# A formal approach to Observables and Measurements

Throughout previous posts, I’ve delivered a way to understand what is a quantum measurement. That way of understanding is simple and may suit our normal intuition, but measurement is a non-trivial physical process and somewhat counter-intuitive for several reasons.

First, a measurement in general results in probabilistic outcomes. No matter how careful you prepare for the measurement procedure, it is inherent that possible outcomes of a measure is distributed according to a certain probability distribution determined by the state space of your quantum system.

Second, once you do a measurement, your quantum state instantly gets rid of its previous state and transforms into a new state. This unavoidable sequence is due to the fact to perform a measurement is also to interfere with the current quantum state. The interaction between your quantum system and the measurement aparatus consequently drives to a general difference between the anterior state and the posterior state.

Third, it is necessary to define a set of theoretical rules that allows us to evaluate the probability distribution determining the result if we perform a measurement. Today we’re going to get a deeper look at what a measurement really is in a formally mathematical fashion.

Recall that quantum computing is the science of computation leveraging quantum properties. Qubits, the protagonists in the field, are merely a computer science-favored name for atomic and sub-atomic particles like photons, electrons, etc.. For that reason, measuring a qubit is just measuring specific characteristics of a particle (including but not limited to position, energy, charge, momentum, angular momentum, and spin) that are collectively known as *observables*. In that way, the outcome of a measurement corresponds to values of the measured qubit’s observables. Mathematically, observables are Hermitian operators, a special kind of operator: if $\hat{A}$ is a linear operator: $\hat{A}: \mathscr{H} \rightarrow \mathscr{H}$ with $\mathscr{H}$ is a finite-dimensional Hilbert space, and if $\hat{A} = \hat{A^{\dagger}}$, then $\hat{A}$ is a Hermitian operator.

It needs some techniques in functional analysis to provide a complete explanation why Hermitian operators associate with observable. This makes the concept convoluted and unappealing (actually because I don’t fully understand it). For simplicity, I would say that this correspondence happens because physical quantities must have real values, and there is a theorem states that Hermitian operators always have real eigenvalues. As a result, it makes sense to use Hermitian operators to represent observables. A quantum postulate summarizes the concepts that

All possible results of a measurement of an observable are limited to the set of eigenvalues of that observable.

Let $\hat{M}$ be an observable (Hermitian operator), which is defined in the state space of your quantum system, that describes a projective measurement. Since the state space has finite dimensions, it is possible to write $$\hat{M} = \sum_i r_i \hat{P}_{r_i}$$ , where $\hat{P}_{r_i}$ associated with the eigenvalue $r_i$ of $\hat{M}$ is the projector on the eigenspace of $\hat{M}$ defined by that eigenvalue. (In case you didn’t know, eigenspace is the space spanned by all eigenvectors corresponding the same eigenvalue, i.e. the space of all vectors that can be written as linear combinations of those eigenvectors.) The possible outcomes of your measurement are the eigenvalues $r_i$ of observable $\hat{M}$.

If the state of your quantum system before the measurement is $|\psi\rangle$, the probability you obtain the result $r_{i}$ is given by $$p(r_i) = \langle\psi| \hat{P}_{r_i} |\psi\rangle$$ , and the posterior state of the quantum system will be $$|\psi\rangle = \frac{\hat{P}_{r_i} |\psi\rangle}{\sqrt{p(r_i)}}$$

#### Example

I’ll use $\displaystyle{|\psi\rangle = \frac{1}{\sqrt{2}}|0\rangle - \frac{i}{\sqrt{2}}|1\rangle}$ as my current qubit to make a clear illustration for the properties above and other widely used results. Suppose that the observable of interest defines a measurement which distinguishes eigenstate $|+\rangle$ with eigenvalue $0$ and eigenstate $|-\rangle$ with eigenvalue $1$.

The projector onto eigenstate $|+\rangle$ is $$\displaystyle{\hat{P}_0 = |+\rangle\langle +| = \begin{bmatrix} \frac{1}{\sqrt{2}}\\ \frac{1}{\sqrt{2}} \end{bmatrix}\begin{bmatrix} \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \end{bmatrix} = \begin{bmatrix} \frac{1}{2} & \frac{1}{2}\\ \frac{1}{2} & \frac{1}{2} \end{bmatrix}}$$ The projector onto eigenstate $|-\rangle$ is handled similarly $$\displaystyle{\hat{P}_1 = |-\rangle\langle -| = \begin{bmatrix} \frac{1}{\sqrt{2}}\\ -\frac{1}{\sqrt{2}} \end{bmatrix}\begin{bmatrix} \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \end{bmatrix} = \begin{bmatrix} \frac{1}{2} & -\frac{1}{2}\\ \frac{1}{2} & -\frac{1}{2} \end{bmatrix}}$$

It’s natural for the observable to have the representation $$\displaystyle{\hat{M} = 0\hat{P}_0 + 1\hat{P}_1 = \begin{bmatrix} \frac{1}{2} & -\frac{1}{2}\\ \frac{1}{2} & -\frac{1}{2}\end{bmatrix}}$$

Keep in mind that the measurement outcome is just an eigenvalue whose corresponding eigenstate is indeed the posterior state. The basic formula for finding the probability for outcome $0$ of the measurement is settled that $$p(0) = \langle\psi| \hat{P}_{0} |\psi\rangle = \langle\psi| |+\rangle\langle +| |\psi\rangle = \langle\psi| +\rangle\langle + |\psi\rangle = \left|\langle + | \psi \rangle \right|^2 = \left|\frac{1-i}{2} \right|^2 = \frac{1}{2}$$

Any state can be arithmetically represented as a superposition of $\hat{M}$ eigenstates: $$ \begin{matrix} |\psi\rangle & = & \displaystyle{\sum_n a_n |e_n\rangle} \\ & = & a_0 |+\rangle + a_1 |-\rangle \end{matrix}$$

The probability amplitudes $a_n$ can also be usefully expressed in terms of the orthogonal eigenstates of $\hat{M}$ $$\begin{matrix} &|\psi\rangle & = & \displaystyle{\sum_n a_n |e_n\rangle} \\ \text{Multiply both sides by } e_j: & \langle e_j |\psi\rangle & = & \displaystyle{\sum_n a_n \langle e_j |e_n\rangle} \\ &&=& a_j \end{matrix}$$ because $\langle e_j |e_n\rangle = \left \{ \begin{matrix} 0 & \text{if } n\neq j\\ 1 & \text{if } n=j \end{matrix} \right .$ by the orthogonality of the eigenstates. I can get the probability amplitude for eigenstates $|+\rangle$ and $|-\rangle$ of my current qubit through $$\begin{matrix}a_0 &=& \displaystyle{\langle + | \left(\frac{1}{\sqrt{2}}|0\rangle - \frac{i}{\sqrt{2}}|1\rangle \right)} \\ &= & \displaystyle{\frac{1}{\sqrt{2}}\langle + | 0\rangle - \frac{i}{\sqrt{2}}\langle + | 1\rangle} \\ & = & \displaystyle{\frac{1}{2}\langle 0 | 0 \rangle + \frac{1}{2}\langle 1 | 0 \rangle - \frac{i}{2}\langle 0 | 1 \rangle} - \frac{i}{2}\langle 1 | 1 \rangle \\ & = & \displaystyle{\frac{1-i}{2}} \\ \\ a_1 &=& \displaystyle{\langle - | \left(\frac{1}{\sqrt{2}}|0\rangle - \frac{i}{\sqrt{2}}|1\rangle \right)} \\ & = & \displaystyle{\frac{1+i}{2}}\end{matrix}$$

From the results above, it makes sense to rewrite the qubit state as $$\begin{matrix}|\psi\rangle &=& \displaystyle{\sum_n \langle e_n |\psi\rangle \cdot |e_n\rangle} \\ &=& \langle + | \psi \rangle \cdot | + \rangle + \langle - | \psi \rangle \cdot | - \rangle\\ & = & \displaystyle{\frac{1-i}{2} |+\rangle + \frac{1+i}{2} |-\rangle} \end{matrix}$$ The notation $\cdot$ is added only for clear differentiation between the complex amplitude $\langle e_n | \psi \rangle$ and the state $| e_n \rangle$.

Suppose that the measurement returns eigenvalue $0$, then we know that the post-measurement state of the qubit is the normalized corresponding eigenstate. $$|\psi\rangle \leftarrow \frac{\hat{P}_0 |\psi\rangle}{\sqrt{p(0)}} = \frac{\frac{1-i}{2\sqrt{2}} |0\rangle + \frac{1-i}{2\sqrt{2}} |1\rangle}{\sqrt{\frac{1}{2}}} = \frac{1-i}{2} |0\rangle + \frac{1-i}{2} |1\rangle = \frac{1-i}{\sqrt{2}} |+\rangle$$ with $\displaystyle{\left | \frac{1-i}{\sqrt{2}}\right | = 1}$ as expected.

After going through this wall of text, you probably have enough background knowledge regarding Quantum Information to understand Quantum Algorithms, the next focus of my posts from now on. If there will be any further post on Quantum Information, it would be at an advanced level and inclined to the Theory of Quantum Mechanics and Quantum Information rather than an essence for Quantum Computing.