Schrödinger eqn. for the atomQuantum Physics Lecture 20B

7/28/2019 Schrdinger eqn. for the atomQuantum Physics Lecture 20B

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PHYS 2040Quantum Physics

The Bohr Model, applicability, pros and cons

Choosing a coordinate system

Converting the Schrdinger equation

The Coulomb potential

Solving the Schrdinger equation

Separation of variables

The and equations Legendre polynomials

The R equation

The radial wavefunct ions

Spherical harmonics

PHYS2040 Lecture 20B Schrdinger eqn. for the atom

Updated: 19/5/2008 8:54 PM

Lecture 20B: Schrdinger eqn. for the atomLecture 20B: Schrdinger eqn. for the atom


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Last week we developed Bohrs model of the atom, where electrons orbit a positivenucleus in orbits that have an angular momentum quantized in integer multiples ofh.We found that this gives us allowed radii and energy levels:

The Bohr modelThe Bohr model

Bohrs model of the atom explained two key experimental findings when it wasdeveloped, these were:

where: a0 = 0h2/me2 = 0.53 is called the Bohr radius(19.6)

202

2

2

0 n

Z

an

mZe

hrn ==

(19.11)2

0

2

222

0

42

8 n

EZ

nh

emZEn ==

E0 = e2/80r0 = me4/80h2 = 13.6eV is called the Rydberg energy

and n is the quantum number of the electrons orbit.

1. Rutherfords experiments showing unexpectedly high backscattering of-particles(4He2+ ions) from a thin gold foil (like a bullet bouncing of tissue paper).

2. Experiments on the optical emission and absorptionspectra of gases showing that the wavelengths ofthe spectral lines depends only on a single constantand two integers (i.e., the Rydberg-Ritz equation)

(19.14)

= 22

111

mnR


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Applicability, pros and consApplicability, pros and cons

Bohrs model works remarkably well for one electron atoms, especially if you correctfor the finite mass of the nucleus (see E&R). Some examples are:

1. The Hydrogen atom one electron orbiting one proton.

2. The Helium He+ ion one electron orbiting two protons and two neutrons.

3. The Lithium atom a helium atom with an electron orbiting it and so on throughthe group I elements.

4. The Beryllium Be+ ion and so on through the group II elements.

It gets systematically less accurate as you work your way down that lis t (and towardshigher mass), and it fails entirely for most of the periodic table, but its a good start.Some pros and cons of Bohrs model:

Pros: Its simple, very simple, but gives remarkably accurate answers for simple

systems.

Cons: Its a little too simple, so it only works for simple systems. Its also purelyradial, which means it cant describe the angular distribution of things like chemicalbonds


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Applicability, pros and consApplicability, pros and cons

Cons: Its a little too simple, so it only works for simple systems. Its also purelyradial, which means it cant describe the angular distribution of things like chemicalbonds

Chemistry at the time had well

established the asymmetry ofcertain atoms in forming chemicalbonds.

This suggests that the Bohrmodel is not really wrong (after

all, it works for simple cases), butits not a complete model of howthe atom works.

Today, were going to start on the

next step, applying theSchrodinger equation to theHydrogen atom, with the intentionof getting a better model of theatom.


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Choosing a coordinate systemChoosing a coordinate system

An important thing in physics is setting up a problem properly. If you do it right, aproblem is rather easy, if you do it wrong, the problem is near impossible.

So far, weve been using the Cartesian or rectangular coordinate system (x,y,z) butthis is a poor choice for a problem such as this one, where we have a clear potentialfor spherical symmetry (i.e., a partic le orbiting another in 3 dimensions). Hence, from

here forth, we will use spherical coordinates (r,,).


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Converting the Schrdinger equationConverting the Schrdinger equation

So far, weve only dealt with the Schrdinger equation in one-dimension, it has theform:

The energies in different directions add directly as scalars, so writ ing the Schrdingerequation in three-dimensions is pretty easy:

(10.1))()()()(

2 2

22

xExxVdx

xd

m

=+

h

(19A.1)),,(),,(),,(),,(2 2

2

2

2

2

22

zyxEzyxzyxVzyxdz

d

dy

d

dx

d

m =+

++

h

It is often convenient to wri te this as:

(19A.2)),,(),,(),,(),,(2

22

zyxEzyxzyxVzyxm

=+h

which is called the Laplacian operator, or del-squared, in rectangular coordinates.

where: (19A.3)2

2

2

2

2

22

dz

d

dy

d

dx

d++=


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Converting the Schrdinger equationConverting the Schrdinger equation

We now need to convert the 3D Schrdinger equation (Eqn 19A.2) into sphericalcoordinates. The main thing we need to do is convert the Laplacian to sphericalcoordinates. You should be able to do this (its just the chain rule, applied numeroustimes if you want to work through it, see Appendix M of Eisberg and Resnick). Theresult is:

(19A.4)),,(),,(),,(),,(2

22

rErrVrm

=+h

which is called the Laplacian operator, or del-squared, in spherical coordinates.

where:

(19A.5)2

2

222

2

2

2

sin

1

sinsin

11

+

+

= rrrrrr

Right now, it might look like Im just making the problem more complicated, but thats

not true, its just a complicated problem over all. If I stayed in rectangular coordinatesthen sure, 2 would look nicer, but Id end up with a lot of sin/cos terms later on, Iwont be able to perform the separation of variables step (coming up) and I wouldhave a very hard time comparing the results directly to Bohrs model also.


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The Coulomb potentialThe Coulomb potential

Of course, the next step is the potential. There is only one force on the electron in thisproblem the Coulomb attraction between the negative electron and the positivenucleus. Because this depends only on the separation between the charges, F andhence V are purely radial (no terms in and ). So, as a force:

(19A.6)( ) ( ) rrZeeF

4 20+=

r

and therefore as a potential:

(19A.7)

r

ZedrrFrV

0

2

4

.)(

==

r

Note that there is a Z in here. Z is the atomic number (each element has a number ofprotons and hence a nucleus charge equal to its atomic number) and it allows us tomake our model applicable to atoms beyond hydrogen, but

This problem stil l only works for one electron (if we have more, they can interact witheach other and we need to account for that), hence the single negative charge. Butyoull see later, we can still infer results about more complex atoms despite thislimitation.


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Solving the Schrdinger equationSolving the Schrdinger equation

So now we have the complete Schrdinger equation for the one-electron atom (with Zprotons in it):

Solving i t is no easy task, but there is one aspect that makes it infinitely easier. This

is that the potential V depends only on r, it has no terms in or. This means we canuse the same trick we did earlier in developing the time-independent Schrdingerequation Separation of variables.

(19A.8)),,(),,(4

),,(2 0

22

2

rErr

Zer

m=

+

h

where:

(19A.5)2

2

222

2

2

2

sin

1sin

sin

11

+

+

=

rrrr

rr

(19A.9))()()(),,( = rRr

In other words, the wavefunction solution to Eqn. 19A.8 is composed of the productof three separate functions, each of which is an independent function of one of thethree spatial variables (n.b., this is the main reason why we switch to spherical vars.)


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Separation of variablesSeparation of variables

If we now substitute our separated solution (Eqn 19A.9) into Eqn 19A.8, and carry outthe three derivatives:

(19A.10)

(19A.11)

r

R

r

=

Notice that the left side has no terms in rand , and the right no terms in . Hence wecan separate this into two equations, just like we did last time in lecture 10. Thecommon value must be a constant, and well designate it as m l

2 for reasons well seelater on.

=

R

2

2

2

2

=

R

we get:

=

+

+

ERR

r

Ze

d

d

r

R

d

d

d

d

r

R

dr

dRr

dr

d

rm 0

2

2

2

222

2

2

2

4sinsin

sin2

h

And if we then multiply both sides by 2mr2sin2/Rh2:

[ ])(sin2

sinsinsin1 22

2

22

2

2

rVErm

d

d

d

d

dr

dRr

dr

d

Rd

d

=

h


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Separation of variablesSeparation of variables

And so we have our first of three equations:

(19A.12)

(19A.13)

And now again we can use the same trick! The equation on the left is independent ofand the equation on the right is independent of r, and so we can separate them andassume both sides are equal to a constant, which we will designate as l (l + 1), againfor reasons that we will soon see. So now we have three equations from which to getthree unknown solutions R(r), () and (), these are:

=

22

2

lm

and:

we can divide through by sin2, and then collect terms as fol lows:

[ ] 2222

22

)(sin2

sinsinsin

lmrVErm

d

d

d

d

dr

dRr

dr

d

R=

h

(19A.14)[ ]

2

2

2

2

2

sinsin

sin

1)(

21 lm

d

d

d

drVEr

m

dr

dRr

dr

d

R+

=+

h


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The three equationsThe three equations

Note that with the last one, Ive multiplied through by R/r2 for convenience.

(19A.12)

(19A.15)

In solving these equations, we will find that the equation for () has acceptablesolutions only for certain values ofm l. Using these values in the equation for(), wefind that this equation has acceptable solutions only for certain values of l. And

finally, using these values in the equation for R(r), this equation is found to haveacceptable solutions only for certain values of the total energy E. In other words, theenergy of the atom is quantised.

= 22

2

lm

(19A.16)[ ] 222

2)1()(

21

r

RllRrVE

m

dr

dRr

dr

d

r+=+

h

+=

+

)1(sinsinsin

12

2

ll

m

d

d

d

d l

And now for the downhill run, lets solve these equations one by one


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The equationThe equation

This ones a piece of cake, weve been solving it (or its kind at least) for weeks, thesolution is:

(19A.12)

Now this obeys the same rules as any wavefunction, namely that its fini te, continuousand single-valued. Now = 0 and = 2 are actually the same angle (in fact it goesdeeper than that, as an angle is modulo 2) therefore:

= 22

2

lm

The quantum numberm l is called the magnetic quantum number.

(19A.17)

llim

m e= )(

(19A.18))2()0( = 20 ll imim ee =

which means m l must be an integer:

,...3,2,1,0=lm,...3,2,1,0,1,2,3..., =lm(19A.19)

or

or

The subscript m l is used to identify the specific form of an acceptable solution.


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The equationThe equation

This equation is not so easy to solve, so I wont go into it deeply. The solutionprocess is s imilar to that used in the simple harmonic osci llator. If youre interested inseeing it all, Appendix N in Eisberg and Resnick discusses it in full detail.

(19A.15)

Firstly, the solutions that are found to be acceptable (i.e., remain finite) are onlyobtained if the quantum number l (called the azimuthal quantum number) is equal toone of the integers:

And the wavefunction solution has the form:,...3,2,1, +++= llll mmmml (19A.20)

)(cossin)( ,, ll

l mlm

ml F= (19A.21)

where Fl,m(cos ) are called Legendre polynomials, which look like

+=+

)1(sin

sinsin

12

2

llm

d

d

d

d l


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Legendre polynomialsLegendre polynomials

Note well, youll sometimessee these with differentnotation, sometimes P,sometimes F, keep youreyes peeled .


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The R equationThe R equation

Same situation forR, the solution is involved and you can find it in detail in AppendixN of Eisberg and Resnick if you want to see it. But ultimately, we find bound-statesolutions that are acceptable (remain finite) only if the constant E (the total energy)has one of the values En, where:

(19A.16)[ ] 222

2)1()(

21

r

RllRrVE

m

dr

dRr

dr

d

r+=+

h

And remarkably, E1 = 13.6eV! In other words, we have exactly the same energyeigenvalues that we got in Bohrs model. The quantum numbern (called the principal

quantum number) is given by:

(19A.22)2 02

222

0

42

2)4( nEZ

nemZEn == h

,...3,2,1 +++= llln (19A.23)


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A quick look at the energy levelsA quick look at the energy levels


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The radial wavefunctionsThe radial wavefunctions

And the R wavefunction solution has the form:

=

00

,0)(

a

ZrG

a

ZrerR nl

l

naZr

ln(19A.24)

where Gn,l(Zr/a0) are polynomials

Note also that a0 = 40h2/me2 =0.53 is the Bohr radius, thesame characteristic radius thatwe found in the Bohr model.


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Spherical harmonicsSpherical harmonics

And so now weve finished the mathematical aspect of the problem. Before I recap thesolution, Ill make a slight tweak to how we present the wavefunctions:

(19A.25)),()()()()(),,( ,,,,,, llll mllnmmllnmln YrRrRr ==

where we combine the angular terms to create what are known as spherical

harmonics Y(,):


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Summary of the solutionsSummary of the solutions

So ultimately, we have eigenfunctions that look like:

(19A.26) ll

l

l

im

ml

mnaZr

nl

l

mln eFea

ZrG

a

Zrr )(cossin),,(

,

00

,,0

=

with energy eigenvalues:

(19A.22)20

2

222

0

42

2)4( n

EZ

n

emZEn ==

h

and three sets of quantum number relations:

,...3,2,1 +++= llln ,...3,2,1, +++= llll mmmml ,...3,2,1,0=lm

Right now, youre probably feeling very mathed out . So well leave it there for now,and in the next lecture, we will have take a more pictorial, physical look at what thesesolut ions really mean.

which determine what are acceptable solutions to the Schrdinger equation.


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The Bohr model works very well in a few very simple cases, and reasonably well inother cases, but its a very simple model so i ts predictive power is rather limi ted. Inthis lecture, we look to (a) test it and (b) extend it by using the Schrdinger equationas a parallel approach to modeling the one-electron atom. (n.b., proper modeling ofmultielectron atoms needs some more advanced techniques/considerations).

For the atomic Schrdinger equation, the potential is due to the Coulomb attractionbetween a nucleus of charge +Ze and an electron of charge e, hence V(r) = Ze2/40r.

This problem is best dealt with in spherical coordinates, mainly because the potentialin this case is spherically symmetric. This then allows us to use separation of

variables to turn a really hard problem into three slightly less hard problems. This is aclassic approach in physics, which is why I really wanted to demonstrate it how asmart choice in setting up the problem can yield good results.

We get three equations that give us eigenfunction solutions R(r), () and (), whichhave acceptable solutions for certain values of three quantum numbers n,l and m l.

Remarkably, the energy eigenvalues for this problem En = 13.6eV/n2 are exactly thesame as those in Bohrs model, which demonstrates that the two theories in principleagree. As well see next week, the Schrdinger model explains some things thatBohrs model cant, namely the geometry of chemical bonds in certain molecules.

SummarySummary

= Brooks/Cole - Thomson

Schrödinger eqn. for the atomQuantum Physics Lecture 20B

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