1\) and compare with the correct rule. What goes wrong, and why is the conjugate needed?
Hint 1
\(\big(\frac{1+i}{2}\big)^2=\frac i2\).
Hint 2
\(|z|^2=zz^*\) is real and nonnegative.
Worked solution
Squaring gives \(\big(\frac{1+i}{2}\big)^2=\frac i2\) and \(\big(\frac{1-i}{2}\big)^2=-\frac i2\): imaginary “probabilities” that add to \(0\). The correct rule gives \(|\frac{1+i}{2}|^2=\frac{1+1}{4}=\frac12\) for each outcome, adding to \(1\). A complex number squared keeps its phase (doubled), while \(zz^*\) removes it and leaves the squared magnitude, which is real and nonnegative. For real amplitudes the two rules agree, which is why the mistake can go unnoticed until complex phases appear.
With equal priors, Bob must decide whether a qubit was prepared as \(\ket0\) or \(\ket+\). He measures in the basis \(\ket{b_0}=\cos\varphi\ket0+\sin\varphi\ket1\), \(\ket{b_1}=-\sin\varphi\ket0+\cos\varphi\ket1\), guessing \(\ket0\) on \(b_0\) and \(\ket+\) on \(b_1\). (a) Show that the success probability is \(\frac12+\frac14(\cos2\varphi-\sin2\varphi)\). (b) Find the best \(\varphi\) and the success. (c) Compare with measuring in the computational basis.
Hint 1
\(p(b_0\mid0)=\cos^2\varphi\) and \(p(b_1\mid+)=\frac12(\cos\varphi-\sin\varphi)^2\).
Hint 2
\(\cos x-\sin x\) is largest, \(\sqrt2\), at \(x=-\pi/4\).
Worked solution
(a) Success \(=\frac12\big[\cos^2\varphi+\frac12(\cos\varphi-\sin\varphi)^2\big]=\frac12\big[\frac{1+\cos2\varphi}{2}+\frac{1-\sin2\varphi}{2}\big]=\frac12+\frac14(\cos2\varphi-\sin2\varphi)\). (b) It is largest at \(2\varphi=-\pi/4\), \(\varphi=-\pi/8\), giving \(\frac12+\frac{\sqrt2}{4}\approx0.854\). (c) The computational basis (\(\varphi=0\)) gives \(\frac12+\frac14=0.75\). The best test uses directions placed symmetrically around the two states. No test reaches \(1\), because \(\ket0\) and \(\ket+\) are not orthogonal; the Helstrom bound shows that \(0.854\) is optimal among all measurements.
The text states that a linear operation preserving the length of every vector must preserve all inner products. Prove it using the identities \(\mathrm{Re}\braket uv=\frac14\big(\|u+v\|^2-\|u-v\|^2\big)\) and \(\mathrm{Im}\braket uv=\frac14\big(\|u-iv\|^2-\|u+iv\|^2\big)\). Then deduce that the columns of such a matrix are orthonormal.
Hint 1
Apply \(U\) to \(u\pm v\) and \(u\pm iv\) and use linearity.
Hint 2
The columns are the images of the orthonormal vectors \(\ket0,\ket1\).
Worked solution
By linearity \(U(u\pm v)=Uu\pm Uv\) and \(U(u\pm iv)=Uu\pm iUv\). Since \(U\) preserves every length, each of the four squared norms in the identities is unchanged when \(u,v\) are replaced by \(Uu,Uv\). Hence both the real and imaginary parts of \(\braket uv\) are preserved: \(\braket{Uu}{Uv}=\braket uv\). The columns of \(U\) are \(U\ket0\) and \(U\ket1\), and since \(\braket00=\braket11=1\) and \(\braket01=0\), the columns are unit vectors with zero inner product. Preserving lengths alone forces the whole geometry to be preserved.
