*©Fernando Caracena* 2013

On a previous blog on the one-dimensional harmonic oscillator, we showed that the algebra of non-commuting operators in quantum mechanics could be exploited systematically to solve for the entire spectrum of state vectors and energy eigenvalues. Algebra sufficed for most of the calculation. In the following series we will develop the techniques of Lie algebra using non-commuting operators to solve for the state vectors and corresponding eigenvalues for the electrons in a hydrogen atom. In doing so, we will cover some techniques for systematically solving problems that have several, simultaneous eigenvalues. In three dimensional problems such a that of atomic structure, angular momentum plays a role in organizing the motion of electrons. We will find that angular momentum is the source of two additional eigenvalues in addition to that of energy.

## The Problem of the Hydrogen Atom

The electron moving in a stationary state in a hydrogen atom satisfies the time independent Schrödinger Equation,

**H**|Φ> =E_{n} |Φ>, (1a)

where the Hamiltonian operator is the following**
H** =

**p**

^{2}/2m – k e

^{2}/

**R**

^{2}, (1b)

and E

_{n}

are its eigenvalues.

### Preliminaries Old Quantum Theory Solution

Bohr solved for the energy eigenvalues as follows by using the old quantum theory, which are negative because they represent bound states of the electron:

E_{n} = –½ (k e^{2})^{2}m_{e}/ħ^{2} (1/n^{2}). (1c)

[Free electron states are represented by positive values.]

The constant k is given by

k=1/(4 πε_{0}).

The expression for energy eigenvalues can be simplified by writing it as

E_{n} = -½ k e^{2}/a_{0} (1/n^{2}), (1d)

where a_{0} is known as the Bohr Radius

a_{0} = ħ/(k e^{2} m_{e})

or

a_{0} = 4πε_{0}ħ/(e^{2} m_{e}). (1e)

Modern quantum theory finds the same energy eigenvalues, but finds that there are two more eigenvalues associated with the angular momentum of the orbital electron.

## Eigenvalues and Casimir Operators

Casimir operators are those of the group of operators that commute among themselves, that is their order of multiplication does not matter. Commutation relations play an important role in modern quantum theory. They are a calculational tools that reduce much of the mathematics of quantum mechanics to algebra.

Commutation relations are written for operators, some of which do not necessarily commute, that is, their product depends on the order that it is taken, such as in the following, in which the commutator is defined as

[**A**, **B**]= **A** **B** – **B** **A**. (2a)

If order does not matter, such as between **X** and **Y**, then

[**X**,**Y**]=0,

their commutator is zero.

Note that we write quantum operators in upper case, bold letters.

The following is a very useful theorem that we will use repeatedly

[**A ****B**, **C**]= [**A**, **C**] **B + ****A**** [B**, **C**]**.** (2b)

Note that the proof falls out just by writing out the operators

**A B****C – C A B = **

**A**

**C**

**B**

**–**C A B + A B**C**

**–****A C B.**

Another almost trivial theorem is left to the reader to verify

[

**A**,

**B**]=

**–**[

**B**,

**A**]. (2c)

Heisenberg's Uncertainty relations apply to expectation values of noncommuting opertors.

If operators, such as **X** and **P**, fail to commute, their values cannot be defined simultaneously with arbitrary precision. For example, Heisenberg's Uncertainty relations apply to components of position and corresponding components of momentum

Δx Δp_{x} ≥ **ħ/2** , (3a)

**Δy Δp**_{y}** ≥ ħ/2** , (3b)

**Δz Δp**_{z}** ≥ ħ/2** , (3c)

where the 'Δ's represent the root mean square (RMS) expectation values defined for x, y and z as

**Δx**^{2}** = <Φ| (X – <Φ|X|Φ>)**

^{2}**|Φ>**, (4a)

**Δy**

^{2}**= <Φ| (Y – <Φ|**|Φ>)

^{2}|Φ>, (4b)

**Δz**

^{2}**= <Φ| (Z**

**–**<Φ|Z|Φ>)

^{2}**|Φ>.**(4c)

Each pair of conjugate variables, such as coordinates and momentum components, are represented by operators that have the commutation relations:

[**X**, **P _{x}**] = i

**ħ**, (5a)

[

**Y**,

**P**] = i

_{y}**ħ**, (5b)

[

**Z**,

**P**] = i

_{z}**ħ.**(5c)

The Heisenberg Uncertainty relations between them result mathematically from the failure of their representative operators to commute, which is not shown here, but perhaps in a later blog.

All of the quantities of a quantum mechanical system that represent variables that can be measured simultaneously with arbitrary precision are represented by operators that commute among themselves, i. e., they are Casimir operators.

### Some further properties involving commutators

Commutators can be used instead of derivatives, thereby reducing a lot of calculation in quantum physics to algebra. For example, (5a) is just the beginning of a series of derivatives that can be taken of the various powers and functions of **X**:

**[X**

^{2},**P**

_{x}**] = i ħ**

**(2X**

**) = i ħ d(**

**X**

^{2}**)/d**(6a)

**X****[**

**X**

^{n},**P**

_{x}**] = i**

**ħ**

**n**

**X**

**=**

^{n-1}

**i ħ**d**X**

^{n}**/d**(6b)

**X****[**f(

**X**

**)**,**P**

_{x}**] =**

**i ħ df(X)/dX.**

**(6c)****Note that (6a) follows directly by applying (2b) then (5a). The reader can prove (6b) by mathematical induction, and (6c) by expanding f(**

**X**) in a Taylor Series in powers of**X**.

*Identifying the Casimir operators for the hydrogen atom*

The energy operator, or Hamiltonian, is a candidate for one of the Casimir operators for the hydrogen atom. What are some other possibilities? A good place to look is at angular momentum, which as an operator is defined as follows

**L** = **R** x **P**.

The components of the angular momentum vector components (**L**** _{x}**,

**L**

_{y, }**L**

**) are each the following operators:**

_{z}**L**

_{x}**=**

**Y**

**P**

_{z}**–**

**Z**

**P**

_{y}**,**(7a)

**L**

_{y}**=**

**Z**

**P**

_{x}**–**

**X**

**P****, (7b)**

_{z}**L**

_{z}**=**

**X**

**P**

_{y}**–**

**Y**

**P****. (7c)**

_{x}Note that the components defined in (7a), (7b) and (7c) do ont commute among themselves, which we will show for **L**_{x}**L**** _{y}**:

[

**L**

_{x}**,**

**L**

**] =**

_{y}

**Y**

**[P**

_{z}**,**

**Z****]**

**P**

_{x}**+**

**P**

_{y}

**X**

**[Z, P**

_{z}**;**

**]**[

**L**

_{x}**,**

**L**

**] = -i**

_{y}**ħ**

**Y**

**P****+**

_{x}

**i****ħ**

**P**

_{y}**;**

**X**[

**L**

_{x}**,**

**L**

**] =**

_{y}

**i****ħ**[

**X**

**P**

_{y}**-**

**Y**

**P****] ;**

_{x}or

[

**L**

_{x}**,**

**L**

**] =**

_{y}

**i****ħ**

**L**

**. (8a)**

_{z} It is left as an exercise for the reader to derive the following two commutation relations:

[**L**_{y}**, ****L**** _{z}**] =

**i****ħ**

**L**

**, (8b)**

_{x}**[L**

_{z}**,**

**L**

_{x}**] =**

**i****ħ**

**L**

**. (8c)**

_{y}Note that each component commutes with itself,

[**L**_{x}**, ****L**** _{x}**] =0 , (8d)

[

**]= 0 , (8e)**

**L**_{y}**,****L**_{y}[

**]= 0 . (8f)**

**L**_{z}**,****L**_{z}It is left as an exercise for the reader to show why.

Nest we show that **L**_{z}** commutes with **** L^{2}**, where

**L**^{2}**=**

**L**_{x}**+**^{2}

**L**_{y}**+**^{2}**L**

_{z}**. (9a)**

^{2}This is easy, by showing that

**L**_{z}_{ }commutes with the square of each component:

[L

[L

_{z}**,**

**L**

_{x}

^{2}**] = [L**

_{z}**,**

**L**

_{x}**] L**

_{x}**+ L**

_{x}**[L**

_{z}**,**

**L**

_{x}**]**; (9b)

**[L**

_{z}**,**

**L**

_{x}

^{2}**] =**

**i****ħ (**

**L**

_{y }**L**

_{x }**+ L**

_{x}**L**

_{y}**)**; (9c)

**[L**

_{z}**,**

**L**

_{y}

^{2}**] = [L**

_{z}**,**

**L**

_{y}**] L**

_{y}**+ L**

_{y}**[L**

_{y}**,**

**L**

_{x}**]**; (9d)

**[L**

_{z}**,**

**L**

_{y}

^{2}**] = -**

**i****ħ (**

**L**

_{x}**L**

_{y }**+ L**

_{y }**L**

_{x }**)**;

[

**L**

_{z}**,**

**L**

_{z}**] = 0 . (9e)**

^{2}It follows that

**[L**

_{z}**,**

**L**

^{2}**] =0 .**(9f)

Finally, we will show that the radius of the orbit (

**R**) commutes with all angular momentum components:

**R= √[X**

^{2}

**+ Y**

^{2}

**+Z**

^{2}**; (10a)**

**]****∂**

**R****/∂**

**X = X****/**

**; (10b)**

**R****∂**

**R****/∂**

**Y**

**= Y****/**

**; (10c)**

**R****∂**

**R****/∂**

**Z**

**= Z****/**

**; (10d)**

**R****[**

**L**

_{z}**,f(**

**R****)] = X [P**

_{y}**,**

**f(**

**R****)]**

**–**

**Y****[**

**P**

_{x}**, f(**

**R)****]**; (10e)

**[**

**L**

_{z}**,f(**

**R****)] =**

**– i**

**ħ**

**X Y****d**

**/ R**

**f****(**

**R****)**

**d**

**/**

**R****+ i**

**ħ**

**Y****d**

**X / R**

**f****(**

**R****)**

**d**

**/****; (10f)**

**R****[**

**L**

_{z}**,f(**

**R****)] = 0**; (10g)

It is left as an exercise for the reader to show that

**[****L**_{x}**,f(****R****)] = 0** ; (10h)

and

**[****L**_{y}**,f(****R****)] = 0** ; (10i)

It follows that

[**f(****R****)****, ****L**** ^{2}**] =0. (10j)

Finally, we show that

**P****commutes with every component of angular momentum**

^{2}**[**

**L**

_{z}**,**

**P**

_{x}

^{2}**] = 2 [**

**X,**

**P**

_{x}**]**

**P**

_{x }

**P****(11a)**

_{y}**[**

**L**

_{z}**,**

**P**

_{z}**] = 0**(11b)

**[**

**L**

_{z}**,**

**P**

_{z}

^{2}**] = 0**(11c)

**[**

**L**

_{z}**,**

**P**

_{x}

^{2}**] = 2 iħ**

**P**

_{x }

**P****(11d)**

_{y}**[**

**L**

_{z}**,**

**P**

_{y}

^{2}**] = [**

**X**

**P**

_{y}**,**

**P**

_{y}

^{2}**]**

**– [**

**Y,**

**P**

_{y}

^{2}**]**

**P****(11e)**

_{x}**[**

**L**

_{z}**,**

**P**

_{y}

^{2}**] = 0 -2 iħ**

**P**

_{y }

**P****(11f)**

_{x}**[**

**L**

_{z}**,**

**P**

_{x}

^{2 }

**+ P**

_{y}

^{2}**] = 2 iħ [**

**P**

_{x }**,**

**P**

_{y}**]**. (11g)

**It follows that**

**[**

**L**

_{z}**,**

**P**

^{2}**] = 0.**(11h)

**It is left as an exercise for the reader t show that**

**[**

**L**

_{x}**,**

**P**

^{2}**] = 0**

**(11j)**

**and**

[

[

**L**

_{y}**,**

**P**

^{2}**] = 0.**(11k)

Using the form of the Hamiltonian (1b) it follows that

[**L**** _{x}**,

**H**] = 0 ; (12a)

[

**L**

**y**,

**H**] = 0 ; (12b)

[

**L**

**,**

_{z}**H**] = 0 ; (12c)

and therefore

[

**L**

**,**

^{2}**H**] = 0. (12d)

The Casimir operators for the hydrogen atom are therefore **H**, **L**** ^{2}**, and

**L**

**. The choice of the latter component of angular momentum is arbitrary, but by convention it is taken as**

_{z}**L**

**.**

_{z}In the next blog in this series on the hydrogen atom, we will derive solutions for three simultaneous eigenvalues of the Casimir operators:

**H**|Φ> =E_{n} |Φ>, (12e)

**L**** ^{2}**|Φ> =α

_{ }|Φ>

**L**

**|Φ> =β**

_{z}_{ }|Φ>. (12g)