How can we measure a quantum system if the sum of amplitudes-squared does not equal one?

For example, if we want to measure $|a\rangle = 0.25|0\rangle + 0.25|1\rangle$, how can we measure it?


2 Answers 2


Normalized states

By the measurement postulate (see e.g. 2.2.3 on page 84 in Nielsen & Chuang), a measurement is described by a collection of operators $M_m$, indexed by the measurement outcomes $m$ satisfying the constraint $\sum_m M_m^\dagger M_m = I$.

The probability of obtaining the outcome $m$ in a measurement on a state represented by a normalized ket $|\psi\rangle$ is

$$ p(m) = \langle\psi|M_m^\dagger M_m|\psi\rangle\tag{1} $$

and the corresponding post-measurement state is

$$ \frac{M_m|\psi\rangle}{\sqrt{\langle\psi|M_m^\dagger M_m|\psi\rangle}}\tag{2}. $$

Non-normalized states

Even though it is often practical to use kets to represent pure quantum states in calculations, it should be remembered that the correspondence between pure states and kets is not one-to-one. Specifically, for any non-zero$^1$ ket $|\phi\rangle$ and a non-zero complex number $\lambda$ the kets $|\phi\rangle$ and $\lambda|\phi\rangle$ represent the same quantum state$^2$.

We can exploit this fact to find the outcome probability and the post-measurement state when the state is specified by a non-normalized ket. Specifically, we first normalize the ket by setting $\lambda=\||\phi\rangle\|^{-1}$ and then use $(1)$ and $(2)$. In this case, the probability of obtaining the outcome $m$ in a measurement on a state represented by a non-normalized ket $|\phi\rangle$ is

$$ p(m) = \frac{\langle\phi|M_m^\dagger M_m|\phi\rangle}{\langle\phi|\phi\rangle}.\tag{1'} $$

and the corresponding post-measurement state is

$$ \frac{M_m|\phi\rangle}{\sqrt{\langle\phi|M_m^\dagger M_m|\phi\rangle}}.\tag{2'} $$

Comparing the formulas we see that the rule $(2)$ for calculating the post-measurement state applies unchanged and the Born rule $(1)$ requires normalization.

$^1$ The zero ket does not represent any quantum state. This is reassuring since the Born rule $(1')$ for non-normalized states does not work when $|\phi\rangle=0$.
$^2$ N.B. this suggests that we will obtain a one-to-one correspondence by modeling pure states using a complex projective space.


A quantum state should be normalized, i.e. Euclidean norm of a vector representing the state should be equal to one. Your state is not normalized. Replacing 0.25 with $1/\sqrt{2}$ leads to normalized states with probability of measurement of both 0 and 1 equal to 50%. Before normalization the probabilities of measuring 0 and 1 were also equal, however, they did not add up to 1. The normalization does not change ratios among probabilities but it ensures that the sum of the probabilities is one. The reason is that only possible results of the measurement are states in the superposition. There is no other possible outcome than 0 or 1 in case of single qubit.


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