Physics 3 for Electrical EngineeringPhysics 3 for Electrical Engineering

Ben Gurion University of the Negevwww.bgu.ac.il/atomchip, www.bgu.ac.il/nanocenter

Lecturers: Daniel Rohrlich, Ron Folman Teaching Assistants: Daniel Ariad, Barukh Dolgin

Week 5. Quantum mechanics – Schrödinger’s equation • superposition principle • time-independent Schrödinger equation • continuity of Ψ and its derivative • probability interpretation of the wave function Ψ • normalization of Ψ Sources: Feynman Lectures I, Chap. 37; Merzbacher (2nd edition) Chap. 4; Merzbacher (3rd edition) Chap. 3 Sects. 1-3, 5;

6,7פרקים בפיסיקה מודרנית, יחידות

Schrödinger’s equation

In 1926, Heisenberg was 25. Schrödinger was 14 years older, and more reasonable.In that year, each of them invented a new mechanics. Schrödinger’s new mechanics was an equation for a continuous wave. Heisenberg’s new mechanics was a matrix algebra – full of “quantum jumps”. Yet they made exactly the same predictions!

We begin with Schrödinger’s equation. In a few weeks, we will see how Heisenberg’s matrix algebra jumps out of it.

Schrödinger made a very reasonable analogy:

Fundamentally, there are no “light rays”. The rule that “the angle of incidence equals the angle of reflection” is only an approximation. So is Snell’s law. Rules about focal length are fictions used by lens-makers. All these “light rays” are actually electromagnetic waves.

Schrödinger made a very reasonable analogy.

Fundamentally, there are no “light rays”. The rule that “the angle of incidence equals the angle of reflection” is only an approximation. So is Snell’s law. Rules about focal length are fictions used by lens-makers. All these “light rays” are actually electromagnetic waves.

So Schrödinger assumed that paths of material objects – trajectories and orbits – are approximations to “matter waves”. He asked, “What equation do these matter waves obey?”

Schrödinger first tried to find a relativistic equation for matter waves, but he failed. (Paul Dirac succeeded two years later.) Then he tried to find a non-relativistic equation.

Schrödinger first tried to find a relativistic equation for matter waves, but he failed. (Paul Dirac succeeded two years later.) Then he tried to find a non-relativistic equation.

Each physical system has its own version of Schrödinger’s equation!

Schrödinger first tried to find a relativistic equation for matter waves, but he failed. (Paul Dirac succeeded two years later.) Then he tried to find a non-relativistic equation.

Each physical system has its own version of Schrödinger’s equation!

Here is Schrödinger’s equation for a one-dimensional particle of mass m in a potential V(x):

where we define . s J 10 3 05457161/2

),( )(),(2

),(

34

2

22

.h

txxVtxxm

txt

i

We don’t know what this equation means…but, as we shall soon see, neither did Schrödinger! So let’s explore: Assume V(x) = 0 (i.e. the particle is free) and solve for Ψ :

If Ψ is of the form Ψ(x,t) = X(x) T(t), then substitution yields

),(2

),(2

22tx

xmtx

ti

)(2)(

1)(

)( 2

22xX

dx

d

mxXtT

dt

d

tT

i

We don’t know what this equation means…but, as we shall soon see, neither did Schrödinger! So let’s explore: Assume V(x) = 0 (i.e. the particle is free) and solve for Ψ :

If Ψ is of the form Ψ(x,t) = X(x) T(t), then substitution yields

),(2

),(2

22tx

xmtx

ti

)(2)(

1constant)(

)( 2

22xX

dx

d

mxXtT

dt

d

tT

i

)(2)(

1constant)(

)( 2

22xX

dx

d

mxXtT

dt

d

tT

i

Try and which together imply

.

We recognize the Planck-Einstein law, ,

and de Broglie’s law, .

So the constant is .

2/constant 22 mk

)( )( ikxti exXetT

hE

khp /

m

p

m

kE

2

2constant

222

We found solutions to Schrödinger’s equation

of the form , where .

These are not all the solutions, but we obtain all the solutions by “superposing” these solutions:

,

where A(k) is any complex function of k and .

),(2

),(2

22tx

xmtx

ti

tiikxk etx ),( 2/2 mk

dkekAtx tiikx )(),(

2/2 mk

Superposition principle

If Ψ1(x,t) and Ψ2(x,t) are solutions of Schrödinger’s equation, then

αΨ1(x,t) + βΨ2(x,t)

is another solution of Schrödinger’s equation, for any complex numbers α and β.

The superposition principle follows directly from the fact that Schrödinger’s equation is always linear in Ψ:

),( )(2

),(2

22txxV

xmtx

ti

Superposition principle

Namely, if

and

then we multiply the first equation by α and the second equation by β and sum them to get the equation that satisfies.

),( )(2

),( 12

22

1 txxVxm

txt

i

),( )(2

),( 22

22

2 txxVxm

txt

i

),(),( 21 txtx

Now let’s try to solve Schrödinger’s equation for V(x) ≠ 0 using the same method of separating out the dependence on t. We already know that we can write

and if we substitute this Ψ(x,t) into Schrödinger’s equation,

we obtain an equation for ψ(x):

)(ψ),( xetx ti

, ),( )(2

),(2

22txxV

xmtx

ti

. )(ψ )(2

)(ψ 2

22xxV

dx

d

mx

Since we write

which is called the time-independent Schrödinger equation.

, )(ψ )(2

)(ψ2

22xxV

dx

d

mxE

E

Since we write

which is called the time-independent Schrödinger equation.

Question: Do solutions of the time-independent Schrödinger equation obey the superposition principle?

Suppose ψ1(x) is a solution for E1 and ψ2(x) is a solution for E2. Does it matter if E1= E2?

What about the solutions ?

E

)(ψ),( / xetx itiE

ii

, )(ψ )(2

)(ψ2

22xxV

dx

d

mxE

Time-independent Schrödinger equation

Let’s try to solve the time-independent Schrödinger equation

for a simple finite “square” potential well V(x):

x

V(x)

V0

0 L/2−L/2

, )(ψ )(2

)(ψ2

22xxV

dx

d

mxE

Time-independent Schrödinger equation

Inside:

the solution is a linear combination of eikx and e−ikx or,

equivalently, of cos kx and sin kx , where k =

; 2/ || , )(ψ2

)(ψ2

22Lxx

dx

d

mxE

x

V(x)

V0

0 L/2−L/2

outside outsideinside

. /2 mE

Time-independent Schrödinger equation

Outside:

the solution is a linear combination of eik′x and e−ik′x if E ≥ V0 ,

or a linear combination of ek′x and e−k′x if E ≤ V0 , where

k′ =

; 2/ || , )(ψ2

)(ψ )(2

22

0 Lxxdx

d

mxVE

x

V(x)

V0

0 L/2−L/2

outside outsideinside

. /||2 0 VEm

Continuity of ψ(x) and its derivative

What about at x = ± L/2? The left side of the equation, Eψ(x) or(E − V0) ψ(x), is finite; but the right side is not finite unless both

ψ(x) and its derivative dψ(x)/dx are both continuous.

Therefore, the solutions of the time-independent Schrödinger equation must be continuous with continuous first derivative.

x

V(x)

V0

0 L/2−L/2

outside outsideinside

Continuity of ψ(x) and its derivative

Let’s guess that because of the symmetry of V(x), the solutions ψ(x) are either symmetric or anti-symmetric. (Afterwards, we will check that this guess is correct.) We’ll save time because we’ll have to match the inside and outside solutions only at one point (let’s say, at x = L/2).

x

V(x)

V0

0 L/2−L/2

outside outsideinside

Probability interpretation of ψ(x)

But before we continue, we note that Schrödinger’s idea of a “matter wave” for electrons turned out to be totally unworkable. Just think about the electric charge of an electron: if an electron is really a matter wave, then we should be able to find bits of its charge in different places. No such bits of charge have ever been seen! On the contrary, the charge of the electron is, as far as we know, indivisible.

Consider a two-slit interference experiment with electrons. If we look at the charges that hit the final screen, we always find wholeelectrons with charge −e, never parts of an electron.

Probability interpretation of ψ(x)

But before we continue, we note that Schrödinger’s idea of a “matter wave” for electrons turned out to be totally unworkable. Just think about the electric charge of an electron: if an electron is really a matter wave, then we should be able to find bits of its charge in different places. No such bits of charge have ever been seen! On the contrary, the charge of the electron is, as far as we know, indivisible.

Indeed, a few months after Schrödinger published his equation, Max Born gave ψ(x) and Ψ(x,t) a new interpretation: they are probability waves, and |ψ(x)|2 = [ψ(x)]*ψ(x) is the probability density to find the electron at the point x. Likewise, |Ψ(x,t)|2 = [Ψ(x,t)]* Ψ(x,t) is the probability density to find the electron at the point x at time t. This interpretation is still accepted today.

Normalization

As long as electrons don’t simply disappear (conservation of charge) the probability to find the electron somewhere must be 1. The sum of all probabilities must be 1. In the case of the continuous variable x, the sum is an integral, so we have the mathematical requirement

This mathematical requirement is called normalization.

dxtxdxx 22 |),(|1|)(ψ|

Let’s return now to the time-independent Schrödinger equation for the finite well and focus on the case E < V0. For this case we

concluded that the outside solutions must be a linear combination

of ek′x and e−k′x, where k′ = But normalization

requires that the outside solution can only be e−k′|x| .

x

V0

0 L/2−L/2

outside outsideinside

. /||2 0 VEm

We considered two types of solutions to the time-independent Schrödinger equation: symmetric and anti-symmetric. For the symmetric solution, the continuity conditions at x = L/2 are

x

V0

0 L/2−L/2

outside outsideinside

; /2at ψ

of continuity , ''2

sin

; /2at ψ of continuity , '2

cos

2/'

2/'

Lxdx

dekA

kLAk

LxeAkL

A

Lk

Lk

kxAx cos)(ψ xkeAx '')(ψ xkeAx '')(ψ

Therefore, the continuity conditions for the symmetric solution imply k tan kL/2 = k′, which in turn implies discrete energy levels! From the definitions of k and k′ we have

x

V0

0 L/2−L/2

outside outsideinside

kxAx cos)(ψ xkeAx '')(ψ xkeAx '')(ψ

22

020222 2)(2

)'(2

tan kmVEVm

kkL

k

and we can solve this equation graphically by setting each side to a variable y2; the solutions are the intersections of the two curves y = k tan kL/2 and y2 + k2 = .

20 /2 mV

20 /2 mV y = k tan kL/2y2 + k2 =

2

yL

2

x

xkeAx '')(ψ xkeAx '')(ψ

outside

0 L/2−L/2

1

kL/22

outside

V0

inside

kxAx cos)(ψ

1

For the anti-symmetric solution, the continuity conditions at x = L/2 are

x

V0

0 L/2−L/2

outside outsideinside

; /2at ψ

of continuity , ''2

cos

; /2at ψ of continuity , '2

sin

2/'

2/'

Lxdx

dekA

kLAk

LxeAkL

A

Lk

Lk

kxAx sin)(ψ xkeAx '')(ψ xkeAx '')(ψ

Therefore, the continuity conditions for the symmetric solution imply k cot kL/2 = –k′ . From the definitions of k and k′ we have

x

V0

0 L/2−L/2

outside outsideinside

22

020222 2)(2

)'(2

cot kmVEVm

kkL

k

and we can solve this equation graphically by setting each side to a variable y2; the solutions are the intersections of the two curves y = k cot kL/2 and y2 + k2 = .

20 /2 mV

kxAx sin)(ψ xkeAx '')(ψ xkeAx '')(ψ

x0 L/2

outsidexkeAx '')(ψ xkeAx '')(ψ

y = k cot kL/2y2 + k2 = 2m

−L/2

outside

2

yL

inside

kxAx sin)(ψ

V0

20 / V

kL/221

2

1

The first three energy levels: the ground state is symmetric, the next state is anti-symmetric, the next state is symmetric, etc.

First excited state

Ground state

0 L/2−L/2

The first three energy levels: the ground state is symmetric, the next state is anti-symmetric, the next state is symmetric, etc.

First excited state

Ground state

0 L/2−L/2

Classically forbidden region

The first three energy levels: the ground state is symmetric, the next state is anti-symmetric, the next state is symmetric, etc.

As long as V0 is finite, there are finitely many energy levels.

For V0 infinite (infinite “square” well), there are infinitely many

levels, and ψ(x) vanishes outside the well (|x| > L/2, classically forbidden region). The derivative of ψ(x) is discontinuous at x = L/2 but this exception to the continuity rule is due to infinite V0. The first three normalized wave functions and energies are

. 2

9 ,

3cos

2)(ψ

, 2

4 ,

2sin

2)(ψ

, 2

,

cos2

)(ψ

2

22

33

2

22

22

2

22

11

mLE

L

x

Lx

mLE

L

x

Lx

mLE

L

x

Lx

Conclusions:

1. There is no physical system exactly like an electron in a one-dimensional square well potential. But it is a model we can solve, and we learn from it some features of quantum mechanics and quantum behavior. We can also use it as an approximate model, e.g. of an electron in a wire segment.

2. We learn that energy levels can be quantized, i.e. energies can take discrete values.

3. We saw that symmetries of the model (here, parity) show up in the symmetries of the solutions.

4. We saw a unique quantum effect: quantum electrons can penetrate regions that are forbidden to classical electrons!