CHAPTER 03

Measurement and changes of basis

Learning goals. Project a state onto an orthonormal basis, retain the conjugation and cancellations, calculate conditional states, and connect probabilities to repeated experiments.

3.1 A basis specifies a question

An orthonormal basis $\{|b_0\rangle,|b_1\rangle\}$ satisfies $\langle b_j|b_k\rangle=\delta_{jk}$. Projective measurement in this basis gives

$$p(b_j)=|\langle b_j|\psi\rangle|^2.$$

For an ideal rank-one measurement, outcome $b_j$ leaves the system in $|b_j\rangle$ up to global phase. The prior state determines probabilities; the observed result determines the conditioned state afterward.

Writing $|\psi\rangle=c_0|b_0\rangle+c_1|b_1\rangle$ and multiplying by $\langle b_j|$ gives $c_j=\langle b_j|\psi\rangle$. A basis change calculates new coordinates of the same vector. Physically applying a gate changes the vector; a gate followed by Z measurement can implement a different measurement basis.

3.2 One state, three full calculations

Our recurring example is

$$|\psi\rangle=\frac{2|0\rangle+3i|1\rangle}{\sqrt{13}}.$$

Its norm squared is

$$\frac{(2\langle0|-3i\langle1|)(2|0\rangle+3i|1\rangle)}{13} =\frac{4+6i(0)-6i(0)-9i^2}{13} =\frac{4+9}{13}=1.$$

3.2.1 Z basis

$$\begin{aligned} \langle0|\psi\rangle &=\frac{2\langle0|0\rangle+3i\langle0|1\rangle}{\sqrt{13}}\\ &=\frac{2(1)+3i(0)}{\sqrt{13}}=\frac2{\sqrt{13}},\\ p(0)&=\left(\frac2{\sqrt{13}}\right)^* \left(\frac2{\sqrt{13}}\right)=\frac4{13}. \end{aligned}$$

Similarly,

$$\langle1|\psi\rangle=\frac{2(0)+3i(1)}{\sqrt{13}} =\frac{3i}{\sqrt{13}},$$ $$p(1)=\frac{(-3i)(3i)}{13}=\frac{-9i^2}{13}=\frac9{13}.$$

3.2.2 X basis

$$\begin{aligned} \langle+|\psi\rangle &=\frac{(\langle0|+\langle1|)(2|0\rangle+3i|1\rangle)}{\sqrt{26}}\\ &=\frac{2\langle0|0\rangle+3i\langle0|1\rangle +2\langle1|0\rangle+3i\langle1|1\rangle}{\sqrt{26}}\\ &=\frac{2+0+0+3i}{\sqrt{26}}=\frac{2+3i}{\sqrt{26}}. \end{aligned}$$

The squared modulus is

$$p(+)=\frac{(2-3i)(2+3i)}{26} =\frac{4+6i-6i-9i^2}{26} =\frac{13}{26}=\frac12.$$

For the other outcome,

$$\begin{aligned} \langle-|\psi\rangle &=\frac{(\langle0|-\langle1|)(2|0\rangle+3i|1\rangle)}{\sqrt{26}}\\ &=\frac{2+0-0-3i}{\sqrt{26}}=\frac{2-3i}{\sqrt{26}},\\ p(-)&=\frac{(2+3i)(2-3i)}{26}=\frac12. \end{aligned}$$

3.2.3 Y basis

Forming the bra conjugates the imaginary coefficient:

$$\langle+i|=\frac{\langle0|-i\langle1|}{\sqrt2}, \qquad\langle-i|=\frac{\langle0|+i\langle1|}{\sqrt2}.$$

Now expand, including the terms that vanish:

$$\begin{aligned} \langle+i|\psi\rangle &=\frac{(\langle0|-i\langle1|)(2|0\rangle+3i|1\rangle)}{\sqrt{26}}\\ &=\frac{2\langle0|0\rangle+3i\langle0|1\rangle -2i\langle1|0\rangle+(-i)(3i)\langle1|1\rangle}{\sqrt{26}}\\ &=\frac{2+0-0-3i^2}{\sqrt{26}}=\frac5{\sqrt{26}},\\ p(+i)&=\frac{25}{26}. \end{aligned}$$

For the negative outcome,

$$\begin{aligned} \langle-i|\psi\rangle &=\frac{(\langle0|+i\langle1|)(2|0\rangle+3i|1\rangle)}{\sqrt{26}}\\ &=\frac{2+0+0+3i^2}{\sqrt{26}}=-\frac1{\sqrt{26}},\\ p(-i)&=\left(-\frac1{\sqrt{26}}\right)^* \left(-\frac1{\sqrt{26}}\right)=\frac1{26}. \end{aligned}$$

A negative amplitude is allowed; a negative probability is not. Each basis has its own normalized distribution:

Do not add all six probabilities. These are three different experimental choices on freshly prepared copies.

3.3 General formulas and a sign check

For normalized $\alpha|0\rangle+\beta|1\rangle$,

$$p(0)=|\alpha|^2,\quad p(1)=|\beta|^2,$$ $$p(\pm)=\frac{|\alpha\pm\beta|^2}{2} =\frac12\pm\operatorname{Re}(\alpha^*\beta),$$ $$p(\pm i)=\frac{|\alpha\mp i\beta|^2}{2} =\frac12\pm\operatorname{Im}(\alpha^*\beta).$$

Check the Y sign rather than memorizing it:

$$|\alpha-i\beta|^2 =(\alpha^*+i\beta^*)(\alpha-i\beta) =|\alpha|^2+|\beta|^2+i(\beta^*\alpha-\alpha^*\beta).$$

If $\alpha^*\beta=u+iv$, the final term is $i((u-iv)-(u+iv))=i(-2iv)=2v$. Divide by two to obtain $1/2+v$. Real coefficients give balanced Y probabilities; an imaginary relative coefficient need not.

3.4 Consecutive measurements

Prepare $|+\rangle$, measure Z, then measure X. Z gives 0 or 1 equally often. Either conditioned state has balanced X probabilities. Therefore

$$\Pr(\text{final }+) =\Pr(0)\Pr(+\mid0)+\Pr(1)\Pr(+\mid1) =\frac12\frac12+\frac12\frac12=\frac12.$$

Without the intermediate Z measurement, the final X outcome would be $+$ with certainty. Ignoring a measurement record does not undo the physical interaction.

For a projector $P_m$, possibly of higher rank,

$$p(m)=\langle\psi|P_m|\psi\rangle,\qquad |\psi_m\rangle=\frac{P_m|\psi\rangle}{\sqrt{p(m)}}.$$

Only outcomes with $p(m)>0$ can be conditioned upon. A projector onto a subspace can retain coherence inside that subspace. Parity checks in quantum error correction exploit this fact.

3.5 Observables and expectation values

If outcomes have numerical values $a_m$, define $A=\sum_ma_mP_m$. Then

$$\langle A\rangle=\sum_ma_mp(m)=\langle\psi|A|\psi\rangle.$$

For Pauli Z, eigenvalues $+1,-1$ correspond to $|0\rangle,|1\rangle$. The ket label 0 is not an eigenvalue zero. Our example gives $\langle Z\rangle=-5/13$, $\langle X\rangle=0$, and $\langle Y\rangle=12/13$.

Variance is $\langle A^2\rangle-\langle A\rangle^2$. Each Pauli squares to identity, so $\operatorname{Var}(Z)=1-\langle Z\rangle^2$. Incompatible observables lack a shared complete eigenbasis. Their uncertainty is not merely a poorly calibrated detector.

3.6 Exercises