# Quantum mechanics/Operators and measurements

Any observable (quantity that can be measured) is associated in quantum mechanics to a mathematical operator. You have seen two examples so far:
${\hat {K}}=-{\frac {\hbar ^{2}}{2m}}{\frac {\partial ^{2}}{\partial x^{2}}}$ (kinetic energy in one dimension)
${\hat {H}}=-{\frac {\hbar ^{2}}{2m}}{\frac {\partial ^{2}}{\partial x^{2}}}+V(x)$ (total energy in one dimension)
To complete the description of the postulates of quantum mechanics, we need to specify:
* how to build the operators corresponding to any observable;
* what the result is of any measure in quantum mechanics.

## How to build all operators

All quantities that can be measured (position, velocity, kinetic energy, total energy, angular momentum, etc.) can be written as the combination of the position x, y, z and the momentum p$_x$, p$_y$, p$_z$ of all particles of a system (but we consider only one particle for simplicity). If we indicate with ${\hat {x}}$ and ${\hat {p}}_{x}$ the quantum mechanical operators corresponding to the classical quantities position and linear momentum along x, one of the postulates of quantum mechanics states that
${\hat {x}}{\hat {p}}_{x}-{\hat {p}}_{x}{\hat {x}}=i\hbar$ (Eq. 1)
One way to satisfy the equation above (see Exercises) is to define
${\hat {x}}\equiv x$ and ${\hat {p}}_{x}=-i\hbar {\frac {\partial }{\partial x}}$ (Eq. 2)
And analogous relations are valid for the other coordinates.
The operator of the position x, y, z is just the coordinate while the operator for the momentum is the partial derivatives along the component of the momentum multiplied by -iħ. This is mysterious but the bottom line is that to build the quantum mechanical operator for a given classical quantity you have to leave unchanged all expressions containing x, y, z and substitute the classical momenta p$_x$, p$_y$, p$_z$ with $-i\hbar {\frac {\partial }{\partial x}},-i\hbar {\frac {\partial }{\partial y}},-i\hbar {\frac {\partial }{\partial z}}$ respectively.
Examples
The kinetic energy for a particle moving only along the x direction is given classically by ${\begin{matrix}{\frac {1}{2}}\end{matrix}}mv_{x}^{2}={\frac {p_{x}^{2}}{2m}}$. The quantum version is simply obtained as:
${\begin{matrix}{\frac {1}{2}}\end{matrix}}mv_{x}^{2}={\frac {p_{x}^{2}}{2m}}\to {\frac {{\hat {p}}_{x}^{2}}{2m}}={\frac {1}{2m}}{\hat {p}}_{x}{\hat {p}}_{x}={\frac {1}{2m}}\left(-i\hbar {\frac {\partial }{\partial x}}\right)\left(-i\hbar {\frac {\partial }{\partial x}}\right)=-{\frac {\hbar ^{2}}{2m}}{\frac {\partial ^{2}}{\partial x^{2}}}$
The quantum version of the potential energy of a harmonic oscillator is:
$V={\begin{matrix}{\frac {1}{2}}\end{matrix}}kx^{2}\to {\hat {V}}={\begin{matrix}{\frac {1}{2}}\end{matrix}}k{\hat {x}}^{2}={\begin{matrix}{\frac {1}{2}}\end{matrix}}kx^{2}$
The total energy operator (Hamiltonian) for the harmonic oscillator is:
$H={\frac {p_{x}^{2}}{2m}}+{\begin{matrix}{\frac {1}{2}}\end{matrix}}kx^{2}\to {\hat {H}}=-{\frac {\hbar ^{2}}{2m}}{\frac {\partial ^{2}}{\partial x^{2}}}+{\begin{matrix}{\frac {1}{2}}\end{matrix}}kx^{2}$

## Eigenvalues and eigenfunctions of an operator

Given an operator ${\hat {A}}$, a function φ such that
${\hat {A}}\varphi =\lambda \varphi$ (Eq. 3)
is called an eigenfunction of the operator ${\hat {A}}$ corresponding to the eigenvalue λ. An equation like 3 is called an eigenvalue equation and usually admits a discrete set of eigenvalues.

## Measure of a quantity

So far we have established that for any classical quantity A there is an associated quantum mechanical operator ${\hat {A}}$. Another postulate of quantum mechanics states that when measuring A, the only possible results of the measure are eigenvalues of the operator ${\hat {A}}$. If the system is in a state that is one of the eigenfunctions of ${\hat {A}}$ corresponding to the eigenvalue λ, the result of the measure will be λ. If the system is prepared in a generic state ψ, the average value of repeated measures of A of a system prepared in the state ψ is given by:
$\left\langle A\right\rangle ={\frac {\int \psi ^{*}{\hat {A}}\psi \,d\tau }{\int \psi ^{*}\psi \,d\tau }}$ (Eq. 4)
This is called the expectation value of A and it is one of the fundamental formulas to remember.
For example, if you want to measure the total energy of a system, the corresponding operator is the Hamiltonian Ĥ and the result of the measure will be one of the eigenvalues of the Hamiltonian. If the system was in an eigenstate of Ĥ with eigenvalue E you can repeat the measure and you will always get E. If the system is not in an eigenstate of Ĥ the result each time will be one of the eigenvalues of E and the average result will be $\left\langle E\right\rangle ={\frac {\int \psi ^{*}{\hat {H}}\psi \,d\tau }{\int \psi ^{*}\psi \,d\tau }}$ (Eq. 5)

## Example: expectation values of the particle in the box

We will calculate now the expectation value of the position, the linear momentum, the kinetic energy and the total energy for a particle in a box in its ground state, i.e. with $\psi ={\sqrt {\frac {2}{L}}}\sin \left({\frac {\pi x}{L}}\right)$.
The denominator of Equation 4 is always 1 since the wavefunction is normalized, i.e.
$\int \psi ^{*}\psi \,d\tau =\int _{0}^{L}{\sqrt {\frac {2}{L}}}\sin \left({\frac {\pi x}{L}}\right){\sqrt {\frac {2}{L}}}\sin \left({\frac {\pi x}{L}}\right)\,dx=1$
The expectation value of the position is:
$\left\langle x\right\rangle ={\frac {\int \psi ^{*}x\psi \,d\tau }{\int \psi ^{*}\psi \,d\tau }}={\frac {\int _{0}^{L}{\sqrt {\frac {2}{L}}}x\sin \left({\frac {\pi x}{L}}\right){\sqrt {\frac {2}{L}}}\sin \left({\frac {\pi x}{L}}\right)\,dx}{1}}={\frac {2}{L}}\int _{0}^{L}x\sin ^{2}\left({\frac {\pi x}{L}}\right)\,dx={\frac {L}{2}}$
The expectation value of the momentum is:
$\left\langle p\right\rangle ={\frac {\int \psi ^{*}{\hat {p}}\psi \,d\tau }{\int \psi ^{*}\psi \,d\tau }}={\frac {\int _{0}^{L}{\sqrt {\frac {2}{L}}}\sin \left({\frac {\pi x}{L}}\right)\left(-i\hbar {\frac {d}{dx}}\right){\sqrt {\frac {2}{L}}}\sin \left({\frac {\pi x}{L}}\right)\,dx}{1}}={\frac {2i\hbar \pi }{L^{2}}}\int _{0}^{L}\sin \left({\frac {\pi x}{L}}\right)\cos \left({\frac {\pi x}{L}}\right)\,dx=0$
The expectation value of the kinetic energy is:
$\left\langle K\right\rangle ={\frac {\int \psi ^{*}{\hat {K}}\psi \,d\tau }{\int \psi ^{*}\psi \,d\tau }}={\frac {{\frac {2}{L}}\int _{0}^{L}\sin \left({\frac {\pi x}{L}}\right)\left(-{\frac {\hbar ^{2}}{2m}}{\frac {\partial ^{2}}{\partial x^{2}}}\right)\sin \left({\frac {\pi x}{L}}\right)\,dx}{1}}={\frac {\hbar ^{2}\pi ^{2}}{2mL^{2}}}{\frac {2}{L}}\int _{0}^{L}\sin ^{2}\left({\frac {\pi x}{L}}\right)\,dx={\frac {\hbar ^{2}\pi ^{2}}{2mL^{2}}}$
A position "on average" is in the middle of the box (L/2). It has equal probability of travelling towards the left or right, so the average momentum and velocity must be zero. The average kinetic energy must be equal to the total energy of the ground state of the particle in the box, as there is no other energy component.
Exercise
Solve explicitly the four integrals in the Section above.
Calculate the expectation value of ${\hat {x}}^{2}$ and ${\hat {p}}^{2}$ for the 1D particle in a box.
Systems do not need to be in an eigenstate of the Hamiltonian in order to make measurements on them. A particle in a box of length L is in a state described by ψ = x(L - x). Calculate the expectation value of the energy.
Solutions
1
Solved in the next Section
$\left\langle E\right\rangle ={\frac {\int _{0}^{L}x(L-x)(-{\frac {\hbar ^{2}}{2m}}{\frac {\partial ^{2}}{\partial x^{2}}})x(L-x)\,dx}{\int _{0}^{L}x(L-x)x(L-x)\,dx}}={\frac {{\frac {\hbar ^{2}}{2m}}\int _{0}^{L}2x(L-x)\,dx}{\int _{0}^{L}x^{2}L^{2}-2Lx^{3}+x^{4}\,dx}}={\frac {\hbar ^{2}10}{2mL^{2}}}$ :
Next: Lesson 9 - Measuring two quantities at the same time: the Heisenberg Principle