Hydrogen Atom inWave Mechanics

The good news is that the Schroedinger equation for the hydrogen atomhas an EXACT ANALYTICAL solution! (this is one of the few problems in Quantum Mechanics that does have such a solution – most problems in QM cannot be solved exactly). The bad news, however, is that the procedure of solving the equationis extremely complicated. In a standard QM textbook it usually spansover about forty pages loaded with math. In order to clearly understandthe solution procedure, one has to be familiar with elements of highly advanced calculus, such as functions called the “spherical harmonics” and things called the “Laguerre polynomials”. This is much beyond the mathematical preparation level required for the Ph314 course. Therefore, we will not go through the details of the solution procedure – we will onlyreview the general properties of the solutions, and you will be asked to“accept without proof” a number of things.

222

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1),,(

),,(2

zyx

ezyxU

EzyxUzyxm

The equation cannot be solved analytically in Cartesian coordinates.However, if we use spherical polar coordinates, the equation beco-mes more complicated, but solvable (although it’s still not an easy task!):

)()()(),,(

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Schroedinger Equation in three dimensions

The solutions of the Schroedinger Equation for the hydrogen atomcontains the variables r, , and , and fundamental constants, such as the the electron mass and charge, the Planck Constant, speed of light, and ε0 . In addition, three parameters emerge in a natural way as “indices” or “labels” for the solutions. They are listed below:

. ,,3 ,2 ,1 ,0

are valuesallowed Its . the

.1 , ,3 ,2 ,1 ,0 are ues val

allowed Its . the

model.Bohr in the as same y theessentiall isIt . 3,2,1

of valuescan take which , the

lm

numberquantummagneticm

nl

numberquantummomentumangularl

nn

numberquantumprincipaln

l

l

Complete with these quantum numbers, the separated solutions ofthe Schroedinger Equation can be wrtiien as:

)()()(),,( ,,,, lll mmllnmln rRr

The principal quantum number n specifies the energy level, justas in the Bohr model. Te quantized energy levels obtained by solving the Schroedinger Equation are given by exactly the same formula as in the Bohr model:

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2

4 eV 6.131

32 nn

meEn

(1, 0, 0)

(2, 0, 0)

-13.6eV

-3.4 eV(2, 1, 1) (2, 1, 0) (2, 1, -1)

-1.5 eV(3, 0, 0) (3, 1, 1) (3, 1, 0) (3, 1, -1) (3, 2, 2) (3, 2, 1) (3, 2, 0) (3, 2, -1) (3, 2, -2)

Each quantum state of the electron in the Wave Mechanics Model of the hydrogen atom can be described by a “triad” of integers: (n, l, ml ). The energy of the state depends only on the first of thesenumbers, but otherwise they are all different quantum states.All states for n=1, 2, and 3 are shown below:

Remember the notion of DEGENERACY we talked aboutsome time ago? What is the degeneracy of the n = 1 state?Of the states with n = 2? Of the states with n = 3?

Radial probability densitiesFirst, let’s shortly recapitulate what we told about probability density.Recently, we talked about the probability of finding a particle in thex region between x and x + dx in a 1-D quantum well. This probabilityIs related to the particle-in-the-well wavefunction as:

dxxdxxP2

)()( In other words, the squared modulus of the wave function gives the probability density.

In the case of a 2-D quantum well, we are interested in the proba-bility of finding the particle in a “box”, or surface area element defined by regions (x, x + dx) and (y, y + dy) . This probability is givenby:

Again, the same rule is valid – the squared modulus of the wave function is the probability density.

dxdyyxdxdyyxP2

),(),(

2

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1

1

1

2

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widthfinite

In the case of the hydrogen atom the “philosophy” is essentially the same.The squared modulus of the wavefun-ction:

gives the probability density of findingthe electron at the location

To compute the actual probability ofFinding the electron in a volume elementdV located atwe multiply the probability density by this volume element. In spherical polar

coordinates the volume element is:

2),,( r

).,,( r

),,,( r

ddrdrdV sin2

)()()(),,( ,,,, lll mmllnmln rRr

A moment ago we said that the solution of the Schroedinger EquationFor the hydrogen atom can be separated into three functions, eachone depending only on a single variable – which can be written as:

ddrdrdVrRrlll mmllnmln sin and )()()(),,(

:page preceding thefrom Repeated2

,,,,

Therefore, the probability is:

ddrdrrRdVrlll mmllnmln sin )()()( ),,( 222

,

2

,

2

,,

Using this expression, one can calculate many “goodies”, or many features of the hydrogen atom.

However, such calculations are not trivial, and we will use them to determine only a single function of interest – namely, the so-called radial probability density.

rdr

Nucleus

Imagine that the nucleus (proton) issurrounded by a thin spherical “shell”of radius r and thickness dr . We want to determine the probabilityof finding the electron within the volumeof this shell.

ddrdrrRdVrlll mmllnmln sin )()()( ),,( 222

,

2

,

2

,,

Repeated from the last slide:

In order to find the probability in question, we have to integrate overthe other two variables, and :

2

0

2

0

2

,

2

,2 )( sin)()( )( dddrrRrdrrP

ll mmlln

This formula looks scary, right? Fortunately, however, it is not scaryat all, because in the separated wavefunction each function is indivi-dually normalized, so that both integrals are equal to unity! Hence, the radial probability density takes a simple form:

drrRrdrrP

rRrrP

ln

ln

2

,2

2

,2

)( )(

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)( )(

:atomhydrogen theof

)( functions waveradial fewFirst rRnl

:hydrogen of stateslowest threefor the

)()( functionsdensity y probabilit radial The2

rRrP nl

Practical example:

(from the final exam in the Fall 2006 Ph314 course).

Since the total wave function of hydrogen is a three-dimensional function, plotting the overall probability density function is a much more complicated task than plotting the radial probability. But the pictures shown here can give you a pretty good idea of how theseFunctions look like for various quantum states of the electron.

Imagine each plotto be rotated by360° about the vertical axis (z).

Another possibleway of illustratinghow the 3-D proba-bility density funct-ions look like is toplot the surfaces ofconstant |ψ(r, , )|2

We already know what role the quantum number n plays:it describes the energy of a given state, and the mathe-matical form of this energy is identical as in the Bohr model.

But what do the other two quantum numbers, l and ml ,

tell us? What is their physical meaning?

The first of these numbers is related to the electron’sangular momentum. Namely, the length of the electron’s angular momentum vector is given by:

)1( llL(sorry, this is another mathematicalequation I have to ask you to “accept without proof”).

12 )13(3 ,3for ; 2 islength vector

momentumangular selectron' the,1 with state ain instance,For

LlL

l

As was said, for states with the quantum number n, l can take values of l = 0, 1, 2, … n-1. As follows from the above, there arestates of the the same energy, but with different angular mo-menta.

This is not a surprising situation in physics. There is a classicalanalog in the planetary motion:

There are many orbits of the same energy, but with differentellipticity. The circular orbit corresponds to the largest angular

momentum. With increasing ellipticity,the angular momentum

decreases.

Circular orbit:Maximum L

Lower ang. momentunthan for circular orbit

Lowest ang. momentumof the three orbits shown

A comet:Very elongated orbit

At this part of theorbit, the comet’smotion is veryslow.

Comets, which are celestial bodies with extremely elongated orbits,spend most time both close to the Sun, or far away from the Sun,in the peripheral region of the Solar System.

Corresponding situation in hydrogen atom

The physical meaning of the ml quantum number: it’s the length ofthe projection of the angular momentum on the z axis.

The projection can only take values that are integer multiplesof ℏ.

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