Physics 3 for Electrical Engineering
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Transcript of Physics 3 for Electrical Engineering
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!