Quantum Mechanics for Engineers |
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© Leon van Dommelen |
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Subsections
A.1 Classical Lagrangian mechanics
Lagrangian mechanics is a way to simplify complicated dynamical
problems. This note gives a brief overview. For details and
practical examples you will need to consult a good book on mechanics.
A.1.1 Introduction
As a trivial example of how Lagrangian mechanics works, consider a
simple molecular dynamics simulation. Assume that the forces on the
particles are given by a potential that only depends on the positions
of the particles.
The difference between the net kinetic energy and the net potential
energy is called the “Lagrangian.” For a system of particles as considered here it
takes the form
where indicates the particle number and the potential of the
attractions between the particles and any external forces.
It is important to note that in Lagrangian dynamics, the Lagrangian
must mathematically be treated as a function of the velocities and
positions of the particles. While for a given motion, the positions
and velocities are in turn a function of time, time derivatives must
be implemented through the chain rule, i.e. by means of total
derivatives of the Lagrangian.
The canonical momentum
of particle
in the direction, (with 1, 2, or 3 for the
, , or components respectively), is defined as
For the Lagrangian above, this is simply the normal momentum
of the particle in the -direction.
The Lagrangian equations of motion are
This is simply Newton’s second law in disguise: the left hand side
is the time derivative of the linear momentum of particle in the
-direction, giving mass times acceleration in that direction;
the right hand side is the minus the spatial derivative of the
potential, which gives the force in the direction on particle
. Obviously then, use of Lagrangian dynamics does not help
here.
A.1.2 Generalized coordinates
One place where Lagrangian dynamics is very helpful is for macroscopic
objects. Consider for example the dynamics of a Frisbee. Nobody is
going to do a molecular dynamics computation of a Frisbee. What you
do is approximate the thing as a solid body,
(or more
accurately, a rigid body). The position of every part of a solid body
can be fully determined using only six parameters, instead of the
countless position coordinates of the individual atoms. For example,
knowing the three position coordinates of the center of gravity of the
Frisbee and three angles is enough to fully fix it. Or you could just
choose three reference points on the Frisbee: giving three position
coordinates for the first point, two for the second, and one for the
third is another possible way to fix its position.
Such parameters that fix a system are called “generalized coordinates.” The word generalized indicates that
they do not need to be Cartesian coordinates; often they are angles or
distances, or relative coordinates or angles. The number of generalized
coordinates is called the number of degrees of freedom. It varies
with the system. A bunch of solid bodies moving around freely will
have six per solid body; but if there are linkages between them, like
the bars in your car’s suspension system, it reduces the number of
degrees of freedom. A rigid wheel spinning around a fixed axis has
only one degree of freedom, and so does a solid pendulum swinging
around a fixed axis. Attach a second pendulum to its end, maybe not
in the same plane, and the resulting compound pendulum has two degrees
of freedom.
If you try to describe such systems using plain old Newtonian
mechanics, it can get ugly. For each solid body you can apply that
the sum of the forces must equal mass times acceleration of the center
of gravity, and that the net moment around the center of gravity must
equal the rate of change of angular momentum, which you then
presumably deduce using the principal axis system.
Instead of messing with all that complex vector algebra, Lagrangian
dynamics allows you to deal with just a single scalar, the Lagrangian.
If you can merely figure out the net kinetic and potential energy of
your system in terms of your generalized coordinates and their time
derivatives, you are in business.
If there are linkages between the members of the system, the benefits
magnify. A brute-force Newtonian solution of the three-dimensional
compound pendulum would involve six linear momentum equations and six
angular ones. Yet the thing has only two degrees of freedom; the
angular orientations of the individual pendulums around their axes of
rotation. The reason that there are twelve equations in the Newtonian
approach is that the support forces and moments exerted by the two
axes add another 10 unknowns. A Lagrangian approach allows you to
just write two equations for your two degrees of freedom; the support
forces do not appear in the story. That provides a great
simplification.
A.1.3 Lagrangian equations of motion
This section describes the Lagrangian approach to dynamics in general.
Assume that you have chosen suitable generalized coordinates that
fully determine the state of your system. Call these generalized
coordinates , , ...and their time
derivatives , , .... The number
of generalized coordinates is the number of degrees of freedom in
the system. A generic canonical coordinate will be indicated as
.
Now find the kinetic energy and the potential energy of your
system in terms of these generalized coordinates and their time
derivatives. The difference is the Lagrangian:
Note that the potential energy depends only on the position
coordinates of the system, but the kinetic energy also depends on how
fast they change with time. Dynamics books give lots of helpful
formulae for the kinetic energy of the solid members of your system,
and the potential energy of gravity and within springs.
The canonical momenta are defined as
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(A.1) |
for each individual generalized coordinate . The equations
of motion are
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(A.2) |
There is one such equation for each generalized coordinate
, so there are exactly as many equations as there are
degrees of freedom. The equations are second order in time, because
the canonical momenta involve first order time derivatives of the
.
The terms are called generalized forces, and are only needed if
there are forces that cannot be modeled by the potential .
That includes any frictional forces that are not ignored. To find the
generalized force at a given time, imagine that the system is
displaced slightly at that time by changing the corresponding
generalized coordinate by an infinitesimal amount
. Since this displacement is imaginary, it is
called a virtual displacement.
During such a
displacement, each force that is not modelled by produces a small
amount of “virtual work.” The net virtual work divided by
gives the generalized force . Note that frictionless
supports normally do not perform work, because there is no
displacement in the direction of the support force. Also,
frictionless linkages between members do not perform net work, since
the forces between the members are equal and opposite. Similarly, the
internal forces that keep a solid body rigid do not perform work.
The bottom line is that normally the are zero if you ignore
friction. However, any collisions against rigid constraints have to
be modeled separately, just like in normal Newtonian mechanics. For
an infinitely rigid constraint to absorb the kinetic energy of an
impact requires infinite force, and would have to be an infinite
spike if described normally. Of course, you could instead consider
describing the constraint as somewhat flexible, with a very high
potential energy penalty for violating it. Then make sure to use an
adaptive time step in any numerical integration.
It may be noted that in relativistic mechanics, the Lagrangian is
not the difference between potential and kinetic energy.
However, the Lagrangian equations of motion (A.1) and
(A.2) still apply.
The general concept that applies both nonrelativistically and
relativistically is that of “action.” The action is defined as the time integral of
the Lagrangian:
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(A.3) |
Here and are suitably chosen starting and ending times
that enclose the time interval of interest. The action is unchanged
by infinitesimal imaginary displacements of the system. It turns
out that that is all that is needed for the Lagrangian equations of
motion to apply.
See {D.3.1} for a derivation of the above claims.
A.1.4 Hamiltonian dynamics
For a system with generalized coordinates the Lagrangian approach
provides one equation for each generalized coordinate .
These equations involve second order time derivatives of the
unknown generalized coordinates . However, if you consider
the time derivatives as additional unknowns, you get
first order equations for these unknowns. An
additional equations are:
These are no longer trivial because they now give the time derivatives
of the first unknowns in terms of the second of them. This
trick is often needed when using canned software to integrate the
equations, because canned software typically only does systems of
first order equations.
However, there is a much neater way to get first order equations
in unknowns, and it is particularly close to concepts in quantum
mechanics. Define the “Hamiltonian” as
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(A.4) |
In the right hand side expression, you must rewrite all the time
derivatives in terms of the canonical momenta
because the Hamiltonian must be a function of the generalized
coordinates and the canonical momenta only. (In case you are not able
to readily solve for the in terms of the
, things could become messy. But in principle,
the equations to solve are linear for given values of the
.)
In terms of the Hamiltonian, the equations of motion are
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(A.5) |
where the , if any, are the generalized forces as before.
If the Hamiltonian does not explicitly depend on time and the
generalized forces are zero, these evolution equations imply that the
Hamiltonian does not change with time at all. For such systems, the
Hamiltonian is the conserved total energy of the system. In
particular for a nonrelativistic system, the Hamiltonian is the sum of
the kinetic and potential energies, provided that the position of the
system only depends on the generalized coordinates and not also
explicitly on time.
See {D.3.2} for a derivation of the above claims.
A.1.5 Fields
The previous subsections discussed discrete mechanical objects like
molecules, Frisbees, and pendulums. However, the Lagrangian and
Hamiltonian formalisms can be generalized to fields like the
electromagnetic field. That is mainly important for advanced physics
like quantum field theories; these are not really covered in this
book. But since it does appear in one advanced addendum,
{A.22}, this subsection will summarize the main points.
The simplest classical field is the electrostatic potential
. However, there may be more than one potential in a
system. For example, in electrodynamics there are also vector
potentials. So the generic potential will be indicated as
, where the index indicates what
particular potential it is. A single potential is
still characterized by infinitely many variables: there is a value of
the potential at each position.
In addition there may be discrete variables. Electromagnetics would
be pretty boring if you would not have some charged particles around.
A generic coordinate of such a particle will be indicated as
. For example, if there is just one charged particle,
, , and could represent the ,
, and components of the position of the particle. If
there are more particles, just keep increasing .
Under the above conditions, the Lagrangian will involve an integral:
Here is called the “Lagrangian density.” It is essentially a Lagrangian per unit
volume. The integral is over all space.
The first part is as before. It will depend on the discrete
variables and their time derivatives:
The dot indicates the time derivative of the variable.
The Lagrangian density will depend on both the fields and the
discrete coordinates:
Here the subscripts on the field indicate partial derivatives:
In principle, there is no reason why the Lagrangian could not contain
higher order derivatives, but fortunately you do not see such things
in quantum field theories.
This brings up one practical point. Consider a contribution such as
the potential energy of a particle called with charge
in an electrostatic field .
Assuming that the particle is a point charge, that potential energy is
where is the
potential evaluated at the position of the particle.
But potentials evaluated at a point are problematic. You would really
want the potentials to always appear inside integrals. To achieve
that, you can assume that the particle is not really a point charge.
That its charge is spread out just a little bit around the nominal
position . In that case, the potential energy
takes the form:
Here is some chosen function
that is zero except within some small distance of
, and that integrates to one. Because this
function is zero except very close to , you can
approximate by and then
take it out of the integral. That gives the original expression for
the potential energy. But the integral is easier to use in the
Lagrangian. Its integrand becomes part of the Lagrangian density.
And you can always take the limit at the end of the
day to get point charges.
The Lagrangian equations for the discrete parameters are exactly the
same as before, but of course now the Lagrangian includes the
integral, {D.3.3}:
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(A.6) |
There is one such equation for each discrete parameter ,
valid at any time.
The Lagrangian equations for the field are based on the Lagrangian
density instead of the Lagrangian itself. That is why you really want
to have the terms involving the field as integrals. The equations are
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(A.7) |
There is one such equation for each field ,
valid at any position and time.
The canonical momenta are now
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(A.8) |
Note that the field momentum is per unit volume.
The Hamiltonian is
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(A.9) |
The time derivatives and must
again be expressed in terms of the corresponding canonical momenta.
Hamilton’s equations for discrete variables are as before:
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(A.10) |
The equations for the fields are a bit tricky. If there are no
discrete variables, there is no problem. Then the Hamiltonian can be
written in terms of a Hamiltonian density as
In that case Hamilton’s equations are
Unfortunately, if there are discrete parameters, products of integrals
will appear. Then there is no Hamiltonian density. So the only thing
you can do do is differentiate the full Hamiltonian instead of a
Hamiltonian density . At the end of every differentiation,
you will then need to drop an and a . In
particular, differentiate the Hamiltonian until you have to start
differentiating inside an integral, like, say,
At that time, make the substitution
This will produce the right answer, although the left hand side above
is mathematically complete nonsense.
See {D.3.3} for a justification of this procedure
and the other claims in this subsection.