Physics 249 Lecture 10, Sep 26th 2012
Reading: Chapter 5 and Chapter 6
HW3: Due Friday, Posted on the course homepage
1) Uncertainty principle and diffraction.
The uncertainty principle only has an effect when making measurements at levels of
precision of order Planck’s constant. As an example, at that level the limitations on
measurement can seen from diffraction effects, a well understood a wave phenomena.
The limitation on measuring positions using light, or any particle since they all diffract:
Dx = L sinq =
l
D
L
Taken from the minimum separation where two central diffractive peaks can be
distinguished. Where the distance Dx is a distance perpendicular to the direction you
have your optical device pointed.
To resolve smaller distances you can use shorter wavelength light. However, when the
light has shorter wavelength it will be higher energy.
𝐸 = ℎ𝑓 =
ℎ𝑐
𝜆
Now simultaneously try to measure the momentum. We can do that by repeating the
experiment a short time later and measuring the new position and seeing how far the
particle moved in that time. We will measure the position again by bounding photons off
the object via Compton scattering.
However, the first photon will transfer momentum to the target changing its momentum
by Δ𝑝. Thus we don’t measure quite the initial momentum.
In a reference frame where the total momentum is zero the photon will have equal and
opposite momentum to the target after the collision. Considering the magnitude of the
targets change in momentum the photon must have lost half of it’s momentum or energy.
The targets momentum is changed by
Δ𝑝 =
1
𝐸
ℎ𝑓
ℎ
𝑝=
=
=
2
2𝑐 2𝑐 2𝜆
In some unknown direction obeying the Compton formula. In the x direction the targets
momentum was changed by Δ𝑝𝑥 . However we can only observed the returning photon if
it has an angle small enough to enter our optical device. Therefore the x momentum can
1
1
𝐷
1 ℎ 𝐷
be anything up to Δ𝑝𝑥 = Δ𝑝𝑥 = 2 𝑝𝑠𝑖𝑛𝜃𝑑 = 2 Δ𝑝 𝐿 = 2 2𝜆 𝐿
So the measurement of the x position changed the x momentum an unknown amount that
is possibly as big as Δ𝑝𝑥 . All you know is you saw the photon the photon so this
represents your measurement error. Note that also that the scattered photon has a
wavelength twice as large since it lost half its energy. Putting these two numbers together
ΔxΔ𝑝𝑥 =
2𝜆𝐿 ℎ 𝐷
1
1
, ΔxΔ𝑝𝑥 = ℎ > ℏ
𝐷 4𝜆 𝐿
2
2
Measuring the x position via diffraction (the limit on how well we can measure any
object optically) by using a shorter wavelength improves the precision for delta x.
However, then we find that gain in precision for delta x comes at the cost of introducing a
larger uncertainty in the momentum delta p.
We find that just from considering quantum the wave nature of the measurement that
there are limitations to our ability to measure objects!
2) An application of the uncertainty principle.
Jumping forward to particle physics. We now understand that the electromagnetic force
is propagated by virtual photons. The concept is that these temporary photons can exist
and even transfer negative (unphysical) momentums as long as they are consistent with
zero within the uncertainty principle. Let’s investigate a consequence of this.
The virtual photons have energy maximum E and exist for a maximum time of t such that
Et~hbar. Or you can say they have maximum momentum p and cross a maximum
distance r such that pr~hbar where r=tc. Under the Heisenberg uncertainty principle if the
energy and momentum are this small their physics properties are consistent with zero
within uncertainty. If they transfer a smaller amount of momentum then they can cross a
larger distance or vice-versa. For a force F= dp/dt = dp/dr dr/dt ~ hbar c/r2. The
functional dependence is a constant governing the strength over r2. The strength of the
interaction depends on the charges involved. F=CqQ/r2 . The photons are transferred
when there are charges the number of photons is going to be proportional to qQ times
some constant that governs the strength of individual electromagnetic interactions.
Some consequences of the uncertainty principle are discussed in the textbook.
3) Schrodinger equation.
If the particles have wave properties and can be characterized by a wave function that
indicates they should be the solutions to a wave equation.
Recall the features of wave functions and wave equations.
Wave functions: Oscillatory functions such as cos, sin or exponentials with an imaginary
exponent.
Wave equations. Second order differential equations constructed such that the solutions
give the same function back after differentiation with an extra sign or constant.
For light the second order differential equation comes from Maxwell’s equations in free
space, i.e. with no sources or currents.
For the electric field
1 𝜕2 𝐸
𝜕2 𝐸
1 𝜕2 𝐸
∇2 𝐸 = 𝑐 2 𝜕𝑡 2 , or in one dimension: 𝜕𝑥 2 𝐸 = 𝑐 2 𝜕𝑡 2
Often you can guess the solution of a wave equation. In this case the above suggested
solutions all work and you would pick among them by considering boundary conditions.
Similarly it is easy to reverse engineer the wave equation given an idea of what type of
solution we want. The wave equation will be an equation that take the wave function and
differentiates in the physics coordinates of space and time to determine how they are
related.
For photons given:
ℎ
ℎ
𝐸 = 𝐸0 𝑐𝑜𝑠(𝑘𝑥 − 𝜔𝑡) and 𝐸 = 𝑝𝑐, where 𝐸 = 𝑓ℎ = 𝜔ℏ and 𝜆 = 𝑝 ⇒ 𝑝 = 𝜆 = 𝑘ℏ
You can hypothesize a reasonable form for the wave equation. E=pc so in guessing our
wave equation we would have the same order of differentiation in space and time.
𝜕 2𝐸
𝜕 2𝐸
𝐸=𝐶 2
𝜕𝑥 2
𝜕𝑡
∇2 𝑘 2 = 𝐶𝜔2 ⇒
𝑝2
𝐸2
1
=
𝐶
⇒ 𝑝2 = 𝐶𝑝2 𝑐 2 ⇒ 𝐶 = 2
2
2
ℏ
ℏ
𝑐
Now for matter particles in free space
𝑝2
ℎ
ℎ
𝜓 = 𝐴𝑒 𝑖(𝑘𝑥−𝜔𝑡) and 𝐸 = 2𝑚, where 𝐸 = 𝜔ℏ and 𝜆 = 𝑝 = 𝑚𝑣 ⇒ 𝑝 =
ℎ
𝜆
= 𝑘ℏ
However, note that the relationship between E and p is such that E is first order and p is
second order. This indicates the wave equation should have one derivative in time that
gives energy/frequency and two in space that give momentum/wavenumber.
𝜕𝜓
𝜕 2𝜓
=𝐶 2
𝜕𝑡
𝜕𝑥
𝑝2
𝐸 𝑝2
𝑝2
2𝑚
−𝑘 = −𝑖𝐶𝜔 ⇒ 2 = 𝑖𝐶 ⇒
=
𝑖𝐶 ⇒ 𝐶 = −𝑖
ℏ
ℏ
ℏ
2𝑚
ℏ
2
𝜕 2𝜓
2𝑚 𝜕𝜓
=
−𝑖
𝜕𝑥 2
ℏ 𝜕𝑡
or more typically write as(Schrodinger equation in free space):
𝜕𝜓
ℏ2 𝜕 2 𝜓
𝑖ℏ
=−
𝜕𝑡
2𝑚 𝜕𝑥 2
This method works but has two limitations. It doesn’t tell us how to treat a potential and
it doesn’t tell us how to treat more complex situations such as dealing with energy and
momentum relationships in relativity.
The solution to this problem is to recognize that the key relationship that defined the
wave equation was the energy momentum relationship. In the presence of a potential this
clearly expands to including the potential energy. The second point is to recognize that
given the form a wave function that you can extract the energy and momentum using
𝜕𝜓
𝜕𝜓
partial derivatives. 𝜕𝑡 for energy and 𝜕𝑥 for momentum.
These ideas can be put together into a general prescription called quantization. In this
case quantization means converting a classical equation into a quantum wave equation.
The prescription is as follows.
1) The total energy relationship including kinetic and potential energy describes the
physics of the system.
2) Quantize the total energy relationship by making the following substitutions.
𝑝 → −𝑖ℏ
𝐸 → 𝑖ℏ
𝜕
𝜕𝑥
𝜕
𝜕𝑡
The total energy relationship becomes:
𝑝2
𝜕𝜓
ℏ2 𝜕 2 𝜓
𝐸=
+ 𝑉 ⇒ 𝑖ℏ
=−
+ 𝑉𝜓 = 𝐻𝜓
2𝑚
𝜕𝑡
2𝑚 𝜕𝑥 2
Where we now have operators p and H (the Hamiltonian operator) that can extract the
momentum or energy.
This prescription can be used with any potential and either non relativistic or relativistic
energy and momentum relationships. In fact for the photon we used 𝐸 2 = 𝑐 2 𝑝2 + 𝑚2 𝑐 4
with mass zero. The photon needs the relativistic version of the total energy relationship
since it moves at c.