# 11: Postulates of Quantum Mechanics (Lecture)

Lecture 10: Expectation values, 2D-PIB and Heisenberg Uncertainty Principle

Last lecture focused on the extension of the 1D particle in a box to the 2-D and 3D cases. This is an easy problem since via a simple application of separation of variables (Worksheet 2) for each dimension (they act independently). From this we identified a few interesting phenomena including multiple quantum numbers ($$n_x$$, $$n_y$$ and $$n_z$$) and degeneracy (if the boxes are of the same length or multiples of each other) where multiple wavefunctions share the identical energy.

We were able to provide a quantitative backing in using the Heisenberg Uncertainty principle from wavefuctions, specifically in terms of the standard deviations of the probability distribution functions (i.e., wavefunction squared).

We ended the lecture on the five postulates of quantum mechanics, which ties together many aspects that we have been discussing.

## 2D Boxes

Let us now consider the Schrödinger Equation for an electron confined to a two dimensional box, $$0 < x < a$$ and $$0 < y < b$$. That is to say, within this rectangle the electron wavefunction behaves as a free particle ($$V(x,y) = 0$$), but the walls are impenetrable so the wavefunction $$\psi(x,y,t)=0$$ at the walls.

Extending the (time-independent) Schrödinger equation for a one-dimensional system

$−\frac{\hbar^2}{2m}\frac{d^ 2\psi(x)}{d x^2}+V(x)\psi(x)=E\psi(x) \label{3.1.3}$

to a two-dimensional system is not difficult

$- \dfrac{\hbar^2}{2m} \left (\dfrac{\partial^2 \psi(x,y)}{\partial x^2} + \dfrac{\partial ^2 \psi(x,y)}{\partial y^2} \right) + V(x,y)\psi(x,y) = E \psi(x,y). \label{e1}$

Equation \ref{e1} can be simplified for the particle in a 2D box since we know that $$V(x,y) =0$$ within the box and $$V(x,y) = \infty$$ outside the box

$- \dfrac{\hbar^2}{2m} \left (\dfrac{\partial^2 \psi(x,y)}{\partial x^2} + \dfrac{\partial ^2 \psi(x,y)}{\partial y^2} \right) = E \psi(x,y). \label{e2}$

Since the Hamiltonian (i.e. left side of Equation \ref{e2}) is the sum of two terms with independent (separate) variables, we try a product wavefunction like in the Separation of Variables approach used to separate time-dependence from the spatial dependence previously. Within this approach we express the 2-D wavefunction as a product of two independent 1-D components

$\psi(x,y) = X(x)Y(y). \label{product}$

After some math building on this (discussed in more detail here), we arrive at the complete normalized 2D wavefunction for a particle in a 2D box

$\psi_{n_x, n_y}(x,y)=\dfrac{2}{L}\sin\left ( \dfrac{n_x \pi x}{L} \right ) \sin\left ( \dfrac{n_y \pi y}{L} \right ) \label{squareWF}$

The wavefunction of a 2D well (Paul Seeburger)

## 3D Boxes

The 1D particle in the box problem can be expanded to consider a particle within a 3D box for three lengths $$a$$, $$b$$, and $$c$$. When there is NO FORCE (i.e., no potential) acting on the particles inside the box.

The potential for the particle inside the box

$V(\vec{r}) = 0$

• $$0 \leq x \leq a$$
• $$0 \leq y \leq b$$
• $$0 \leq z \leq c$$
• $$a < x < 0$$
• $$b < y < 0$$
• $$c < z < 0$$

$$\vec{r}$$ is the vector with all 3 components along the 3 axis of the 3-D box: $$\vec{r} = a\hat{x} + b\hat{y} + c\hat{z}$$. When the potential energy is infinite, then the wave function equals zero. When the potential energy is zero, then the wavefunction obeys the Schrödinger equation.

$\psi(r) = \sqrt{\dfrac{8}{v}}\sin \left( \dfrac{n_{x} \pi x}{a} \right) \sin \left(\dfrac{n_{y} \pi y}{b}\right) \sin \left(\dfrac{ n_{z} \pi z}{c} \right)$

$v = a \times b \times c = volume \; of \; box$

with Total Energy levels:

$E_{n_x,n_y,x_z} = \dfrac{h^{2}}{8m}\left(\dfrac{n_{x}^{2}}{a^{2}} + \dfrac{n_{y}^{2}}{b^{2}} + \dfrac{n_{z}^{2}}{c^{2}}\right) \label{3.9.10}$

As with 1-D boxes, the quantum numbers $$n_x$$, $$n_y$$, and $$n_z$$ go from 1 to infinity.

Exercise $$\PageIndex{1}$$

What is the energy of the ground state of a 3D box that is $$a$$ by $$a$$ by $$a$$?

This is when $$n_x=1$$, $$n_y=1$$, and $$n_z=1$$. So from Equation $$\ref{3.9.10}$$, this is

$E_{1,1,1}= \dfrac{h^{2}}{8m}\left(\dfrac{1}{a^{2}} + \dfrac{1}{a^{2}} + \dfrac{1}{a^{2}}\right) = \dfrac{3h^{2}}{8ma^{2}}$

Exercise $$\PageIndex{2}$$

What is the energy of the 1st excited state of a 3D box that is $$a$$ by $$a$$ by $$a$$?

This is trickier, since there is a degeneracy in the system with three wavefunctions having the same energy:

• $$\Psi_{2,1,1}$$
• $$\Psi_{1,2,1}$$
• $$\Psi_{1,1,2}$$

Each of these wavefunctions have the same energy (via Equation $$\ref{3.9.10}$$) of

$E= \dfrac{6h^{2}}{8ma^{2}}$

## Postulates of Quantum Mechanics in a Nutshell (you know all these already)

There are five assumptions made for the purpose of reasoning in quantum mechanics. These assumptions are called postulates, and they set the groundwork for quantum mechanics.

### Postulate 1

The state of the system is completely specified by $$\psi$$. All possible information about the system can be found in the wavefunction $$\psi$$. For valid, well-behaved wavefunctions, normalized probability holds true, such that the integral over all space is equal to 1.

$\int_{-\infty}^\infty \psi^*(x)\psi(x)\;dx=1.$

What this means is that the chance to find a particle is 100% over all space. Wavefunctions must be

• single valued,
• continuous, and
• finite.

That means that the function must be a function, it cannot have an undefined slope, and the function cannot go to $$-\infty$$ or $$+\infty$$.

### Postulate 2

For every observable in classical mechanics, there is a corresponding operator in quantum mechanics.

Name Observable Symbol Operator Symbol Operation
Table: Some common Operators in Quantum Mechanics
Position $$x$$ $$\hat{X}$$ Multiply by $$x$$
$$r$$ $$\hat{R}$$ Multiply by $$r$$
Momentum $$p_{x}$$ $$\hat{P_{x}}$$ -$$\imath \hbar \dfrac{d}{dx}$$
$$p$$ $$\hat{P}$$ -$$\imath$$$$\hbar$$[$$\text{i}$$$$\dfrac{d}{dx}$$+$$\text{j}$$$$\dfrac{d}{dy}$$+$$\text{k}$$$$\dfrac{d}{dz}$$]
Kinetic Energy $$K_{x}$$ $$\hat{K_{x}}$$ $$\dfrac{-(\hbar^{2})}{2m}$$$$\dfrac{d^{2}}{dx^{2}}$$
$$K$$ $$\hat{K}$$

$$\dfrac{-(\hbar^{2})}{2m}$$[$$\dfrac{d^{2}}{dx^{2}}$$+$$\dfrac{d^{2}}{dy^{2}}$$+$$\dfrac{d^{2}}{dz^{2}}$$]

Which can be simplified to
$$\dfrac{-(\hbar^{2})}{2m}$$$$\bigtriangledown^{2}$$

Potential Energy $$V(x)$$ $$\hat{V(x)}$$ Multiply by $$V(x)$$
$$V(x,y,z)$$ $$\hat{V(x,y,z)}$$ Multiply by $$V(x,y,z)$$
Total Energy $$E$$ $$\hat{E}$$ $$\dfrac{-(\hbar^{2})}{2m}$$$$\bigtriangledown^{2}$$ + $$V(x,y,z)$$
Angular Momentum $$L_{x}$$ $$\hat{L_{x}}$$ -$$\imath$$$$\hbar$$[$$\text{y}$$$$\dfrac{d}{dz}$$ - $$\text{z}$$$$\dfrac{d}{dy}$$]
$$L_{y}$$ $$\hat{L_{y}}$$ -$$\imath$$$$\hbar$$[$$\text{z}$$$$\dfrac{d}{dx}$$ - $$\text{x}$$$$\dfrac{d}{dz}$$]
$$L_{z}$$ $$\hat{L_{z}}$$ -$$\imath$$$$\hbar$$[$$\text{x}$$$$\dfrac{d}{dy}$$ - $$\text{y}$$$$\dfrac{d}{dx}$$]

### Postulate 3

If $$\hat{A}$$$$\psi(x)$$ produces eigenvalues $$\{a\}$$, where

$\hat{A} \psi_n(x) = a_n \psi_n(x)$

then the set $$\{a_n\}$$ is the set of values (and only those values) that can be observed.

### Postulate 4

For a normalized wavefunction, the average value of any observable with operation $$\hat{A}$$ is given by

$\langle a \rangle = \int_{-\infty}^\infty \psi^*(x) \hat{A}\ \psi(x) dx$

### Postulate 5

Wavefunctions evolve in time, given by the Time-dependent Schrödinger Equation

$\hat{H}\Psi(x,t) = \imath \hbar \dfrac{d(\Psi(x,t))}{dt}$

where

$\Psi(x,t) = \psi(x)f(t)$

Using the Separation of Variables technique results in

$f(t)= e^\frac {-iEt}{\hbar} = e^{-it\omega}$

## Commuting Operators

One important property of operators is that the order of operation matters. Thus:

$\hat{A}{\hat{E}f(x)} \not= \hat{E}{\hat{A}f(x)} \label{4.6.3}$

unless the two operators commute. Two operators commute if the following equation is true:

$\left[\hat{A},\hat{E}\right] = \hat{A}\hat{E} - \hat{E}\hat{A} = 0 \label{4.6.4}$

To determine whether two operators commute first operate $$\hat{A}\hat{E}$$ on a function $$f(x)$$. Then operate$$\hat{E}\hat{A}$$ the same function $$f(x)$$. If the same answer is obtained subtracting the two functions will equal zero and the two operators will commute.

Example $$\PageIndex{1}$$

Do $$\hat{A}$$ and $$\hat{E}$$ commute if

$\hat{A} = \dfrac{d}{dx}$

and

$\hat{E} = x^2$

Solution

This requires evaluating $$\left[\hat{A},\hat{E}\right]$$, which requires solving for $$\hat{A} \{\hat{E} f(x)\}$$ and $$\hat{E} \{\hat{A} f(x)\}$$ for arbitrary wavefunction $$f(x)$$ and asking if they are equal.

$\hat{A} \{\hat{E} f(x)\} = \hat{A}\{ x^2 f(x) \}= \dfrac{d}{dx} \{ x^2 f(x)\} = 2xf(x) + x^2 f'(x)$

From the product rule of differentiation.

$\hat{E} \{\hat{A}f(x)\} = \hat{E}\{f'(x)\} = x^2 f'(x)$

Now ask if they are equal

$\left[\hat{A},\hat{E}\right] = 2x f(x) + x^2 f'(x) - x^2f'(x) = 2x f(x) \not= 0$

Therefore the two operators do not commute.

Example $$\PageIndex{2}$$

Do $$\hat{B}$$ and $$\hat{C}$$ commute if

$\hat{B}= \dfrac {h} {x}$

and

$\hat{C}\{f(x)\} = f(x) +3$

Solution
This requires evaluating $$\left[\hat{B},\hat{C}\right]$$ like in Example $$\PageIndex{1}$$.

$\hat{B} \{\hat{C}f(x)\} = \hat{B}\{f(x) +3\} = \dfrac {h}{x} (f(x) +3) = \dfrac {h f(x)}{x} + \dfrac{3h}{x}$

$\hat{C} \{\hat{B}f(x)\} = \hat{C} \{ \dfrac {h} {x} f(x)\} = \dfrac {h f(x)} {x} +3$

Now ask if they are equal

$$\left[\hat{B},\hat{C}\right] = \dfrac {h f(x)} {x} + \dfrac {3h} {x} - \dfrac {h f(x)} {x} -3 \not= 0$$

The two operators do not commute.

## Heisenberg Uncertainty Principle Redux (Redux Redux)

Although it will not be proven here, there is a general statement of the uncertainty principle in terms of the commutation property of operators. If two operators $$\hat {A}$$ and $$\hat {B}$$ do not commute, then the uncertainties (standard deviations σ) in the physical quantities associated with these operators must satisfy

$\sigma _A \sigma _B \ge \left| \int \psi ^* [ \hat {A} \hat {B} - \hat {B} \hat {A} ] \psi d\tau \right| \label{4-52}$

where the integral inside the square brackets is called the commutator, and ││signifies the modulus or absolute value. If $$\hat {A}$$ and $$\hat {B}$$ commute, then the right-hand-side of equation $$\ref{4-52}$$ is zero, so either or both $$σ_A$$ and $$σ_B$$ could be zero, and there is no restriction on the uncertainties in the measurements of the eigenvalues $$a$$ and $$b$$.