The Quantum Theory of Atoms Schrodinger Equation 2009

8/3/2019 The Quantum Theory of Atoms Schrodinger Equation 2009

1/19

The Quantum Theory of

Atoms and Molecules

The Schrdinger equation and

how to use wavefunctions

Dr Grant Ritchie


2/19

An equation for matter waves?

De Broglie postulated that every particles has an associated wave of wavelength:

ph/

Wave nature of matter confirmed by electron diffraction studies etc (see earlier).If matter has wave-like properties then there must be a mathematical function that is the

solution to a differential equation that describes electrons, atoms and molecules.

The differential equation is called the Schrdinger equation and its solution is called the

wavefunction,.

What is the form of the Schrdinger equation ?


3/19

The classical wave equation

2

2

22

21

tvx

We have seen previously that the wave equation in 1d is:

Where v is the speed of the wave. Can this be used for matter waves in free space?

Try a solution: e.g.)(

),(tkxi

etx

Not correct! For a free particle we know thatE=p2/2m.


4/19

An alternative.

tx

2

2

)(

),(tkxi

etx

ti

xm

2

22

2

Try a modified wave equation of the following type:

(is a constant)

Now try same solution as before: e.g.

Hence, the equation for matter waves infree space is:

For )(),(tkxi

etx ),(),(

2

22

txtxm

k

then we have

which has the form: (KE) wavefunction = (Total energy) wavefunction


5/19

The time-dependent Schrdinger equation

),(2

2

txVm

pE For a particle in a potential V(x,t) then

and we have (KE + PE) wavefunction = (Total energy) wavefunction

titxV

xm

),(

2 2

22

TDSE Points of note:

1. The TDSE is one of the postulates of quantum mechanics. Though the SE

cannot be derived, it has been shown to be consistent with all experiments.2. SE is first order with respect to time (cf. classical wave equation).

3. SE involves the complex number i and so its solutions are essentially complex.

This is different from classical waves where complex numbers are used imply

for conveniencesee later.


6/19

The Hamiltonian operator

HtxV

xmtxV

xm),(

2),(

2 2

22

2

22

)(2

)(2

2

2

22

xVm

pxV

xmH x

xipx

LHS of TDSE can

be written as:

whereis called theHamiltonian operator which is the differential operator that

represents thetotal energy of the particle.

Thus

where themomentum operator is

Thus shorthand for TDSE is:t

iH


7/19

Solving the TDSEAaargh!

)(

2 2

22

xVxmt

i

)()(),( tTxtx

Suppose the potential is independent of

time i.e. V(x, t) = V(x) then TDSE is:

LHS involves variation of with twhile RHS involves variation of withx. Hence

look for a separated solution:

t

TiTxV

xT

m

)(

2 2

22then

Now divide byT:t

T

TixV

xm

1)(

1

2 2

22

LHS depends only uponx, RHS only on t. True for allx and tso both sides must equal a

constant,E(E= separation constant).

Thus we have:

ExVxm

Et

T

Ti

)(1

2

1

2

22


8/19

Time-independent Schrdinger equation

Solving the time equation:

/)(

1 iEtAetTdt

iE

T

dTE

dt

dT

Ti

This is exactly like a wave e-it withE = . Therefore T(t) depends upon the energyE.

To find out what the energy actually is we must solve the space part of the problem....

The space equation becomes

EHExVxm

)(

2 2

22

or

This is the time independentSchrdinger equation(TISE) .

** All Prelims problems are concerned with solving TISE rather than the

TDSE!

The TISE can often be very difficult to solveit depends upon V(x)!


9/19

Eigenvalue equations

The Schrdinger Equation is the form of anEigenvalue Equation: EH

where is the Hamiltonian operator,

is the wavefunction and is an eigenfunctionof;

Eis the total energy (T+ V) and an eigenvalue of.Eis just a constant!

)(2

2

22

xVdx

d

mVTH

Later in the course we will see that the eigenvalues of an operator give the possible

results that can be obtained when the corresponding physical quantity is measured.


10/19

TISE for a free-particle

Exm

2

22

2

)/()()(),(

EtkxietTxtx

For a free particle V(x) = 0 and TISE is:

and has solutions

Thus the full solution to the full TDSE is:

m

kEwhereeore

ikxikx

2

22

Corresponds to waves travelling in either x direction with:

(i) an angular frequency, =E/ E= !

(ii) a wavevector, k= (2mE)1/2 / = p /p = h /!

WAVE-PARTICLE DUALITY!


11/19

Interpretation of(x,t)

As mentioned previously the TDSE has solutions that are inherently complex (x,t)

cannotbe a physical wave (e.g. electromagnetic waves). Therefore how can (x,t)

relate to real physical measurements on a system?

The Born Interpretation

dxtxPdxtxdxtxtx ),(),(),(),(2*

* is real as required for a probability distribution and is the probabilityper unitlength (or volume in 3d).

The Born interpretation therefore calls theprobability amplitude, * (= P(x,t) )

theprobability density and * dx theprobability.

Probability of finding a particle in a small length dx at position x and time t is equal to


12/19

Expectation values

xxPxxi

ii )(

dxtxxdxxxPxx2

),()(

Thus if we know (x, t) (a solution of TDSE), then knowledge of* dx allows the

average position to be calculated:

In the limit that x 0 then the summation becomes:

dxtxxdxxPxx2222

),()(Similarly

The average is also know as the expectation value and are very important in quantum

mechanics as they provide us with the average values of physical properties because inmany cases precise values cannot, even in principle, be determinedsee later.


13/19

Normalisation

1),()(2dxtxdxxP

Total probability of finding a particle anywhere must be 1:

This requirement is known as theNormalisation condition. (This condition arises

because the SE is linear in and therefore if is a solution of TDSE then so is c

where c is a constant.)

Hence if original unnormalised wavefunction is (x,t), then the normalisation integral is:

dxtxN22

),(

And the (re-scaled) normalised wavefunction norm= (1/N) .

Example 1: What value ofNnormalises the functionN x (x L) of 0 xL?

Example 2: Find the probability that a system described by the function 21/2sin (x) where

0 x 1 is found anywhere in the interval 0 x 0.25.


14/19

Boundary conditions for

In order for to be a solution of the Schrdinger equation to represent a physicallyobservable system, must satisfy certain constraints:

1. Must be a single-valued function ofx and t;

2. Must be normalisable; This implies that the 0 asx;

3. (x) must be a continuous function ofx;

4. The slope ofmust be continuous, specifically d(x)/dx must be

continuous (except at points where potential is infinite).

(x)

x

(x)

x

(x)

x

(x)

x


15/19

Stationary states

/)()()(),(

iEtextTxtx

Earlier in the lecture we saw that even when the potential is independent of time the

wavefunction still oscillates in time:

Solution to the full TDSE is:

But probability distribution is static:

2//2)()()(*),(),( xexextxtxP

iEtiEt

Thus a solution of the TISE is known as a Stationary State.


16/19

What other information can you get from ?

(and how!)

)(

2

2

2

2

xV

dx

d

m

H

.xi

px

nnnnnnnnn

nnn

ExExExH

EH

ddd

We have seen how we can use the probability distribution to calculate the

average position of a particle. What happens if we want to calculate the average

energy or momentum because they are represented by the following differential

operators:

Do the operators work on , or on , or on alone?

Take TISE and multiply from

left by and integrate:

NB is normalised.

Suggest that in order to calculate the

average value of the physical quantity

associated with the QM operator we

carry out the following integration:

xnn d


17/19

Momentum and energy expectation values

dxtxxi

txpx ),(),(*

where .

xipx

is the operator for thex component of momentum.

Example: Derive an expression for the average

energy of a free particle. m

pE

2

2

then

m

pE

2

2

Since V= 0 the expectation value for energy for a particle moving in one dimension is

dxtxxm

txE ),(2

),(*2

22

The expectation value ofmomentum involves the representation of momentum as a

quantum mechanical operator:

Our definition of the expectation value is one of the postulates of QMsee later

lectures.


18/19

Summary

titxV

xm

),(2 2

22

dxtxPdxtxdxtxtx ),(),(),(),(2*

1),()(2dxtxdxxP

TDSE:

Born interpretation:

Normalisation:

TISE:

EHExVxm

)(

2 2

22

or

/)()()(),(

iEtextTxtx

Boundary conditions on : single-valued, continuous, normalisable, continuous first derivative.

Expectation value of operator:

dxtxtx ),(),(


19/19

Its never as bad as it seems.

The Quantum Theory of Atoms Schrodinger Equation 2009

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