Math

Independent Variables

For example, $\psi(x)$ contains the independent variable $x$ .

Bases

Quantum mechanics is complicated. A basis is something that can be observed in an experiment.

Bases are very important in quantum mechanics. This is probably the most important section of this chapter. There are different bases in quantum mechanics. For example, there is the position basis; the energy basis; the momentum basis; the spin basis; the z basis; and many other bases. In fact, can you guess how many bases there are in quantum mechanics? 5? 10? Quantum mechanics uses Hilbert space to describe its functions. A Hilbert space is an abstract vector space. So there are an infinite number of bases one can use in quantum mechanics. However, we typically stick to ones most useful to the physics of the system, such as energy, momentum, spin, etc.

Position basis

\ket{\psi} = \psi(x) =

\ket{\psi} = \psi(x) =

Energy basis

\ket{\psi} = \ket{n} = \ket{4}

\ket{\psi} = \ket{E_n} = \ket{-10 \, eV}

Momentum basis

Spin basis

\ket{\psi} = \ket{+}

\ket{\psi} = \ket{-}

Eigenstates

Energy eigenstates

\ket{\psi} = \psi_E = \ket{-9 \,eV}

\ket{\psi} = \psi(x)= \sqrt{\frac{2}{L}} \sin\big(\frac{5\pi x}{L}\big)

\ket{\psi} = \psi(p)=

\ket{\psi}= \psi(t) = \ket{-3 \, eV}

Position eigenstates

\ket{\psi} = \psi(x) =

\ket{\psi} = \psi(p) =

\ket{\psi} = \psi(t) =

Momentum eigenstates

\ket{\psi} = \psi(x) =

\ket{\psi} = \psi(p) =

\ket{\psi} = \psi(t) =

Spin eigenstates

\ket{\psi} = \psi(x) =

\ket{\psi} = \psi(p) =

\ket{\psi} = \psi(t) =

For energy, the energy eigenstates are of the form $\ket{n}$ and $\ket{E_n}$ . For example, $\ket{\psi} = \ket{3}$ where $n = 3$ . However, we can write an energy eigenstate in the position basis by $\braket{x|n} =\ket{\psi} = \sqrt{\frac{2}{L}} \sin\big(\frac{3\pi x}{L}\big)$ .

Representations

Content coming soon.

Kets

Consider the state $\ket{\psi} = \frac{1}{2} \ket{+} + \frac{\sqrt{3}}{2} \ket{-}$

Operators

Position operator

The position operator is $\hat{x}$ .

The momentum operator is $\hat{p} = \frac{\partial^2}{\partial x^2}$

The spin operator is

There are many different kinds of energy. Kinetic energy, potential energy, sound energy, mass energy, total energy, kinetic minus potential (Lagrangian), etc. Which of these have quantum mechanical manifestations? All of them? Some of them? How do we know?

The energy operator is

Is there a Lagrangian operator?

The kinetic energy operator is

The potential energy operator is $\hat{V} =$

The mass energy operator is

The sound energy operator is

There are an infinite number of operators in quantum mechanics, as any observable from an experiment can be an operator.

Name	Operator symbol	Time representation	Position basis	Momentum basis	Spin basis	Energy basis
$\text{Position}$	$\hat{x}$		$x$	$i \hbar \frac{\partial}{\partial p}$	$\hat{x} \otimes \hat{S}$	$\text{Ill defined}$
$\text{Momentum}$	$\hat{p}$		$- i \hbar \frac{\partial}{\partial x}$	$p$	$\hat{p} \otimes \hat{S}$	$\text{Ill defined}$
$\text{Total energy}$	$\hat{H}$	$i \hbar \frac{\partial}{\partial t}$	$\frac{- \hbar^2}{2m} \frac{\partial^2}{\partial x^2} + V(x)$	$\frac{p^2}{2m} + \int_{-\infty}^\infty \tilde{V}(p - p') \psi(p')dp'$	$\text{Ill defined}$	$\text{Ill defined}$
$\text{Spin}$	$\hat{S}$		$\text{not possible}$	$\text{not possible}$	$\hat{x} \otimes \hat{S}$	$S_x, S_y, S_z$
$\text{Orbital angular momentum}$	$\hat{L}$				$\hat{x} \otimes \hat{S}$
$\text{Orbital angular momentum, z}$	$\hat{L}_z$		$- i \hbar(y \frac{\partial}{\partial x} - x \frac{\partial}{\partial y})$

Probability

When we make a measurement on an unknown state $\psi$ , the possible results must be one of the eigenvalues of the state $\psi$ . For example, consider the state $\psi(x) = a\ket{\uparrow} + b\ket{\downarrow}$ The eigenstates of $\psi$ are $\ket{\uparrow}$ and $\ket{\downarrow}$ , and as will be shown later, the eigenvalue associated with $\ket{\uparrow}$ is $\frac{\hbar}{2}$ and with $\ket{\downarrow}$ $-\frac{\hbar}{2}$ . Now it is said that $\psi$ must contain all of the information about a system, and that is true; however there are advanced cases where things like interference phenomena can be measured which might not obviously be deduced from the state $\psi$ . We leave this topic to more advanced treatments; for now, the only possible results of a measurement of $\psi$ are one of the eigenvalues. It must be the case that $|a|^2 + |b|^2 = 1$ .

For example, consider again the state $\ket{\psi} = \frac{1}{2} \ket{+} + \frac{\sqrt{3}}{2} \ket{-}$ . We know that a measurement of the state must produce either $\frac{\hbar}{2}$ or $-\frac{\hbar}{2}$ , so $|\frac{1}{2}|^2$ + $|\frac{\sqrt{3}}{2}|^2 = 1$ .

Commutator

I want you to put a book on the table in front of you, front cover facing up so you can read it. The commutator comes the commutative property: AB = BA. But is this always true? We define the commutator as [A,B] = AB - BA (you might also see {A,B}). For example, with regular numbers, $7 * 3 - 3 * 7 = 0$ ; however, with something like spin matrix operators [ $S_x, S_y$ ] = $S_x S_y - S_y S_x = i \hbar S_z$ .

Worked Example 1.1: Determining Coefficients

Problem

Determine the coefficient $a$ in the state $\ket{\psi} = a\ket{0} - \frac{\sqrt{5}}{3} \ket{1}$ .

Solution

The long way to do this is this way... but we can also do this in our head. We know that $|a|^2 + |-\frac{\sqrt{5}}{3}|^2=1$ , so $|a|^2 = 1 - \frac{5}{9} \rightarrow |a|^2 = \frac{4}{9} \rightarrow a = \pm \frac{2}{3}$ . But how do you know if $a = + \frac{2}{3}$ or $-\frac{2}{3}$ ?

1 of 27

A particle has wave function ψ(x) = A for 0 ≤ x ≤ L = 3.0 nm. Drag the slider to find the normalization constant A so that ∫₀ᴸ |ψ|² dx = 1.

∫|ψ|²dx =

0.270 / 1.000

A = 0.3000 nm⁻¹/²

2 of 27

For the normalized box wave function (L = 3.0 nm), drag the boundary to find the probability of the particle being in 0 ≤ x ≤ 1.0 nm.

Drag the yellow handle to set the right boundary at 1.0 nm, then submit the probability.

P(0 → 1.50 nm) = 0.5000

3 of 27

A particle is in state ψ(x) = A sin(πx/L) for 0≤x≤L. Find A in terms of L. For L=1.0 nm, A in nm^(-1/2) is:

Unlock Exercise 3