Thanks to Zarah

Works for light (photons), why doesn’t it work for electrons?

Getting to Schrödinger’s wave equation

We found that solutions to this equation are

)cos()sin()sin()cos( tkxBtkxAy )sin()sin( tkxDtkxCy or

with the constraint ck which can be written kc

Multiplying by ħ we get kc which is just E=pc

But we know that E=pc only works for massless particles so this equation can’t work for electrons.

Equal numbers of derivatives result in E=pc

doesn’t work for electrons. What does?

Getting to Schrödinger’s wave equation

Note that each derivative of x gives us a k (momentum) while each derivative of t gives us an (energy).

)cos()sin()sin()cos( tkxBtkxAy )sin()sin( tkxDtkxCy or

For massive particles we need kinetic energy

So we need two derivatives of x for p2 but only one derivative of t for K.

If we add in potential energy as well we get the Schrödinger equation…

The Schrodinger equation for a The Schrodinger equation for a matter wave in one dimension matter wave in one dimension

(x,t):(x,t):

The Schrödinger equation

Kinetic energy

Potential energy

Total energy=+

This is the time dependent Schrödinger equation (TDSE) (discussed in 7.11) and is also the most general form.

This potential energy is a function of x and t. It gives the potential energy of the particle for any x and t. It is not intrinsic to the particles but something from the problem at hand.

Time independent Schrödinger equation

In most physics situations (like hydrogen atom) the potential function U does not change in time so can write U(x,t) = U(x).

In this case, we can separate(x,t) into (x)(t): )()(),( txtx

We will then use the time independent Schrödinger equation (TISE) for the x-component of the wavefunction (lower case psi):

Given a potential energy function U(x), where would a particle naturally want to be?

A. Where U(x) is highestB. Where U(x) is lowestC.Where U(x) < kinetic energyD.Where U(x) > kinetic energyE. Does not depend on V(x)

U(x)

x

Particles want to go to position of lowest potential energy, like a ball going downhill.

0 Pot

entia

l Ene

rgy

x

)()(

2 2

22xE

dxxd

m

0 a

The potential energy function is

This is called the infinite square well (referring to the potential energy graph) or particle in a box (since the particle is trapped inside a 1D box of length a.

x < 0: U(x) ≈ ∞x > a: U(x) ≈ ∞0 < x < a: U(x) = 0

We are interested in the region 0 < x < a where U(x) = 0 so

The infinite square well (particle in a box)

Becomes (for the states in the box)

which gives

)()(2

22xEx

mk

Guess a solution to )()(

2 2

22xE

dxxd

m

How about (x) = Acos(kx)?

or

The total energy E is completely kinetic energy

(because we set the potential energy U=0)

LHS: x = 0:

na2 1

2

The functional form of the solution is

Now we apply the boundary conditions00 a

∞ ∞

V(x

)

We also know so that

Infinite square well solution

or

For an infinite square well, what are the possible values for E?

00 a

∞ ∞

V(x

)

)()(

2 2

22xE

dxxd

m n

a2ank A.

B.

C.

D.

E. Any value (E is not quantized)

2

22

nmaEn

anhcEn 2

2

222

2manEn

2

22

manEn

)sin()( kxBx

Putting into the TISE )sin()( kxBx )()(

2 2

22xE

dxxd

m

gives )()(2

22xEx

mk so m

kE2

22

Putting in the k quantization condition gives 2

222

2manE

(just kinetic energy)

2

22222

22 man

mkEn

After applying boundary conditions we found )sin()( kxBx and a

nk which gives us an energy of

Infinite square well (particle in a box) solution

What is the lowest energy possible?

A)

B)

C)Something else

2

22222

22 man

mkEn

After applying boundary conditions we found )sin()( kxBx and a

nk which gives us an energy of

xU=0 a0

E1

4E1

9E1

16E1 n=4

n=3

n=2

n=1

Energy

Minimum energy E1 is not zero. This is a general principle of QM.

Energies are quantized.

Things to notice:

Infinite square well (particle in a box) solution

2

22222

22 man

mkEn

After applying boundary conditions we found )sin()( kxBx and a

nk which gives us an energy of

xV=0 a0

E1

4E1

9E1

16E1 n=4

n=3

n=2

n=1

Energy

Minimum energy E1 is not zero.

Energies are quantized.

When a is large, energy levels get closer so energy becomes more like continuum (like classical result).

Consistent with uncertainty principle. x is between 0 and a so x~a/2. Since xp≥ħ/2, must be uncertainty in p. But if E=0 then p=0 so p=0, violating the uncertainty principle.

Things to notice:

Infinite square well (particle in a box) solution

Suzy gently places a tiny grain of sand at the bottom of a very narrow and deep well. She says: “Because of the laws of QM, this grain of sand have a finite energy so it must be floating off the ground”. Liz says: “That is ridiculous – of course it is not levitating off the ground. Therefore sand must not be quantum.”. Who is correct? xU=0 a0

E1

4E1

9E1

16E1 n=4

n=3

n=2

n=1

Energy

A grain of sand

A) SuzyB) LizC) Neither – sand is “quantum” but

“finite energy” does not mean that the sand is levitating, which would mean U is larger.