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Operators
•A function is something that turns numbers into numbers
f  x   sin  kx 
•An operator is something that turns functions into functions
d
•Example: The derivative operator
O=
d
O f (x ) =
f (x ) = d sin (kx) = k cos (kx)
dx
dx
dx
•In quantum mechanics, x cannot be the position of a particle
•Particles don’t have a definite position
•Instead, think of x as something you multiply a wave function by to get a new wave
function
xopy (x)= xy (x)
•x is an operator, sometimes written as xop or X
•There are lots of other operators as well, like momentum

p k
i x

pop 
i x
Expectation Values
•Suppose we know the wave function (x) and we measure x.
What answer will we get?
•We only know probability of getting different values
•Let’s find the average value you get
•Recall |(x)|2 tells you the probability density that it is at x
•We want an expectation value
x  P  x x
•It is denoted by x
x



x     x  xdx    *  x  x  x  dx
2


•For any operator, we can similarly get an average measurement

O    *  x  O   x  dx

  x 
p    *  x  pop  x  dx    *  x 
dx

i 
x


 0  x  4
Sample Problem
A

exp   12 Ax 2 
A
A particle is in the ground state of a harmonic oscillator.
What is the expectation value of the operators x, x2, and p?




x    * x dx  A   e
x
2

 Ax2

   * x  dx  A   e
2



p   *
dx 
i 
x
i

Note: x2  x2
More on this later
A





e


Note: Always use normalized wave
functions for expectation values!
x 0
xdx
 Ax2
x2  1 2 A
2
x dx
 Ax2 2
  Ax  e
n  Ax 2
xe
m
 Ax2 2
 
dx  
 0
p 0
dx
n 1
2
A
 n21
if n is even
if n is odd
  12    ,   23   12 
The Hamiltonian Operator
•In classical mechanics, the Hamiltonian is the formula
for energy in terms of the position x and momentum p
•In quantum, the formula is the same, but x and p are
reinterpreted as operators
•Schrodinger’s equations rewritten with the
Hamiltonian:
E  H
2
p
H
V  x
2m
H
•The expectation value of the Hamiltonian is the
average value you would get if you measure the energy
E H
Advanced Physics:
•The Hamiltonian becomes much more complicated
•More dimensions, Multiple particles, Special Relativity
•But Schrodinger’s Equations in terms of H remain the same
2
op
p
2m
 V  xop 
2

 V  xop 
2
2m x
2

i
 H
t
Sample Problem
A particle is trapped in a 1D infinite
square well 0 < x < L with wave function
given at right. If we measure the energy,
what is the average value we would get?
30
  x   5  Lx  x 2 
L
2
2 L
 pop

d 2
 V  x   dx  
E  H    * 
 * 2 dx


2m 0
dx

 2m

L
2
2 L
30 2
d
15
2
2
2

Lx

x
Lx

x
dx


Lx

x
 2  dx



5 
2 
5 
2mL 0
dx
mL 0

30 2 1 2 1 3 L
30 2 1 3 1 3
5 2

Lx  3 x  

2 L 3L  
5 2
5
0
mL
mL
mL2
•Compare to ground state:
•Often gives excellent approximations

2 2
4.935
E

2
2mL
mL2
2
Tricks for Finding Expectation Values
•We often want expectation values of x or x2 or p or p2
•If our wave function is real, p is trivial

d
p   *
dx 
i 
dx
i


d
d
2 
2



dx


dx


 dx
 dx

2i
2i 
p

0
2
•To find p , we will use integration by parts

d
d
 
2
     * 2 dx    *
dx
dx
 i  

2 
p
2
2

p
2

2


d
dx
dx
2
d * d 

dx 
dx dx 




Uncertainty
•Recall: x2  x2. Why?
•The difference between these is a measure
of how spread out the wave function is
•Define the uncertainty in x:
 x 
2
 x  x
2
4 6
 46
 26  
  25
2
 2 
2
2
2
2
•We can similarly define the uncertainty in any operator:
 p 
2
 p  p
2
2
 O 
2
 O
2
 O
2
Heisenberg Uncertainty Principle
 x  p   12
 0  x  4
Sample Problem
A particle is in the ground state of a harmonic
oscillator. Find the uncertainty in x and p,
and check that it obeys uncertainty principle
exp   12 Ax 2 

m
A
A
x  0 x2  1 2 A p  0
•Much of the work was done five slides ago
•We even found p, but since  is real, it is trivial anyway
•Now work out p2:
2

p2 
2

2


d
dx 
dx

2
A x e
2
A

1



Ax
exp

Ax
dx


2


  
A
2  Ax2

2
p2 
dx
2
A
p  p
2
1
2
2
•Now get the uncertainties
x 
x  x
2
2

1
2A
p 
xp 
1
2
2

A2