The Quantum Theory ofAtoms and Molecules

The Schrödinger equation and

how to use wavefunctions

Dr Grant Ritchie

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 Schrödinger equation and its solution is called the wavefunction, .

What is the form of the Schrödinger equation ?

The classical wave equation

2

2

22

2 1

tvx

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

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

Try a solution: e.g. )(),( tkxietx

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

An alternative….

tx

2

2

)(),( tkxietx

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 in free space is:

For )(),( tkxietx ),(),(2

22

txtxm

k then we have

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

The time-dependent Schrödinger 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 convenience – see later.

The Hamiltonian operator

HtxV

xmtxV

xmˆ),(

2),(

2 2

22

2

22

)(ˆ2

ˆ)(

2ˆ

2

2

22

xVm

pxV

xmH x

xipx

ˆ

LHS of TDSE can be written as:

where Ĥ is called the Hamiltonian operator which is the differential operator that represents the total energy of the particle.

Thus

where the momentum operator is

Thus shorthand for TDSE is:t

iH

ˆ

Solving the TDSE – Aaargh!

)(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 t while RHS involves variation of with x. Hence look for a separated solution:

t

TiTxV

xT

m

)(

2 2

22then

Now divide by T:t

T

TixV

xm

1

)(1

2 2

22

LHS depends only upon x, RHS only on t. True for all x and t so both sides must equal a constant, E (E = separation constant).

Thus we have:

ExVxm

Et

T

Ti

)(1

2

1

2

22

Time-independent Schrödinger equation

Solving the time equation:

/)(

1 iEtAetTdtiE

T

dTE

dt

dT

Ti

This is exactly like a wave e-it with E = ћ. Therefore T(t) depends upon the energy E.

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

The space equation becomes EHExV

xm

ˆ)(2 2

22

or

This is the time independent Schrödinger equation (TISE) .

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

The TISE can often be very difficult to solve – it depends upon V(x)!

Eigenvalue equations

The Schrödinger Equation is the form of an Eigenvalue Equation: EH ˆ

where Ĥ is the Hamiltonian operator,

is the wavefunction and is an eigenfunction of Ĥ;

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

)(2

ˆˆˆ2

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.

TISE for a free-particle

E

xm

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 / ћ E = ћ !

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

WAVE-PARTICLE DUALITY!

Interpretation of (x,t)

As mentioned previously the TDSE has solutions that are inherently complex (x,t) cannot be 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 probability per unit length (or volume in 3d).

The Born interpretation therefore calls the probability amplitude, * (= P(x,t) ) the probability density and * dx the probability.

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

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 in many cases precise values cannot, even in principle, be determined – see later.

Normalisation

1),()(2dxtxdxxP

Total probability of finding a particle anywhere must be 1:

This requirement is known as the Normalisation 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 of N normalises the function N x (x L) of 0 x L?

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.

Boundary conditions for

In order for to be a solution of the Schrödinger equation to represent a physically observable system, must satisfy certain constraints:

1. Must be a single-valued function of x and t;2. Must be normalisable; This implies that the 0 as x ;3. (x) must be a continuous function of x;4. The slope of must be continuous, specifically d (x)/dx must be

continuous (except at points where potential is infinite).

(x)

x

(x)

x

(x)

x

(x)

x

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.

What other information can you get from ? (and how!)

)(2

2ˆ

2

2

xVdx

d

mH

.ˆ

xipx

nnnnnnnnn

nnn

ExExExH

EH

dddˆ

ˆ

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 dˆ

Momentum and energy expectation values

dxtxxi

txpx ),(),(*

where .ˆxi

px

is the operator for the x component of momentum.

Example: Derive an expression for the average energy of a free particle. m

pE

2

2

thenm

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 of momentum involves the representation of momentum as a quantum mechanical operator:

Our definition of the expectation value is one of the postulates of QM – see later lectures.

Summary

titxV

xm

),(

2 2

22

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

1),()(2dxtxdxxP

TDSE:

Born interpretation:

Normalisation:

TISE: EHExV

xm

ˆ)(2 2

22

or

/)()()(),( iEtextTxtx

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

Expectation value of operator Ω:

dxtxtx ),(ˆ),(

It’s never as bad as it seems….