QM Lecture 7: Quantum Mechanics in 3D

**By: Tao Steven Zheng (鄭濤)**

*This article was originally posted on Brilliant.org by the same author.*

## **3D Coordinates**

The transition towards three dimensions mimics classical mechanics, where we express our equations in the

x
,
y
,
z

{\displaystyle x, y, z}

 rectangular coordinates. We shall let

r
→

=
x

i

y

j

z

k

{\displaystyle \vec{r} = x \;\mathbf{i} + y \;\mathbf{j} + z\;\mathbf{k} }

 denote a vector in three-dimensional space. In classical mechanics, the Hamiltonian of a system is prescribed by

# H

1

2
m

(

p

x

2

p

y

2

p

z

2

)

V
(
x
,
y
,
z
)
.

{\displaystyle H=\frac{1}{2m}\left({{p}\_{x}}^{2}+{{p}\_{y}}^{2}+{{p}\_{z}}^{2}\right) +V(x,y,z). }

In quantum mechanics, the Hamiltonian operator

H
^

=

[

−

ℏ
2

2
m

∇
2

V
(

r
→

)

]

{\displaystyle \hat{H} = \left[-\frac{\hbar^2}{2m} \nabla^2 + V(\vec{r}) \right] }

acts on the wavefunction. Note that

∇

2

=

∂

2

∂

x

2

∂

2

∂

y

2

∂

2

∂

z

2

{\displaystyle {\nabla}^{2}=\frac{{\partial}^{2}}{\partial {x}^{2}}+\frac{{\partial}^{2} }{\partial {y}^{2}}+\frac{{\partial}^{2} }{\partial {z}^{2}} }

 is the Laplacian operator in rectangular coordinates. Consequently, the time-dependent Schrödinger equation becomes

i
ℏ

∂

∂
t

Ψ
(

r
→

,
t
# )

[

−

ℏ
2

2
m

∇
2

V
(

r
→

)

]

Ψ
(

r
→

,
t
)
.

{\displaystyle i\hbar \frac{\partial}{\partial t}\Psi(\vec{r},t) = \left[-\frac{\hbar^2}{2m} \nabla^2 + V(\vec{r}) \right]\Psi(\vec{r},t). }

Linearity still applies in three dimensions, so all the linear operators used in one dimension are valid:

p
^

x

=
−
i
ℏ

∂

∂
x

,

p
^

y

=
−
i
ℏ

∂

∂
y

,

p
^

z

=
−
i
ℏ

∂

∂
z

.

{\displaystyle {\hat{p}}\_{x} = -i\hbar \frac{\partial}{\partial x}, \qquad {\hat{p}}\_{y} = -i\hbar \frac{\partial}{\partial y}, \qquad {\hat{p}}\_{z} = -i\hbar \frac{\partial}{\partial z}.
 }

It is customary in physics to break a higher-dimensional problem into its directional components, then solve each component, and finally reassemble the solutions. Thus, wavefunctions are now expressed as

Ψ
(

r
→

,
t
# )
∑

c

n

ψ

n

(

r
→

)

e

−
i

E

n

t

/

ℏ

.

{\displaystyle \Psi(\vec{r} , t) = \sum{ {c}\_{n}{\psi}\_{n}(\vec{r}) {e}^{-i{E}\_{n}t/\hbar}}. }

As always, our wavefunctions must be continuous and normalizable. In three dimensions, we integrate the probability density of the wavefunction throughout all space using volume integrals:

∫

V

Ψ

∗

Ψ

d
# V

{\displaystyle \int\_{V} {\Psi}^{\*}{\Psi}dV =1. }

Note that many problems in physics exhibit different symmetries, in which certain coordinate systems are more natural to work with (e.g. **cylindrical coordinates** or **spherical coordinates**). Keep in mind that the geometry of a physical system dictates the appropriate methods used.

## **Schrödinger Equation in Spherical Coordinates**

Now we will briefly investigate the 3D Schrödinger equation in spherical coordinates, which is the basis for solving the **hydrogen atom** and **atomic orbitals**. Separation of variables is a common and powerful method when confronting higher-dimensional differential equations. This is valid in whichever coordinate system

r
,
θ
,
ϕ

{\displaystyle r, \theta, \phi}

 we use. In spherical coordinates, consider the ansatz:

Ψ
(
r
,
θ
,
ϕ
# )
R
(
r
)
Y
(
θ
,
ϕ
)
.

{\displaystyle \Psi(r, \theta, \phi) = R(r)Y(\theta, \phi). }

We use the Laplacian operator in spherical coordinates

∇

2

=

1

r

2

∂

∂
r

(

r

2

∂

∂
r

)

1

r

2

sin
⁡

θ

∂

∂
θ

(

sin
⁡

θ

∂

∂
θ

)

1

r

2

sin

2

⁡

θ

(

∂

2

∂

ϕ

2

)

{\displaystyle {\nabla}^{2} = \frac{1}{{r}^{2}}\frac{\partial}{\partial r}\left({r}^{2} \frac{\partial}{\partial r}\right) + \frac{1}{{r}^{2}\sin{\theta}} \frac{\partial}{\partial \theta} \left(\sin{\theta} \frac{\partial}{\partial \theta}\right) + \frac{1}{{r}^{2}\sin^{2}{\theta}}\left(\frac{{\partial}^{2}}{\partial {\phi}^{2}}\right) }

thus the Schrödinger equation is

−

ℏ

2

2
m

[

1

r

2

∂

∂
r

(

r

2

∂
Ψ

∂
r

)

1

r

2

sin
⁡

θ

∂

∂
θ

(

sin
⁡

θ

∂
Ψ

∂
θ

)

1

r

2

sin

2

⁡

θ

(

∂

2

Ψ

∂

ϕ

2

)

]

V
# Ψ
E
Ψ

{\displaystyle \frac{-{\hbar}^{2}}{2m}\left[\frac{1}{{r}^{2}}\frac{\partial}{\partial r}\left({r}^{2} \frac{\partial \Psi}{\partial r}\right) + \frac{1}{{r}^{2}\sin{\theta}} \frac{\partial }{\partial \theta} \left(\sin{\theta} \frac{\partial \Psi}{\partial \theta}\right) + \frac{1}{{r}^{2}\sin^{2}{\theta}}\left(\frac{{\partial}^{2} \Psi}{\partial {\phi}^{2}}\right)\right] + V\Psi = E\Psi }

Equivalently

−

ℏ

2

2
m

[

Y

r

2

∂

∂
r

(

r

2

∂
R

∂
r

)

R

r

2

sin
⁡

θ

∂

∂
θ

(

sin
⁡

θ

∂
Y

∂
θ

)

R

r

2

sin

2

⁡

θ

(

∂

2

Y

∂

ϕ

2

)

]

V
R
# Y
E
R
Y

{\displaystyle \frac{-{\hbar}^{2}}{2m}\left[\frac{Y}{{r}^{2}}\frac{\partial}{\partial r}\left({r}^{2} \frac{\partial R}{\partial r}\right) + \frac{R}{{r}^{2}\sin{\theta}} \frac{\partial }{\partial \theta} \left(\sin{\theta} \frac{\partial Y}{\partial \theta}\right) + \frac{R}{{r}^{2}\sin^{2}{\theta}}\left(\frac{{\partial}^{2} Y}{\partial {\phi}^{2}}\right)\right] + VRY = ERY }

We can rearrange the above equation as

[

1
R

d

d
r

(

r

2

d
R

d
r

)

−

2
m

r

2

ℏ

2

(
V
−
E
)

]

1
Y

[

1

sin
⁡

θ

∂

∂
θ

(

sin
⁡

θ

∂
Y

∂
θ

)

1

sin
2

⁡

θ

∂

2

Y

∂

ϕ

2

]

=
0

{\displaystyle \left[\frac{1}{R}\frac{d}{dr}\left({r}^{2}\frac{dR}{dr}\right)-\frac{2m{r}^{2}}{{\hbar}^{2}}(V-E) \right]+ \frac{1}{Y}\left[\frac{1}{\sin{\theta}}\frac{\partial}{\partial \theta}\left(\sin{\theta}\frac{\partial Y}{\partial \theta}\right) + \frac{1}{\sin^2{\theta}} \frac{{\partial}^{2}{Y}}{\partial{\phi}^{2}}\right] = 0 }

Now we can separate the above equation as follows:

**Radial equation**

1
R

d

d
r

(

r

2

d
R

d
r

)

−

2
m

r

2

ℏ

2

[
V
−
E
# ]
l
(
l
1
)

{\displaystyle \frac{1}{R}\frac{d}{dr}\left({r}^{2}\frac{dR}{dr}\right)-\frac{2m{r}^{2}}{{\hbar}^{2}}[V-E] = l(l+1) }

**Angular equation**

1
Y

[

1

sin
⁡

θ

∂

∂
θ

(

sin
⁡

θ

∂
Y

∂
θ

)

1

sin
2

⁡

θ

∂

2

Y

∂

ϕ

2

]

=
−
l
(
l
1
)

{\displaystyle \frac{1}{Y}\left[\frac{1}{\sin{\theta}}\frac{\partial}{\partial \theta}\left(\sin{\theta}\frac{\partial Y}{\partial \theta}\right) + \frac{1}{\sin^2{\theta}} \frac{{\partial}^{2}{Y}}{\partial{\phi}^{2}}\right] = -l(l+1) }

But how do we solve these monsters? Well, I will not write out the solutions because they are long and quite advanced. What you should know is that we can rearrange these equations and hope that existing solutions for the differential equations can be found. For a good exposition of the solutions to the radial and angular wavefunctions, and its applications to the hydrogen atom, click here. The full solutions of the radial equation and angular equation are extremely complex, because they require the knowledge of **Legendre polynomials** and **Laguerre polynomials**.

What is important to know is that the solutions of the angular equation contribute to the shape of atomic orbitals (angular probability density), and the solutions of the radial equation give the radial probability density.

## **The Hydrogen Spectrum**

One of the greatest triumphs of quantum theory is the mathematical confirmation of the **hydrogen spectrum**. The hydrogen spectrum phenomena can be explained by electrons gaining or losing energy. In doing so, the electron changes from one stationary state to another stationary state. Usually, when an electron loses energy, it gives off a photon and moves down an energy state. This difference in energy is not only observable, but quantifiable by the equation

E

γ

=

E

i

−

E

f

=
−
13.6

(

1

n

i

2

−

1

n

f

2

)

e
V

{\displaystyle {E}\_{\gamma} = {E}\_{i} - {E}\_{f} = -13.6 \left( \frac{1}{{n}\_{i}^{2}}-\frac{1}{{n}\_{f}^{2}}\right) \; eV }

where

−
13.6

e
V

{\displaystyle -13.6 \; eV}

(electronvolts) is the ground state, or lowest energy state, of the hydrogen atom. But where did this equation come from? Historically, the original equation looks like this

1
λ

=
R

(

1

n

f

2

−

1

n

i

2

)

{\displaystyle \frac{1}{\lambda} = R \left( \frac{1}{{n}\_{f}^{2}}-\frac{1}{{n}\_{i}^{2}}\right) }

which we now know can be rearranged with a bit of algebra using the Planck relation

E

γ

=
h
ν

{\displaystyle {E}\_{\gamma} = h \nu }

.

This is the famous **Rydberg formula**, which was found empirically based on findings by Balmer (1885), Lyman (1906), and Paschen (1908). In the early 20th century, Niels Bohr formulated his theory of the Hydrogen atom and discovered that the Rydberg constant can in fact be calculated with fundamental constants

# R

m

e

e

4

8

ϵ

0

2

h

3

c

=
1.097
×

10

7

m

−
1

.

{\displaystyle R = \frac{{m}\_{e}{e}^{4}}{8{\epsilon}\_{0}^{2}{h}^{3}c} = 1.097 \times {10}^{7} \; {m}^{-1}. }

## **Problems**

### **Problem 1:** **3D Quantum Particle in a Box**

Imagine a box with zero potential enclosed in dimensions

(

0
<
x
<
a

)

,

(

0
<
y
<
b

)

,

(

0
<
z
<
c

)

{\displaystyle \left(0 < x < a \right), \left(0 < y < b \right), \left(0 < z < c \right) }

. Outside the box is the region where the particle’s wavefunction does not exist. Hence, the potential outside the box be infinite. Obtain the wavefunction of the particle in the box where the potential function is

# V

{

0

Inside the box

∞

Outside the box

{\displaystyle V = \begin{cases} 0\quad \quad \text{Inside the box} \\ \infty \quad \text{Outside the box} \end{cases} }

subject to the boundary conditions

{

Ψ
(
0
,
y
,
z
# )
Ψ
(
a
,
y
,
z
# )
0

Ψ
(
x
,
0
,
z
# )
Ψ
(
x
,
b
,
z
# )
0

Ψ
(
x
,
y
,
0
# )
Ψ
(
x
,
y
,
c
# )
0

.

{\displaystyle
\begin{cases}
\Psi(0,y,z) = \Psi(a,y,z) = 0 \\[5 pt]
\Psi(x,0,z) = \Psi(x,b,z) = 0 \\[5 pt]
\Psi(x,y,0) = \Psi(x,y,c) = 0 \\[5 pt]
\end{cases}.
}

Obtain the time-independent wavefunction of the particle in the box.

Solution