2. Solving Schr

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2. Solving Schrödingers Equation
Superposition

• Given a few solutions of Schrödingers equation,
  we can make more of them
• Let ?1 and ?2 be two solutions of Schrödingers
  equation
• Let c1 and c2 be any pair of complex numbers
• Then the following is also a solution of
  Schrödingers equation

• We can generalize this to an arbitrary number of
  solutions

• We can even generalize to a continuum of
  solutions

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2A. The Free Schrödingers Equation
We know a lot of solutions

• Consider the free Schrödinger equation
• We know a lot of solutions of this equation

• By combining these, we can make a lot more
  solutions

• The function c(k) can be (almost) any function of
  k
• This actually includes all possible solutions

The factor of (2?)3/2 is arbitrary and inserted
here for convenience. In 1D, it would be (2?)1/2
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We can (in principle) always solve this problem
Goal Given ?(r,t 0) ?(r), find ?(r,t) when
no potential is present

• Set t 0

• This just says that c(k) is the Fourier transform
  of ?(r)

• Substitute it in the formula above for ?(r,t) and
  we are done
• Actually doing the integrals may be difficult

4
Sample Problem
A particle in one dimension with no potential has
wave function at t 0 given by What is the wave
function at arbitrary time?

• Need 1D versions of these formulas

From Appendix Find c(k)
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Sample Problem (2)
A particle in one dimension with no potential has
wave function at t 0 given by What is the wave
function at arbitrary time?

• Now find ?(x,t)

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2B. The Time Independent Schrödinger Eqn.
Separation of Variables

• Suppose that the potential is independent of time

• Conjecture solutions of the form
• Substitute it in

• Divide by ?(r)?(t)

• Left side is independent of r
• Right side is independent of t
• Both sides are independent of both! Must be a
  constant. Call it E

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Solving the time equation

• The first equation is easy to solve
• Integrate both sides

• By comparison with e-i?t, we see that E ?? is
  the energy
• Substitute it back in

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The Time Independent Schrödinger Equation

• Multiply the other equation by ?(r) again

The Strategy for solving

• Given a potential V(r) independent of time, what
  is most general solution of Schrödingers
  time-dependent equation?
• First, solve Schrödingers time-independent
  equation
• You should find many solutions ?n(r) with
  different energies En
• Now just multiply by the phase factor
• Then take linear combinations

• Later well learn how to find cn

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Why is time-independent better?

• Time-independent is one less variable
  significantly easier
• It is a real equation (in this case), which is
  less hassle to solve
• If in one dimension, it reduces to an ordinary
  differential equation
• These are much easier to solve, especially
  numerically

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2C. Probability current
Probability Conservation

• Recall the probability density is
• This can change as the wave function changes
• Where does the probability go as it changes?
• Does it flow just like electric charge does?
• Want to show that the probability moves around
• Ideally, show that it flows from place to place
• A formula from E and M can we make it like this?

• To make things easier, lets make all functions
  of space and time implicit, not write them

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The derivation (1)

• Start with Schrödingers equation
• Multiply on the left by ?
• Take complex conjugate of this equation
• Subtract
• Rewrite first term as a total derivative
• Cancel a factor of i?
• Left side is probability density

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The derivation (2)

• Consider the following expression
• Use product rule on the divergence
• Substitute this in above
• Define the probability current j
• Then we have

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Why is it called probability current?

• Integrate it over any volume V with surface S
• Left side is P(r ? V)
• Use Gausss law on right side
• Change in probability is due to current flowing
  out

V

• If the wave function falls off at infinity (as it
  must) and the volume V becomes all of space, we
  have

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Calculating probability current

• This expression is best when doing proofs
• Note that you have a real number minus its
  complex conjugate
• A quicker formula for calculation is
• Lets find ? and j for a plane wave

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Sample Problem
A particle in the 1D infinite square well has
wave function For the region 0 lt x lt a. Find ?
and j.
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Sample Problem (2)
A particle in the 1D infinite square well has
wave function For the region 0 lt x lt a. Find ?
and j.

• In 1D

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Sample Problem (3)
A particle in the 1D infinite square well has
wave function For the region 0 lt x lt a. Find ?
and j.

• After some work

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2D. Reflection from a Step Boundary
The Case E gt V0 Solutions in Each Region
I

• A particle with energy E impacts a step-function
  barrier from the left

incident
transmitted
reflected
II

• Solve the equation in each of the regions
• Assume E gt V0
• Region I
• Region II

• Most general solution
• A is incident wave
• B is reflected wave
• C is transmitted wave
• D is incoming wave from the right set D 0

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Step with E gt V0 The solution
I
incident
transmitted
reflected

• Schrödingers equation second derivative finite
• ?(x) and ?(x) must be continuous at x 0

II

• We cant normalize wave functions
• Use probability currents!

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Summary Step with E gt V0
I
incident
transmitted
reflected
II
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Step with E lt V0

• What if V0 gt E?
• Region I same as before
• Region II we have

I
II
incident
reflected
evanescent

• Most general solution
• A is incident wave
• B is reflected wave
• C is damped evanescent wave
• D is growing wave, cant be normalized

• ?(x) and ?(x) must be continuous at x 0

• No transmission since evanescent wave is damped

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Step Potential All cases summarized

• For V0 gt E, all is reflected

• Note that it penetrates, a little bit into the
  classically forbidden region, x gt 0
• This suggests if barrier had finite thickness,
  some of it would bet through

• Reflection probability

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2E. Quantum Tunneling
Setting Up the Problem
V(x)

• Barrier of finite height and width

V0

• Solve the equation in each of the regions
• Particle impacts from left with E lt V0
• General solution in all three regions

III
II
I
- d/2
d/2
x

• Match ? and ? at x -d/2 and x d/2

Why didnt I include e-ikx in ?III? Why did I
skip letter E?

• Solve for F in terms of A

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Skip this Slide Solving for F in terms of A

• Multiply 1 by ik and add to 2
• Multiply 3 by ? and add to 4
• Multiply 3 by ? and subtract 4
• Multiply 5 by 2? and substitute from 6 and 7

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Barrier Penetration Results

• We want to know transmission probability

• For thick barriers,

• Exponential suppression of barrier penetration

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Unbound and Bound State

• For each of the following, we found solutions for
  any E
• No potential
• Step potential
• Barrier
• This is because we are dealing with unbound
  states, E gt V(??)
• Our wave functions were, in each case, not
  normalizable
• Fixable by making superpositions

• We will now consider bounds states
• These are when E lt V(??)
• There will always only be discrete energy values
• And they can be normalized
• Usually easier to deal with real wave functions

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2F. The Infinite Square Well
Finding the Modes
V(x)

• Infinite potential implies wave function must
  vanish there
• In the allowed region, Schrödingers equation is
  just

x
a
0

• The solution to this is simple
• Because potential is infinite, the derivative is
  not necessarily continuous
• But wave functions must still be continuous

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Normalizing Modes and Quantized Energies

• We can normalize this wave function

• Note that we only get discrete energies in this
  case
• Note that we can normalize these
• Most general solution is then

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The 3D Infinite Square Well

• In allowed region
• Guess solution
• Normalize it
• This is product of 1D functions
• Energy is
• This is sum of 1D energies

c
b
a
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2G. The Double Delta-Function Potential
Finding Bound States
V(x)
a/2
-a/2
I
III
II

• First, write out Schrödingers Equation

x

• Bound states have E lt V(?) 0
• Within each region we have

• General solution (deleting the parts that blow up
  at infinity)

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Dealing with Delta Functions
V(x)
a/2
-a/2
I
III
II

• To deal with the delta functions, integrate
  Schrödingers equation over a small region near
  the delta function
• For example, near x a/2

• Do first term on right by fundamental theorem of
  calculus
• Do second term on right by using the delta
  functions

• Take the limit ? ? 0
• Left side small in this limit

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Simplifying at x ½a
V(x)
a/2
-a/2
I
III
II

• Since there is a finite discontinuity in ?, ?
  must be continuous at this boundary

On the right side of the equation above, is that
?I, ?II, or ?III?

• Write these equation out explicitly
• Substitute first into second

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Repeating at x ½a

• Repeat the steps we did, this time at x ½a

• Note these equations are nearly identical
• The only numbers equal to their reciprocal are ?1

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Graphical Solution

• Right side is two curves, left side is a straight
  line

• Black line always crosses red curve, sometimes
  crosses green curve, depending on parameters
• Sometimes two solutions, sometimes one

• Normalize to finish the problem
• Note one solution symmetric, one anti-symmetric