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Physics 2170 – Spring 2009 1http://www.colorado.edu/physics/phys2170/
Getting to the Schrödinger equation
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Announcements:
Erwin Schrödinger (1887 – 1961)
Physics 2170 – Spring 2009 2http://www.colorado.edu/physics/phys2170/
Classical waves obey the wave equation:
Where we go from here
We will finish up classical waves
Then we will go back to matter waves which obey a different wave equation called the time dependent Schrödinger equation:
2
2
22
2 1ty
vxy
tiV
xm
2
22
2
On Friday we will derive the time independent Schrödinger equation: EV
xm
2
22
2
Physics 2170 – Spring 2009 3http://www.colorado.edu/physics/phys2170/
Solving the standard wave equation
1. Guess the functional form(s) of the solution
2. Plug into differential equation to check for correctness, find any constraints on constants
3. Need as many independent functions as there are derivatives.
4. Apply all boundary conditions (more constraints on constants)
2
2
22
2 1ty
vxy
The standard wave equation is
Generic prescription for solving differential equations in physics:
Physics 2170 – Spring 2009 4http://www.colorado.edu/physics/phys2170/
Claim that is a solution to
Step 2: Check solution and find constraints
)cos()sin( CtBxAy 2
2
22
2 1ty
vxy
Time to check the solution and see what constraints we have
LHS: )cos()sin(22
2
CtBxABxy
RHS: )cos()sin(12
2
2
2
2 CtBxvAC
ty
v
Setting LHS = RHS: )cos()sin()cos()sin( 2
22 CtBx
vACCtBxAB
This works as long as 2
22
vCB
We normally write this as )cos()sin( tkxAy
so this constraint just means 2
22
vk or f
kv
Physics 2170 – Spring 2009 5http://www.colorado.edu/physics/phys2170/
and we have the constraint that
Since the wave equation has two derivatives, there must be two independent functional forms.
Constructing general solution from independent functions
2
2
22
2 1ty
vxy
)sin()cos( tkxAy
2
22
vk
)cos()sin( tkxBy
The general solution is )sin()cos()cos()sin( tkxBtkxAy
Can also be written as )sin()sin( tkxDtkxCy
x
yt=0
We have finished steps 1, 2, & 3 of solving the differential equation.
Last step is applying boundary conditions. This is the part that actually depends on the details of the problem.
Physics 2170 – Spring 2009 6http://www.colorado.edu/physics/phys2170/
Boundary conditions for guitar string
0 L
Guitar string is fixed at x=0 and x=L.
2
2
22
2 1ty
vxy
Wave equation
Functional form:
)sin()cos()cos()sin( tkxBtkxAy
Boundary conditions are that y(x,t)=0 at x=0 and x=L.
Requiring y=0 when x=0 means
)sin()0cos()cos()0sin(0 tBtA which is )sin(00 tB
This only works if B=0. So this means )cos()sin( tkxAy
Physics 2170 – Spring 2009 7http://www.colorado.edu/physics/phys2170/
Clicker question 1 Set frequency to DABoundary conditions require y(x,t)=0 at x=0 & x=L. We found for y(x,t)=0 at x=0 we need B=0 so our solution is . By evaluating y(x,t) at x=L, derive the possible values for k.
A. k can have any value B. /(2L), /L, 3/(2L), 2/L … C. /LD. /L, 2/L, 3/L, 4/L … E. 2L, 2L/2, 2L/3, 2L/4, ….
)cos()sin( tkxAy
To have y(x,t) = 0 at x = L we need0)cos()sin( tkLA
This means that we need 0)sin( kL
This is true for kL = n. That is, Lnk
n=1
n=2
n=3
So the boundary conditions quantize k. This also quantizes
because of the other constraint we have: 2
22
vk
Physics 2170 – Spring 2009 8http://www.colorado.edu/physics/phys2170/
Summary of our wave equation solution1. Found the general solution to the wave equation
2
2
22
2 1ty
vxy
)cos()sin()sin()cos( tkxBtkxAy )sin()sin( tkxDtkxCy or
2. Put solution into wave equation to get constraint 2
22
vk
3. Have two independent functional forms for two derivatives
4. Applied boundary conditions for guitar string. y(x,t) = 0 at x=0 and x=L. Found that B=0 and k=n/L.
Our final result:)sin()cos( tkxAy
2
22
vk L
nk with and
n=1
n=2
n=3
Physics 2170 – Spring 2009 9http://www.colorado.edu/physics/phys2170/
Standing waves
Standing wave
Standing wave constructed from two traveling waves moving in opposite directions
Physics 2170 – Spring 2009 10http://www.colorado.edu/physics/phys2170/
Examples of standing waves
Same is true for electromagnetic waves in a microwave oven:
For standing waves on violin string, only certain values of k and are allowed due to boundary conditions (location of nodes).
We also get only certain waves for electrons in an atom.
We will find that this is due to boundary conditions applied to solutions of Schrödinger equation.
Physics 2170 – Spring 2009 11http://www.colorado.edu/physics/phys2170/
Clicker question 2 Set frequency to DA
x
y
x
y
x
y
Case I: no fixed ends
Case II: one fixed end
Case III: two fixed ends
For which of the three cases do you expect to have only certain frequencies and wavelengths allowed? That is, in which cases will the allowed frequencies be quantized?
A. Case IB. Case IIC.Case IIID.More than one case
After applying the 1st boundary condition we found B=0 but we did not have quantization. After the 2nd boundary condition we found k=n/L. This is the quantization.
Physics 2170 – Spring 2009 12http://www.colorado.edu/physics/phys2170/
Electron bound in atom Free electron
Only certain energies allowedQuantized energies
Any energy allowed
E
Boundary Conditions standing waves
No Boundary Conditions traveling waves
Boundary conditions cause the quantization
Physics 2170 – Spring 2009 13http://www.colorado.edu/physics/phys2170/
2
2
22
2 1
t
E
cx
E
Works for light (photons), why doesn’t it work for electrons?
Getting to Schrödinger’s wave equation
Physics 2170 – Spring 2009 14http://www.colorado.edu/physics/phys2170/
Clicker question 3 Set frequency to DA
The equation E = hc/ is…
A. true for photons and electrons
B. true for photons but not electrons
C.true for electrons but not photons
D.not true for either electrons or photons
hfE works for photons and electrons
khp works for photons and electrons
hcpcE only works for massless particles (photons)
Physics 2170 – Spring 2009 15http://www.colorado.edu/physics/phys2170/
2
2
22
2 1
t
E
cx
E
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 pcE
But we know that E=pc only works for massless particles so this equation can’t work for electrons.
Physics 2170 – Spring 2009 16http://www.colorado.edu/physics/phys2170/
Equal numbers of derivatives result in
2
2
22
2 1
t
E
cx
E
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
pcE
For massive particles we need mp
K2
2
So we need two derivatives of x for p2 but only one derivative of for K.
If we add in potential energy as well we get the Schrödinger equation…
Physics 2170 – Spring 2009 17http://www.colorado.edu/physics/phys2170/
The Schrodinger equation for a The Schrodinger equation for a matter wave in one dimension matter wave in one dimension
(x,t):(x,t):
ttx
itxtxVxtx
m
),( ),(),(
),(2 2
22
The Schrödinger equation
Kinetic energy
Potential energy
Total energy=+
This is the time dependent Schrödinger equation (discussed in 7.11) and is also the most general form.
Physics 2170 – Spring 2009 18http://www.colorado.edu/physics/phys2170/
If we plug this into the Schrödinger equation, what do we get?
ttx
itxtxVxtx
m
),( ),(),(
),(2 2
22
The Schrödinger equation applied to a plane wave
A plane matter wave can be written as )(),( tkxietx
),()()())((),( 22
2
2
txktkxiektkxieikikxtx
),()(),(txitkxiei
ttx
),( ),(),( ),(2
22txtxtxVtx
mk
EtxVmp
),( 2
2
E=K+U seems to make sense