Classical Mechanics notes (4 of 10)

7/30/2019 Classical Mechanics notes (4 of 10)

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Chapter 4. Hamiltonian Dynamics(Most of the material presented in this chapter is taken from Thornton and Marion, Chap.

7)

4.1 IntroductionHamiltonian mechanics is based rather firmly on Lagranges mechanics but adds a fewtwists that make it extremely useful in certain situations. Lagranges mechanics is a

method for finding the equations of motion of a system, which then must be solved tofind the actual motion. Lagranges mechanics helps with the first part but not the second.

Hamiltonian mechanics is similar in that it provides the equations of motion, but has twopossible advantages. First, Hamiltons formalism provides first-order differential

equations (that is, containing only d/dt), while Lagranges produces second-order

equations (that is, with 2 2/d dt ). Since lower-order equations are easier to solve (usually)

this is an advantage of Hamiltonian mechanics. However we should point out thatHamiltons method produces twice as many equations to solve, so there is something of a

downside. Secondly, through Hamilton-Jacobi theory, a mechanism even exists fordetermining the best coordinate system to solve a given problem in. However, we will see

that this has its own price to pay as well.

4.2 The Canonical Equations of MotionWe define the generalized momentum by

.j

j

Lp

q

=

(4.1)

The units of this may turn out to be linear momentum or angular momentum or

something else entirely because the coordinate qi is a generalized coordinate and could beanything (hence the name generalized momentum).

We can rewrite the Lagrange equations of motion with substitution and rearrangement

0j j

L d L

q dt q

=

(4.2)

as follows

.j

j

Lp

q

=

(4.3)

We can also express our earlier equation for the Hamiltonian (from Chapter 3)


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.j j

H p q L= (4.4)

Equation (4.4) is expressed as a function of the generalized momenta and velocities, andthe Lagrangian, which is, in turn, a function of the generalized coordinates and velocities,

and possibly time. There is a certain amount of redundancy in this since the generalized

momenta are obviously related to the other two variables through equation (4.1). We can,in fact, invert this equation to express the generalized velocities as

( ), , .j j k kq q q p t = (4.5)

This amounts to making a change of variables from ( ) ( ), , to , ,j j j jq q t q p t ; we,therefore, express the Hamiltonian as

( ) ( ), , , , ,k k j j k k H q p t p q L q q t= (4.6)

where it is understood that the generalized velocities are to be expressed as a function ofthe generalized coordinates and momenta through equation (4.5). An important point

(which wont be worked out in detail here) is that, if 1) the equations defining thegeneralized coordinates dont depend explicitly on time and 2) if the forces are derived

from a conservative potential, then

H T U= + (4.7)

Thus the Hamiltonian is often just the energy of the system. However it is important to

note that the Hamiltonian must be expressed as a function of ( ), ,k kq p t and not ( ), ,k kq q t .

The Hamiltonianis always considered a function of the ( ), ,k kq p t set, whereas the

Lagrangian is a function of the ( ), ,k kq q t set:

( )

( )

, ,

, ,

k k

k k

H H q p t

L L q q t

=

=(4.8)

So weve taken the Hamiltonian, which is closely related to the total energy but whoseactually utility to us we have yet to explain, and changed variables, removing any

generalized velocities and replaced them with generalized momenta. Still a few moresteps to go before we can see why this is useful.

Lets now consider the following total differential

,k kk k

H H HdH dq dp dt

q p t

= + +

(4.9)


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(Note: Einstein summation convention!) but from equation (4.6) simply using the chain

rule we can also write

.k k k k k k

k k

L L LdH q dp p dq dq dq dt

q q t

= +

(4.10)

Replacing the third and fourth terms on the right-hand side of equation (4.10) by insertingequations (4.3) and (4.1), respectively, we find

( ) .k k k k L

dH q dp p dq dt t

=

(4.11)

Equating this last expression to equation (4.9) we get the so-called Hamiltons or

canonical equations of motion

k

k

k

k

Hq

p

Hp

q

=

=

(4.12)

and

.H L

t t

=

(4.13)

Furthermore, if we insert the canonical equations of motion in equation (4.9) we find

(after dividing by dt on both sides)

,k k k k

dH H H H H H

dt q p p q t

= +

(4.14)

or again

.dH H

dt t

=

(4.15)

Thus, the Hamiltonian is a conserved quantity if it does not explicitly contain time.

We can therefore choose to solve a given problem in two different ways: i) solve a set of

second order differential equations with the Lagrangian method or ii) solve twice asmany first order differential equations with the Hamiltonian formalism. It is important to

note, however, that it is sometimes necessary to first find an expression for the


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Lagrangian and then use equation (4.6) to get the Hamiltonian when using the canonicalequations to solve a given problem. It is not always possible to find straightforwardly an

expression for the Hamiltonian as a function of the generalized coordinates and momenta.

Example

Use the Hamiltonian method to find the equations of motion for a spherical pendulum of

mass m and length b .

Solution. The generalized coordinates are and . The kinetic energy is therefore

Figure 4.1 A spherical pendulum with generalized coordinates and .

( )

( ) ( )( ) ( ) ( )( ) ( )( )

( ) ( ) ( ) ( ){( ) ( ) ( ) ( ) ( ) }

( )( )

2 2 2

2 2 2

22

2 2

2 2 2 2

1

2

1sin cos sin sin cos

2

1cos cos sin sin

2

cos sin sin cos sin

1 sin ,2

T m x y z

d d dm b b b

dt dt dt

mb

mb

= + +

= + +

=

+ + +

= +

(4.16)

with the usual definition for , , andx y z . The potential energy is

( )cos .U mgb = (4.17)


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The generalized momenta are then

( )

2

2 2

sin ,

Lp mb

L

p mb

= =

= =

(4.18)

and using these two equations we can solve for and as a function of the generalized

momenta.

Since the system is conservative and that the transformation from Cartesian to sphericalcoordinates does not involve time, we can use (4.7) and so have

( ) ( ) ( )

( )( )

22

2 2 2

2 2 2

22

2 2 2

1 1

sin cos2 2 sin

cos .2 2 sin

H T U

pp

mb mb mgbmb mb

ppmgb

mb mb

= +

= +

= +

(4.19)

Finally, the equations of motion are

( )

( )

( )( )

2

2 2

2

2 3

sin

cossin

sin

0.

pH

p mb

pH

p mb

pHp mgb

mb

Hp

= =

= =

= =

= =

(4.20)

Because is cyclic (or ignorable), the generalized momentum p about the symmetry

axis is a constant of motion. p is actually the component of the angular momentum

along the -axisz . Since 0p = , the z-component of the angular momentum will beconserved, or constp = .

4.3 Canonical TransformationsAs we saw in our previous example of the spherical pendulum, there is one case when thesolution of Hamiltons equation is trivial. This happens when a given coordinate is cyclic;


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in that case the corresponding momentum is a constant of motion. This makes that part ofthe solution particularly easy eh? Wouldnt it be nice if you could cleverly figure out a

coordinate system where ALL the coordinates are cyclic? Remember were usinggeneralized coordinates which could be almost any crazy thing we can imagine. Can we

figure out a coordinate system to use where the solution is easy? Is this possible? Lets

look a little more closely, however to do this we have to swim out into some deepwaters

Consider the case where the Hamiltonian is a constant of motion (ie. doesnt explicitly

contain t) , and where all coordinatesj

q are cyclic. The momenta are then constants

,j j

p = (4.21)

and since the Hamiltonian is not be a function of the coordinates or time, we can write

( ) ,jH H

=(4.22)

where 1, ... ,j n= with n the number of degrees of freedom. Applying the canonical

equation for the coordinates, we find

,j j

j

Hq

= =

(4.23)

where thej

s can only be a function of the momenta (i.e., thej

s) and are, therefore,

also constant. Equation (4.23) can easily be integrated to yield

,j j j

q t = + (4.24)

where thej

s are constants of integration that will be determined by the initial

conditions.

It may seem that such a simple example can only happen very rarely, since usually not

every generalized coordinate of a given set is cyclic. But there is, however, a lot offreedom in the selection of the generalized coordinates. Going back, for example, to the

case of the spherical pendulum, we could choose the Cartesian coordinates to solve theproblem, but we now know that the spherical coordinates allow us to eliminate one

coordinate (i.e., ) from the expression for the Hamiltonian. We could have used x,yCartesian coordinates and gotten a perfectly valid set of equations to solve, but neither

coordinate would turn out to be cyclic and the solution of these equations would prove tobe much more difficult. The solution would correctly expressed the motion in x,y

coordinates of course but would be mathematically more complex. The root cause of thisis simply that x,y coordinates do not provide a very natural frame for expressing the

motion of the pendulum, and so we have to fight the coordinate system to a certainextent.


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We can, therefore, surmise that there exists a way to optimize our choice of coordinatesfor maximizing the number of cyclic variables. There may even exist a choice for which

all coordinates are cyclic.

Is there a way beyond intuition to select the best coordinate system? We now try to

determine the procedure that will allow us to transform form one set of variable to

another set. We start out just by considering what constitutes a valid transformation,called a canonical transformation.

In the Hamiltonian formalism the generalized momenta and coordinates are considered

independent, so a general coordinate transformation from the ( ),j jq p to, say, ( ),j jQ P sets can be written as

( )

( )

, ,

, , .

j j k k

j j k k

Q Q q p t

P P q p t

=

=(4.25)

Since we are strictly interested in sets of coordinates that are canonical, we require thatthere exits some function ( ), ,k kK Q P t such that

.

j

j

j

j

KQ

P

KP

Q

=

=

(4.26)

The new function K plays the role of the Hamiltonian. Basically, for our new coordinate

system to be canonical, we need Hamiltons equations to apply in this new coordinatesystem, or in other words, that the new generalized coordinates and momenta be properlyrelated to each other.

It is important to note that the transformation defined by equations (4.25)-(4.26) must be

independent of the problem considered. That is, the pair ( ),j jQ P must be canonicalcoordinates in general, for any system considered.

From the relationship between H and L expressed in equation (4.4) we can rewrite

Hamiltons principle as

( )2

1

, , 0,t

k k j jt

p q H q p t dt = (4.27)

which we can similarly write for the new coordinates and momenta of equations (4.26)

( )2

1

, , 0.t

k k j jt

P Q K Q P t dt = (4.28)

Equations (4.27)-(4.28) are pretty complicated. Were basically saying that the integral ofthe Lagrangian has to be equal in both coordinate systems. But since both coordinate


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systems can be chosen arbitrarily, it turns out mathematically that the only way both canbe true is if

( ) ( ), , , ,k k j j k k j jp q H q p t P Q K Q P t = (4.29)

Well, thats not quite true. It can still hold if we multiply one or the other by a constant

value (which well call here) or if we add a term dF/dt to the equation (why, seebelow). So the formal relationship between the old qi, picoordinates and the new Qi, Picoordinates looks like this

( ) ( ), , , , ,k k j j k k j jdF

p q H q p t P Q K Q P tdt

= + (4.30)

A quick look at why the dF/dt term appears. Consider an arbitrary function F (with acontinuous second order derivative), which can be dependent on any mixture of

, , , andj j j j

q p Q P , and time (e.g., ( ), ,j jF F q P t = , or ( ), ,j jF F Q p t = , or

( ), ,j jF F q Q t = , etc.). Such a function has the following characteristic

( )( ) ( )

2 2

1 1

2

1

2 1

0.

t t

t t

t

t

dFdt dF dt

dt

F

F t F t

=

=

=

=

(4.31)

The last step is valid because of the fact the no variations of the generalized coordinates

are allowed at the end points of the systems trajectory. So the dF/dt term playssomething like the role of a constant of integration. For example

2

2

xxdx C= +

where C can be any constant because when you differentiate it, it disappears.

Nevertheless, the function F actually has a somewhat more important role as it isimportant in specifying the transformation and is called the generating function of the

transformation between the old qi, picoordinates and the new Qi, Pi coordinates.

Ok, back to

( ) ( ), , , , ,k k j j k k j jdF


= + (4.32)

where is some constant. But since it will always be possible to scale the new variables

andj j

Q P , it is sufficient to only consider the case when 1 = . We, therefore, rewrite

equation (4.32) as


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( ) ( ), , , , .k k j j k k j jdF


= + (4.33)

This ugly equation specifies the relationship between the old and new coordinate systems

such that the transformation be canonical. However it doesnt specify what thetransformation should be. It just specifies a condition that must hold if the coordinate

transformation is canonical (that is, appropriate for our use). Since there are a infinitenumber of canonical coordinates systems (Cartesian coordinates, polar coordinates,

cylindrical coordinates, spherical coordinates to name just a few) we still have a lot ofleeway in our choices.

Ok, how do we actually put any of this to use? Lets consider the case where

( ), , .j jF F q Q t = (4.34)This means weve chosen a generating function (yet to be specified) that is only afunction of qj, Qj and t. Why have we chosen the generating function to only be a

function of the old coordinates, new coordinates and t, and not to include the old or newmomenta? Because we can! Is there a particular reason to make this choice? Perhaps, but

the choice of the generating function is something we wont step too deeply into here (seeHamilton-Jacobi theory, later)

Ok, how does our equation (4.33) restrict our choice of the specific form ofF?

Well, equation (4.33) then takes the form

.

k k k k

k k k k

k k

dFp q H P Q K

dtF F F

P Q K q Qt q Q

= +

= + + +

(4.35)

(where weve used the chain rule on dF/dt). This equation will be satisfied (and the

transformation specified by Fwill be canonical) if

.

j

j

j

j

Fp

q

FP

Q

FK H

t

=

=

= +

(4.36)

The first of equations (4.36) represent a system of n equations that define thej

p as

functions of the , , andk k

q Q t. If these equations can be inverted, we then have


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established relations for thej

Q in terms of the , , andk k

q p t, which can then be inserted

in the second of equations (4.36) to find thej

P as functions of , , andk k

q p t. Finally, the

last of equations (4.36) is used to connect K with the old Hamiltonian H (where

everything is expressed as functions of the , andk k

Q P ).

This is an extremely arduous way of making a coordinate transformation! Normally wecan just implicitly use the fact that the transformation between Cartesians and spherical

coordinates (say) is canonical and avoid all this pain and suffering. Why do we worryabout this? Well, first of all, someone had to verify once that the most commonly used

coordinate systems are canonical. More importantly, were interested in the questions ofIs there a best coordinate system to use? (that is, one where all or many coordinates are

cyclic) and How do you find this system?, both questions which require us tounderstand which transformations are allowed. This will become clearer near the end of

this chapter.

Example

Ok, lets do an example where we choose a coordinate system where all the coordinatesare cyclic. In this case we wont yet consider how to choose this system, well just show

that it can be done.

Lets consider the problem of the simple harmonic oscillator. The Hamiltonian for thisproblem can be written as

( )

2 2

2 2 2 2

2 2

1,

2

H T U

p kq

m

p m qm

= +

= +

= +

(4.37)

with 2k m= . The form of the Hamiltonian (a sum of squares) suggests (to a very cleverperson) that the following transformation

( ) ( )

( )( )

cos

sin

p f P Q

f Pq Q

m

=

=

(4.38)

might render the Q cyclic if we make this transformation (here f(P) is some as yet

undefined function only of the new momentum) . Indeed, by substitution we find that


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( )( ) ( )( )

( )

2

2 2

2

cos sin2

,2

f PH Q Q

m

f P

m

= +

=

(4.39)

and Q is indeed cyclic. Now, is this a canonical transformation? And what form should

F(p) have?

We must now find the form of the function ( )f P that makes the transformation

canonical (assuming one exists). Consider the generating function

( )2

cot .2

m qF Q

= (4.40)

How did we find this? Cleverness at this point (Hamilton-Jacobi theory later will tell us

about this process). Nevertheless, using equations (4.36) we get

( )

( )

2

2

cot

,2sin

Fp m q Q

q

F m qP

Q Q

= =

= =

(4.41)

and we can solve for , andq p using these two equations

( )

( )

2 sin

2 cos .

Pq Qm

p Pm Q

=

=

(4.42)

We can now identify ( )f P with the help of equations (4.38)

( ) 2 .f P m P= (4.43)

It follows that

,K H P= = (4.44)

and once again, Q is indeed cyclic. Identifying the Hamiltonian with the energy E,

which is a constant, we find that the generalized momentum P is also a constant

.E

P

= (4.45)


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The equations of motion for Q is given by

,K

QP

= =

(4.46)

which yields

,Q t = + (4.47)

where is an integration constant determined by the initial conditions. Finally, we havefrom equation (4.42)

( )

( )

2

2sin

2 cos .

Eq t

m

p mE t

= +

= +

(4.48)

Although the solution of such a simple problem does not require the use of canonical

transformations, this example shows how the Hamiltonian can be brought into a formwhere all generalized coordinated are cyclic.

4.4 The Hamilton-Jacobi TheoryOk, lets actually work out how to find the best coordinate system (that is, one where

all the coordinates are cyclic). Referring back to section 4.3 on canonical transformations,where

( )

( )

, ,

, , , , 1, ... , ,

j j k k

j j k k

Q Q q p t

P P q p t j k n

=

= =(4.49)

we consider the case where the generating function F is now a function the old

coordinates and the new momenta , andj jq P , respectively. Previously (4.34) we had

selected a generating function that was a function of the qj, Qj and t. Why now use qj, Pj

and t? Because were allowed to and were very clever.

We write

( ), , .j j k kF S q P t Q P= (4.50)

Now what we want to find is a generating function S so that all the coordinates are cyclic.This amounts to find a canonical transformation to a coordinate system where all the

generalized coordinates are cycle (recall the generating function is closely tied to thecoordinate transformation, indeed it specifies it).


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So we follow a procedure similar to what was done for our analysis when

( ), ,j jF F q Q t = above, we can write for the Lagrangian

.

k k k k

k k k k k

k k k

k k k k k k k k

k k

k k k k

k k

dFp q H P Q K

dtF F F F P Q K q Q P

t q Q P

S S SP Q K q P Q P Q P

t q P

S S SQ P K q P

t q P

= +

= + + + +

= + + +

= + + +

(4.51)

Equation (4.51) will be satisfied if

.

j

j

j

j

Sp

q

SQ

P

SK H

t

=

=

= +

(4.52)

The above simply expresses the fact that the coordinate transformation we want to make

must be canonical. We now must add the additional conditions that the transformation

ensures that both the new coordinates and momenta are constant in time. From thecanonical equations we then have

0

0.

j

j

j

j

KQ

P

KP

Q

= =

= =

(4.53)

Equations (4.53) will be satisfied if we simply force the transformed Hamiltonian K to

be identically zero, or alternatively

( ), , 0.j jS

H q p tt

+ =

(4.54)

Hmmm, our new Hamiltonian is identically zero? Isnt that a problem? Well, it could be

but turns out not to be because we dont actually need the new Hamiltonian to find theequations of motion. So we press on


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We can use the first of equations (4.52) to replace the old momentaj

p in equation (4.54)

, , 0,j

j

S SH q t

q t

+ =

(4.55)

with , 1, ... ,j k n= . Equation (4.55), know as the Hamilton-Jacobi equation, constitutes

a partial differential equation in ( )1n + variables, , andjq t, for the generating function

S. The solution S is commonly called Hamiltons principal function. If we can find S,

it gives us the generating function that transforms to a coordinate system where all thecoordinates are cyclic. Too easy eh?

So to find a coordinate system where all the coordinates are cyclic, we have to solve apartial differential equation in n+1 variables? Since partial differential equations are

generally harder than ordinary differential equations have we really gained much? In fact,though it may not be all that clear here, in computing the Hamiltons principal function

(or rather , computing the transformation to the coordinate system where all coordinatesare cyclic) we effectively have to solve the problem of the motion of the particle.

The Hamilton-Jacobi equation still retains a lot of value though as it provides us with yetanother approach for tackling the problem, one which may prove fruitful if other

approaches prove mathematically difficult. But our quest for the perfect coordinatesystem, while successful, generally leaves us with a problem which is not much easier to

solve than our original one.

For those who are interested, more details are below, but lets skip ahead to an example

(below) of how this procedure might work for a simple harmonic oscillator.

Details

The integration of equation (4.55) only provides the dependence on the old coordinates

jq and time; it does not appear to give information on how the new momenta

jP are

contained in S. We do know, however, from equations (4.53) that they are constants.

Mathematically, once equation (4.55) is solved it will yield ( )1n + constants of

integration1 1, ... ,

n

+. But one of the constants of integration, however, is irrelevant to

the solution since S itself does not appear in equation (4.55); only its partial derivatives

to andj

q t are involved. Therefore, if S is a solution, so is S + , where is anyone of

the ( )1n+

constants. We can then write for the solution

( ), , , 1, ... , ,j jS S q t j n= = (4.56)

and we choose to identify the new momenta to the n (non-additive) constants, that is

.j jP = (4.57)


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An interesting corollary is that the new coordinates and momenta can now be expressed

as functions of the initial conditions for the old coordinates and momenta, i.e.,

( ) ( )0 0andj jq t p t . This is because we have in the first two equations of (4.52)

( )

( )

, ,

, ,,

k k

j

j

k k

j j

j

S q t

p q

S q tQ

=

= =

(4.58)

relations that can be evaluated at time0

t . The first equation allows us to connect the

constantsk

(which are, in fact, the new momenta) with the initial conditions when

0t t= is set in equation (4.58), and the second equation completes the connection by

defining the new coordinates in terms of the same initial conditions (through, in part, the

k s). The new coordinates

jQ are also identified with a new set of constants

j (see

equation (4.53).

The problem can finally be solved by, first inverting the second of equations (4.58) to get

( ), , ,j j k kq q t = (4.59)

and using the first of equations (4.58) to replace in the relation forj j

q p to get

( ), , .j j k kp p t = (4.60)

To get a better understanding of the physical significance of Hamiltons principal

function, lets calculate it total time derivative

,

j j

j j

j

j

dS S S S q

dt q t

S Sq

q t

= + +

= +

(4.61)

since the new momentaj j

P = are constants. But from the first of equations (4.52) and

the fact that we set 0K= , we can rewrite equation (4.61) as

.j jdS

p q H Ldt

= = (4.62)

Alternatively, we have


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,S L dt cste= + (4.63)

where the integral is now indefinite. We know that by applying Hamiltons Principle tothe definite integral form of the action, we could solve a problem through Lagranges

equations of motion. We now further find that the indefinite form of the same actionintegral furnishes a different way to solve the same problem through the formalism

presented in this section.

A simplification arises when the Hamiltonian is not a function of time. We can then write

( ), ,j jH q p a= (4.64)

where a is a constant. From equation (4.55) we can write

( ) ( ), , , ,j j j jS q t W q at = (4.65)

and

, .j

j

WH H q

q

=

(4.66)

The function ( ),j jW q is called Hamiltons characteristic function. Its physicalsignificance is understood from its total time derivative

.j j j

j j j

dW W W W q q

dt q q

= + =

(4.67)

But since

,j

j j

S Wp

q q

= =

(4.68)

we find

.j jdW

p qdt

= (4.69)

Equation (4.69) can be integrated to give

,j j j j

W p q dt p dq= = (4.70)


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a quantity usually called the abbreviated action.

Example

We go back to the one-dimensional harmonic oscillator problem. The Hamiltonian for

this problem is

( )2 2 2 21

,2

H p m q Em

= + (4.71)

with

.k

m = (4.72)

We now write the Hamlton-Jacobi equation for Hamiltons principal function S bysetting p S q= in the Hamiltonian

2

2 2 21 0.2

S Sm q

m q t

+ + =

(4.73)

Since the HamiltonianH(4.71) is not a function of time, we can write

2

2 2 21 .2

Sm q

m q

+ =

(4.74)

where is a constant of integration. It is clear that, in this case, the constant of

integration is to be identified with the total energy E (this part is covered in thedetails section we skipped above).

Equation (4.74) can be rearranged into

2 2

2 1 .

2

m qS mE dq Et

E

= (4.75)

Although we could solve equation (4.75) for S, we are mostly interested in its partial

derivatives such as


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2 2

2

'2

12

1

arcsin .2

S S m dqt

E E m q

E

m

q tE

= = =

=

(4.76)

Equation (4.76) is easily inverted to give

( )22

sin ,E

q tm

= + (4.77)

where

'. = (4.78)

The momentum can be evaluated with

( )

( )

2 2 2

2

2

2 1 sin

2 cos .

S Wp mE m q

q q

mE t

mE t

= = =

= +

= +

(4.79)

Finally, we can connect the constants (or ) andE to the initial conditions 0 0andq p

at 0t= . Therefore, setting 0t= in equations (4.77) and (4.79), and rearranging them we

have

2 2 2 2

0 02 .mE p m q= + (4.80)

Using the same equations, we can also easily find that

( ) 0

0

tan .q

mp

= (4.81)

Thus, Hamiltons principal function is the generator of a canonical transformation to a

new coordinate that measures the phase angle of the oscillation and to a new momentumidentified as the energy. Rather strange new coordinates and momenta, but they work. At

the same time, we see that though we have a new approach to take, the road may be arocky one. Nevertheless, Hamiltons equations (4.12) often provide a valuable starting

point even if the full blown Hamilton-Jacobi theory turns out to be too much to handle.


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4.5 The route to quantum mechanics: The Poisson Bracket(This section is optional and will not be subject to examination. It will not be found

in Thornton and Marion. The treatment presented below closely follows that of

Goldstein, Chapter 9 .)

4.5.1 The Poisson Bracket and Canonical TransformationsThe Poisson bracket of two functions andu v with respect to the canonical variables

andj j

q p is defined as

{ },

, ,q p

j j j j

u v u vu v

q p p q

=

(4.82)

where a summation on the repeated index is implied. Suppose that we choose the

functions andu v from the set of generalized coordinates and momenta, we then get thefollowing relations

{ } { }, ,

, , 0,j k j k

q p q pq q p p= = (4.83)

and

{ } { }, ,

, , ,j k j k jk

q p q pq p p q = = (4.84)

wherejk

is the Kronecker delta tensor. Alternatively, we could choose to express

andu v as functions of a set andj jQ P , which are canonical transformations of the

original generalized coordinates andk k

q p , and calculate the Poisson bracket in this

coordinate system. To simplify calculations, we will limit our analysis to so-called

restricted canonical transformations where time is not involved in the coordinatetransformations. That is,

( )

( )

,

, ,

j j k k

j j k k

Q Q q p

P P q p

=

=(4.85)

or, alternatively

( )

( )

,

, .

j j k k

j j k k

q q Q P

p p Q P

=

=(4.86)

In order to carry our calculation more effectively, we will introduce a new notation. Lets

construct a column vector , which for a system with n degrees of freedom is written as


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110

, , 1, ... , .j j j n j

q p j n+

= = = (4.87)

Similarly, another column vector involving the derivatives of the Hamiltonian relative to

the component of is given by

, , 1, ... , .j jj j n

H H H Hj n

q p+

= = =

(4.88)

Finally, we define a 2 2n n matrix J as

,=

0 1J

1 0(4.89)

where the matrices and0 1 are the n n

zero and unit matrices, respectively. Usingequations (4.88) and (4.89), we can write the canonical equations as

.H

=

J

(4.90)

It can be easily verified that equation (4.82) for the Poisson bracket can also be rewrittenas

{ }, ,T

u vu v

=

J

(4.91)

where ( )T

u is the transpose of u and, therefore, a row vector. We can also

compactly write the results of equations (4.83) and (4.84) as

{ }, .

= J (4.92)

Equations (4.92) are called thefundamental Poisson brackets.

Just as we did for the generalized coordinates andk kq p with equation (4.87), we

introduce a new column vector such that

, , 1, ... , .j j j n jQ P j n += = = (4.93)

Relations similar to equations (4.88), (4.90) and (4.91) exist for just as they do for .

We now introduce the Jacobian matrix M relating the time derivatives between the twosystems


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111

,= M (4.94)

where

.j

jk

k

M

=

(4.95)

Transformation of derivatives of the Hamiltonian can also be expressed as a function of

the Jacobian matrix

,kkj

j k j k

H H HM

= =

(4.96)

or

.TH H

=

M

(4.97)

Using equations (4.94), (4.90), and (4.97) we can write

.T

H

H H

=

=

=

M

MJ

MJM J

(4.98)

This last result implies that

.T =MJM J (4.99)

We are now in a position to obtain a few important results. First, from equations (4.91),(4.95), and (4.99) we have

{ },

.

T

T

=

=

=

J

M JM

J

(4.100)

But me must also have a similar result for the fundamental Poisson bracket in thej

system, that is


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112

{ }, .

= J (4.101)

We have, therefore, the result that thefundamental Poisson brackets are invariant undercanonical transformations. We will now show that this result also extends to any

arbitrary functions andu v . We first note that

.Tu u

=

M

(4.102)

Inserting this equation in (4.91) we have

{ },

,

T

T

T T

T

T

T

u vu v

u v

u v

u v

=

=

=

=

J

M J M

MJM

J

(4.103)

and therefore

{ } { } { }, , , .u v u v u v = (4.104)

We thus find that all Poisson brackets are invariant to canonical transformations. This is

the reason why we dropped any index in the last expression in equation (4.104). Just asthe main characteristic of canonical transformations is that the leave the form of

Hamiltons equations invariant, the canonical invariance of the Poisson brackets impliesthat equations expressed in term of Poisson brackets are also invariant in form to

canonical transformations. It is important to note that, although we have limited ouranalysis to restricted canonical transformations, the results obtained here can be shown to

be equally valid for coordinate transformations that explicitly involve time.

The algebraic properties of the Poisson bracket are therefore of considerable importance.Here are some important relations


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113

{ }

{ } { }

{ } { } { }

{ } { } { }

{ } { } { }

{ }{ } { }{ } { }{ }

, 0

, ,

, , ,

, , ,

, , ,

, , , , , , 0.

u u

u v v u

au bv w a u w b v w

uv w u v w u w v

u vw v u w u v w

u v w v w u w u v

=

=

+ = +

= +

= +

+ + =

(4.105)

The last of equations (4.105) is calledJacobis identity.

It is important to note that although we restricted our analysis to canonical

transformations where time was not involved, it can be shown that the results obtainedapply equally well to the more general case.

4.5.2 The Equations of MotionPractically the entire Hamiltonian formalism can be restated in terms of Poisson brackets.Because of their invariance to canonical transformations, the relations expressed withPoisson brackets will be independent of whichever set of generalized coordinates and

momenta are used. Lets consider, for example the total time derivative of a function u

,

j j

j j

j j j j

du u u uq p

dt q p t

u H u H u

q p p q t

= + +

= +

(4.106)

or

{ },du u

u Hdt t

= +

(4.107)

Equation (4.107) is indeed independent of the system of coordinates. Special cases for

equation (4.107) are that of Hamiltons equations of motion

{ }

{ }

,

, .

j j

j j

q q H

p p H

=

=

(4.108)

Equations (4.108) can be combined into a single vector equation using the notation

defined earlier

{ }, .H= (4.109)


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114

We can also verify that

{ }, .dH H H

H Hdt t t

= + =

(4.110)

Finally, if a function u is a constant of motion, that is 0u = , then

{ }, ,u

H ut

=

(4.111)

and for function that are independent of time, we find that

{ }, 0.H u = (4.112)

4.5.3 The Transition to Quantum MechanicsIn quantum mechanics, simple functions, such as andu v , must be replaced by

corresponding operators andu v , so we write

.

u u

v v

(4.113)

For example, if we have a quantum system that we represent by the ket on which

we want to measure the position x of a certain particle that is part of the system, the

action of the measurement of x is represented by x . In cases where is an

eigenvector of the operator x (it is often useful to think of the operators and the kets asmatrices and vectors, respectively), we can write

.x x = (4.114)

In other words, x can be thought of some measuring apparatus for x , and the

physical system on which the measurement is made. Quantum mechanics also requiresthat for any quantity that can be measured experimentally, a so-called observable,

corresponds a linear and hermitian operator. A hermitian operator is defined such that

,u u= (4.115)

where the sign denotes the adjoint operation. For example, when a matrix is used to

represent the operator u we get

* ,jk kju u= (4.116)

and when the operator is hermitian we have


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115

* .jk kj jk

u u u= = (4.117)

Hermitian operators also have the property that their eigenvalues are real. This is to be

expected if they are to adequately represent experimentally measurable quantities.

There are at least two other important facts that characterize measurements in quantummechanics:

1. The quantization of certain physical quantities is an experimental realityincompatible with classical mechanics.

2. The sequence with which successive measurements are made has consequencesthat will affect their results.

Although both statements are extremely important, we will now concentrate on the last

and write its mathematical equivalent. For example, lets assume that we want to measure

two quantities andu v on the system

using their corresponding operators andu v .We could do this in two different ways. That is, we can first measure and thenu v , or

proceed in the opposite sequence. If we denote by 1 and 2 the state of the system

after the two sets of measurements, we can write

( )

( )

1

2 .

v u

u v

=

= (4.118)

According to statement 2 above, in quantum mechanics we have the (extremely) non-

classical result that, in general,

1 2 . (4.119)

We can express the difference in the results obtained between the two sets of

measurements as

( )

[ ]

2 1

, ,

uv vu

u v

=

= (4.120)

where the quantity

[ ] ,u v uv vu= (4.121)

is called the commutatorof andu v . The reason for the importance of the sequence with

which measurements are made in quantum mechanics lays in the fact that quantum, ormicroscopic, systems can be affected by the action of measurement itself. That is, a

system can be changed by a measurement. Of course, this behavior is alien to classical


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116

mechanics, i.e., it does not matter what order we use to make a series of measurements ona macroscopic system.

Starting from equation (4.121) we can calculate the following properties for thecommutators of different operators

[ ]

[ ] [ ]

[ ] [ ] [ ]

[ ] [ ] [ ]

[ ] [ ] [ ]

[ ] [ ] [ ]

, 0

, ,

, , ,

, , ,

, , ,

, , , , , , 0.

u u

u v v u

au bv w a u w b v w

uv w u v w u w v

u vw v u w u v w

u v w v w u w u v

=

=

+ = +

= +

= +

+ + =

(4.122)

Comparison with equations (4.105) will show that the commutator in quantum mechanics

behave just like the Poisson bracket in classical mechanics. It is, therefore, tempting totry to make the transition from classical to quantum mechanics using the Poisson bracket

as our tool. That is, we will attempt to find a relationship between the Poisson bracket

and the commutator. To do so, lets start by evaluating the bracket { }1 2 1 2,u u v v . This can

be done in two different ways using the fourth and fifth of equations (4.105). That is

{ } { } { }

{ } { }( ) { } { }( )

{ } { } { } { }

1 2 1 2 1 2 1 2 1 1 2 2

1 1 2 2 2 1 2 1 1 2 1 1 2 2

1 1 2 2 1 2 1 2 1 1 2 2 1 1 2 2

, , ,

, , , ,

, , , , ,

u u v v u u v v u v v u

u v u v u v v v u v u v v u

u v u v u u v v v u v u u v v u

= +

= + + +

= + + +

(4.123)

or

{ } { } { }

{ } { }( ) { } { }( )

{ } { } { } { }

1 2 1 2 1 1 2 2 1 2 1 2

1 1 2 2 1 2 2 1 2 1 1 1 2 2

1 1 2 2 1 1 2 2 1 2 1 2 1 1 2 2

, , ,

, , , ,

, , , , .

u u v v v u u v u u v v

v u u v u v u u u v u v u v

v u u v v u v u u u v v u v u v

= +

= + + +

= + + +

(4.124)

Equating the last of equations (4.123) and (4.124) we get

{ } { } { } { }1 1 2 2 1 1 2 2 1 1 2 2 1 1 2 2, , , , ,u v u v u v v u v u u v u v u v+ = + (4.125)

or again

( ){ } { }( )1 1 1 1 2 2 1 1 2 2 2 2, , .u v v u u v u v u v v u = (4.126)


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But since the functions1 1

andu v can be chosen to be completely independent from the

2 2andu v functions, we must have

( ) { }

( ) { }

1 1 1 1 1 1

2 2 2 2 2 2

,

, ,

u v v u g u v

u v v u g u v

=

=

(4.127)

where g is some universal constant.

Now, when making a transition from the classical to the quantum world we must come to

a domain intermediate to the two, where both descriptions are valid. Therefore, becauseof the resemblance of the terms on the left side of equations (4.127) to commutators, we

take the step of identifying these terms as such, and make the following correspondence

[ ] { } , , ,u v g u v (4.128)

for any classical functions andu v and their corresponding hermitian operators andu v .

We now have in equation (4.128) a relation that is both quantum and classicalmechanical.

Applying the adjoint operation on both sides of equation (4.128) we get

[ ] ( )

( ) ( )

[ ]

,

, ,

u v uv vu

uv vu

v u u v

vu uvu v

=

=

=

= =

(4.129)

and

{ }( ) { }( )

{ }

{ }

*

**

*

, ,

,

,

g u v g u v

g u v

g u v

=

=

=

(4.130)

since the Poisson bracket is real when andu v are real. Inserting equations (4.129) and(4.130) into equation (4.128) we have

[ ] { }* , , .u v g u v (4.131)

The combination of equations (4.128) and (4.131) implies that the constant g must be a

complex quantity. We therefore further define g as


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118

,g i= (4.132)

with some new real universal constant, commonly called Plancks constant. We finallyhave from equation (4.128)

{ } [ ]1

, , .u v u vi

(4.133)

We are now in a position where we can use the classical equation for the time evolution

of function v , given by equation (4.107), to obtain its quantum mechanical equivalent by

making the substitution suggested in equation (4.133). That is,

1 ,du u

u Hdt i t

= +

(4.134)

with H the quantum mechanical Hamiltonian operator. Equation (4.134) is famouslyknown for representing Heisenbergs picture of quantum mechanics. It is often called

Heisenbergs equation.

The Schrdinger and Heisenberg pictures

In the Schrdinger picture of quantum mechanics the operators are often (but not always)

independent of time, while the evolution of the system is contained in the ketrepresenting it. For example, the time dependent Schrdinger equation is written as

( ) ( ) ( ) ,S Si t H t t t

= (4.135)

where the subscript S specifies that we are dealing with the Schrdinger representation.

On the other hand, in the Heisenberg picture the kets are constant in time and theevolution of the system is contained in the operators (as specified by equation (4.134)).

We can certainly choose the representation of the Schrdinger ket at time0

t for the

Heisenberg ket, since0

t is constant. That is,

( )0 .H S t = (4.136)

Lets consider the operator ( )0,U t t that allows the passage of the Schrdinger ket from

time0

t to time t such that

( ) ( ) ( )0 0, .S St U t t t = (4.137)


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119

It can also be shown that ( )0,U t t is unitary, that is

( ) ( ) 10 0, , ,U t t U t t

= (4.138)

or alternatively

( ) ( ) 0 0, ,U t t U t t = 1 (4.139)

with 1 the unit operator. Inserting equation (4.137) into equation (4.135) we find that

( ) ( ) ( )0 0, , .Si U t t H t U t t t

=

(4.140)

Now, given an observable operator, it is clear that the evolution of its mean value must be

the same irrespective of which picture is used, and we must have

( ) ( ) ( ) ( ) ( ) ,H H H S S Su t u t t u t t = = (4.141)

and therefore upon insertion of equation (4.137) into this last equation, we have

( ) ( ) ( ) ( )

( ) ( ) ( ) ( ) ( ) 0 0

, , .

H H H S S S

S S S

u t t u t t

t U t t u t U t t t

=

= (4.142)

It follows that the representation of operatorH

u , i.e., using Heisenbergs picture, is

related to Su , the same operator in Schrdingers representation, by the following relation

( ) ( ) ( ) ( ) 0 0 , , .H Su t U t t u t U t t = (4.143)

If we now calculate the time derivative of the last equation we get

( ) ( ) ( ) ( )

( )( )

( ) ( ) ( ) ( ) ( )

( ) ( ) ( )

0 0 0 0

0

0

0 0 0 0

0 0

, , , ,

,,

1, , , ,

1, , ,

SHS

S

SS S

S S

uduU t t u U t t U t t U t t

dt t t

U t tU t t u

tu

U t t H t u U t t U t t U t t i t

U t t u H t U t t i

= +

+

= +

+

(4.144)


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where we used equation (4.140) to obtain the last equation. If we now insert

( ) ( )0 0, ,U t t U t t = 1 between ( ) andS SH t u in the first and third term on the right-hand

side of the last of equations (4.144) we have

( ) ( ) ( ) ( ) ( ) ( ) ( )

( ) ( ) ( ) ( ) ( )

0 0 0 0 0 0

0 0 0 0

1

, , , , , ,

1, , , , ,

SH

S S

S S

udu

U t t H t U t t U t t u U t t U t t U t t dt i t

U t t u U t t U t t H t U t t i

= +

+

(4.145)

and finally upon using equation (4.143) (and a similar one for the Hamiltonian) we have

( ) ( ) ( )

( )

0 0

1 , ,

1 ,

SHH H

H H

uduH t u U t t U t t

dt i t

u H ti

= +

+

(4.146)

or

[ ] 1

, SHH H

H

uduu H

dt i t

= +

(4.147)

Comparison of equation (4.147) with equation (4.134) shows that the two pictures of

quantum mechanics are equivalent if we set

( ) ( ) 0 0

, , .S SH

H

u uuU t t U t t

t t t

=

(4.148)

The Schrdinger and Heisenberg are equivalent representations of quantum mechanics.

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Classical Mechanics notes (4 of 10)