### Lecture6.QM.to.Lagrangian.Densities

```From Quantum Mechanics to
Lagrangian Densities
Just as there is no “derivation” of quantum mechanics
from classical mechanics, there is no derivation of
relativistic field theory from quantum mechanics. The
“route” from one to the other is based on physically
reasonable postulates and the imposition of Lorentz
invariance and relativistic kinematics.
The final
“theory” is a model whose survival depends
absolutely on its success in producing “numbers”
which agree with experiment.
The ten minute course in QM.
Summary: Quantum Mechanics
Momentum
Becomes an
operator
and use
The Hamiltonian
becomes an operator.
Physical interpretation of the wave function.
This condition places a strong mathematical condition on the wave function.
Note that the Schrodinger
equation reflects this
relationship
Quantization arises from placing boundary conditions on
the wave function. It is a mathematical result!
L =rxP =rx
A “toy” model postulate approach to quantum field theory
-p
-
-
+
p
Note that *(r,t) (r,t) does not represent the probability
per unit volume density of the particle being at (r,t).
-
-
The resulting “wave equation”:
The “negative energy” states arose from
This emerges from starting out with
We know the energy of a real particle can’t be < 0.
.
Suppose the mass, m, is zero:
This is the same equation we derived from Maxwell’s equations for
the A vector (except, of course, above we have a scalar, ).
Solving the wave equation
Following the same derivation used for the A vector we have:
We try a solution of the form
This is just like the E & M
equation – except for the
mass term.
We can see that after taking the partial derivatives there is a condition
on the components of k and k0.
Note, now the 4-dimensional
dot product cannot = 0.
The k and k0 (with p= k and p0 = k0 )
must satisfy the same conditions as a
relativistic particle with rest mass, m.
We have the following two linearly independent
solutions to the “wave equation”:
The most general (complete) solution to the wave equation is
The field operator for a neutral, spin =0, particle is
creates a
single
particle with
momentum
p= k and
p0 = k0
at (r,t)
Destroys a
single
particle with
momentum
p= k and
p0 = k0
at (r,t)
Lagrangians and the Lagrangian Density
Recall that,
and the Euler-Lagrange equations give F = ma
In quantum field theory, the Euler-Lagrange
equations give the particle wave function!
This calls for a different kind of “Lagrangian” -- not like the one used
in classical or quantum mechanics. So, we have another postulate,
defining what is meant by a “Lagrangian” – called a Lagrangian density.
d/dt in the classical theory
Note that the Lagrangian density is quadratic in (r,t) and the
Lorentz invariance is satisfied by using µ and µ
We can apply the Euler Lagrange equations to the above L:
This part is
easy.
This part has
a very simple
result
but
it is hard to
carry out.
Finally,
with this Lagrangian
density
The Euler-Lagrange
Equations
give the wave equation
for the neutral spin
= 0 particle.
Summary for neutral (Q=0) scalar
(spin = 0) particle, , with mass, m.
Lagrangian density
wave equation
field operator
Charged (q = ±e) scalar (spin =0) particle with mass, m
8 terms
cancel
charged scalar particle
From the Lagrangian density
and the Euler-Lagrange equation
we can derive the
wave equation
creates positively
charged particle
with momentum
p= k and
p0 = k0 at (r,t)
destroys negatively
charged particle
with momentum
p= k and
p0 = k0 at (r,t)
destroys negatively
charged particle
with momentum
p= k and
p0 = k0 at (r,t)
creates negatively
charged particle
with momentum
p= k and
p0 = k0 at (r,t)
The scalar field (which represents a boson) must also
satisfy a special boson commutation property:
Creation and
annihilation
operators with
the same k don’t
commute.
Everything else
commutes.
Example of how the commutation relation is used
We will use this when we calculate the charge of a particle.
Later we will find that fermions satisfy a different commutation
relation.
What follows is for the graduate students.