This Quantum World/Feynman route/From quantum to classical

From quantum to classical edit

Action edit

Let's go back to the propagator

 

For a free and stable particle we found that

 

where   is the proper-time interval associated with the path element  . For the general case we found that the amplitude   is a function of   and   or, equivalently, of the coordinates  , the components   of the 4-velocity, as well as  . For a particle that is stable but not free, we obtain, by the same argument that led to the above amplitude,

 

where we have introduced the functional  , which goes by the name action.

For a free and stable particle,   is the proper time (or proper duration)   multiplied by  , and the infinitesimal action   is proportional to  :

 


Let's recap. We know all about the motion of a stable particle if we know how to calculate the probability   (in all circumstances). We know this if we know the amplitude  . We know the latter if we know the functional  . And we know this functional if we know the infinitesimal action   or   (in all circumstances).

What do we know about  ?

The multiplicativity of successive propagators implies the additivity of actions associated with neighboring infinitesimal path segments   and  . In other words,

 

implies

 

It follows that the differential   is homogeneous (of degree 1) in the differentials  :

 

This property of   makes it possible to think of the action   as a (particle-specific) length associated with  , and of   as defining a (particle-specific) spacetime geometry. By substituting   for   we get:

 

Something is wrong, isn't it? Since the right-hand side is now a finite quantity, we shouldn't use the symbol   for the left-hand side. What we have actually found is that there is a function  , which goes by the name Lagrange function, such that  .


Geodesic equations edit

Consider a spacetime path   from   to   Let's change ("vary") it in such a way that every point   of   gets shifted by an infinitesimal amount to a corresponding point   except the end points, which are held fixed:   and   at both   and  

If   then  

By the same token,  

In general, the change   will cause a corresponding change in the action:   If the action does not change (that is, if it is stationary at   ),

 

then   is a geodesic of the geometry defined by   (A function   is stationary at those values of   at which its value does not change if   changes infinitesimally. By the same token we call a functional   stationary if its value does not change if   changes infinitesimally.)

To obtain a handier way to characterize geodesics, we begin by expanding


 
 


This gives us


 


Next we use the product rule for derivatives,


 
 


to replace the last two terms of (*), which takes us to


 


The second integral vanishes because it is equal to the difference between the values of the expression in brackets at the end points   and   where   and   If   is a geodesic, then the first integral vanishes, too. In fact, in this case   must hold for all possible (infinitesimal) variations   and   whence it follows that the integrand of the first integral vanishes. The bottom line is that the geodesics defined by   satisfy the geodesic equations


 


Principle of least action edit

If an object travels from   to   it travels along all paths from   to   in the same sense in which an electron goes through both slits. Then how is it that a big thing (such as a planet, a tennis ball, or a mosquito) appears to move along a single well-defined path?

There are at least two reasons. One of them is that the bigger an object is, the harder it is to satisfy the conditions stipulated by Rule   Another reason is that even if these conditions are satisfied, the likelihood of finding an object of mass   where according to the laws of classical physics it should not be, decreases as   increases.

To see this, we need to take account of the fact that it is strictly impossible to check whether an object that has travelled from   to   has done so along a mathematically precise path   Let us make the half realistic assumption that what we can check is whether an object has travelled from   to   within a a narrow bundle of paths — the paths contained in a narrow tube   The probability of finding that it has, is the absolute square of the path integral   which sums over the paths contained in  

Let us assume that there is exactly one path from   to   for which   is stationary: its length does not change if we vary the path ever so slightly, no matter how. In other words, we assume that there is exactly one geodesic. Let's call it   and let's assume it lies in  

No matter how rapidly the phase   changes under variation of a generic path   it will be stationary at   This means, loosely speaking, that a large number of paths near   contribute to   with almost equal phases. As a consequence, the magnitude of the sum of the corresponding phase factors   is large.

If   is not stationary at   all depends on how rapidly it changes under variation of   If it changes sufficiently rapidly, the phases associated with paths near   are more or less equally distributed over the interval   so that the corresponding phase factors add up to a complex number of comparatively small magnitude. In the limit   the only significant contributions to   come from paths in the infinitesimal neighborhood of  

We have assumed that   lies in   If it does not, and if   changes sufficiently rapidly, the phases associated with paths near any path in   are more or less equally distributed over the interval   so that in the limit   there are no significant contributions to  

For a free particle, as you will remember,   From this we gather that the likelihood of finding a freely moving object where according to the laws of classical physics it should not be, decreases as its mass increases. Since for sufficiently massive objects the contributions to the action due to influences on their motion are small compared to   this is equally true of objects that are not moving freely.

What, then, are the laws of classical physics?

They are what the laws of quantum physics degenerate into in the limit   In this limit, as you will gather from the above, the probability of finding that a particle has traveled within a tube (however narrow) containing a geodesic, is 1, and the probability of finding that a particle has traveled within a tube (however wide) not containing a geodesic, is 0. Thus we may state the laws of classical physics (for a single "point mass", to begin with) by saying that it follows a geodesic of the geometry defined by  

This is readily generalized. The propagator for a system with   degrees of freedom — such as an  -particle system with   degrees of freedom — is

 

where   and   are the system's respective configurations at the initial time   and the final time   and the integral sums over all paths in the system's  -dimensional configuration spacetime leading from   to   In this case, too, the corresponding classical system follows a geodesic of the geometry defined by the action differential   which now depends on   spatial coordinates, one time coordinate, and the corresponding   differentials.

The statement that a classical system follows a geodesic of the geometry defined by its action, is often referred to as the principle of least action. A more appropriate name is principle of stationary action.


Energy and momentum edit

Observe that if   does not depend on   (that is,   ) then

 

is constant along geodesics. (We'll discover the reason for the negative sign in a moment.)

Likewise, if   does not depend on   (that is,   ) then

 

is constant along geodesics.

  tells us how much the projection   of a segment   of a path   onto the time axis contributes to the action of     tells us how much the projection   of   onto space contributes to   If   has no explicit time dependence, then equal intervals of the time axis make equal contributions to   and if   has no explicit space dependence, then equal intervals of any spatial axis make equal contributions to   In the former case, equal time intervals are physically equivalent: they represent equal durations. In the latter case, equal space intervals are physically equivalent: they represent equal distances.

If equal intervals of the time coordinate or equal intervals of a space coordinate are not physically equivalent, this is so for either of two reasons. The first is that non-inertial coordinates are used. For if inertial coordinates are used, then every freely moving point mass moves by equal intervals of the space coordinates in equal intervals of the time coordinate, which means that equal coordinate intervals are physically equivalent. The second is that whatever it is that is moving is not moving freely: something, no matter what, influences its motion, no matter how. This is because one way of incorporating effects on the motion of an object into the mathematical formalism of quantum physics, is to make inertial coordinate intervals physically inequivalent, by letting   depend on   and/or  

Thus for a freely moving classical object, both   and   are constant. Since the constancy of   follows from the physical equivalence of equal intervals of coordinate time (a.k.a. the "homogeneity" of time), and since (classically) energy is defined as the quantity whose constancy is implied by the homogeneity of time,   is the object's energy.

By the same token, since the constancy of   follows from the physical equivalence of equal intervals of any spatial coordinate axis (a.k.a. the "homogeneity" of space), and since (classically) momentum is defined as the quantity whose constancy is implied by the homogeneity of space,   is the object's momentum.

Let us differentiate a former result,

 

with respect to   The left-hand side becomes

 

while the right-hand side becomes just   Setting   and using the above definitions of   and   we obtain

 

  is a 4-scalar. Since   are the components of a 4-vector, the left-hand side,   is a 4-scalar if and only if   are the components of another 4-vector.

(If we had defined   without the minus, this 4-vector would have the components  )

In the rest frame   of a free point mass,   and   Using the Lorentz transformations, we find that this equals

 

where   is the velocity of the point mass in   Compare with the above framed equation to find that for a free point mass,

 


Lorentz force law edit

To incorporate effects on the motion of a particle (regardless of their causes), we must modify the action differential   that a free particle associates with a path segment   In doing so we must take care that the modified   (i) remains homogeneous in the differentials and (ii) remains a 4-scalar. The most straightforward way to do this is to add a term that is not just homogeneous but linear in the coordinate differentials:

 

Believe it or not, all classical electromagnetic effects (as against their causes) are accounted for by this expression.   is a scalar field (that is, a function of time and space coordinates that is invariant under rotations of the space coordinates),   is a 3-vector field, and   is a 4-vector field. We call   and   the scalar potential and the vector potential, respectively. The particle-specific constant   is the electric charge, which determines how strongly a particle of a given species is affected by influences of the electromagnetic kind.

If a point mass is not free, the expressions at the end of the previous section give its kinetic energy   and its kinetic momentum   Casting (*) into the form

 

and plugging it into the definitions

 

we obtain

 

  and   are the particle's potential energy and potential momentum, respectively.

Now we plug (**) into the geodesic equation

 

For the right-hand side we obtain

 

while the left-hand side works out at

 

Two terms cancel out, and the final result is

 

As a classical object travels along the segment   of a geodesic, its kinetic momentum changes by the sum of two terms, one linear in the temporal component   of   and one linear in the spatial component   How much   contributes to the change of   depends on the electric field   and how much   contributes depends on the magnetic field   The last equation is usually written in the form

 

called the Lorentz force law, and accompanied by the following story: there is a physical entity known as the electromagnetic field, which is present everywhere, and which exerts on a charge   an electric force   and a magnetic force  

(Note: This form of the Lorentz force law holds in the Gaussian system of units. In the MKSA system of units the   is missing.)


Whence the classical story? edit

Imagine a small rectangle in spacetime with corners

 

Let's calculate the electromagnetic contribution to the action of the path from   to   via   for a unit charge ( ) in natural units (   ):

 
 

Next, the contribution to the action of the path from   to   via  :


 
 

Look at the difference:

 

Alternatively, you may think of   as the electromagnetic contribution to the action of the loop  


 


Let's repeat the calculation for a small rectangle with corners

 


 
 
 
 
 

Thus the electromagnetic contribution to the action of this loop equals the flux of   through the loop.

Remembering (i) Stokes' theorem and (ii) the definition of   in terms of   we find that

 

In (other) words, the magnetic flux through a loop   (or through any surface   bounded by   ) equals the circulation of   around the loop (or around any surface bounded by the loop).

The effect of a circulation   around the finite rectangle   is to increase (or decrease) the action associated with the segment   relative to the action associated with the segment   If the actions of the two segments are equal, then we can expect the path of least action from   to   to be a straight line. If one segment has a greater action than the other, then we can expect the path of least action from   to   to curve away from the segment with the larger action.


 


Compare this with the classical story, which explains the curvature of the path of a charged particle in a magnetic field by invoking a force that acts at right angles to both the magnetic field and the particle's direction of motion. The quantum-mechanical treatment of the same effect offers no such explanation. Quantum mechanics invokes no mechanism of any kind. It simply tells us that for a sufficiently massive charge traveling from   to   the probability of finding that it has done so within any bundle of paths not containing the action-geodesic connecting   with   is virtually 0.

Much the same goes for the classical story according to which the curvature of the path of a charged particle in a spacetime plane is due to a force that acts in the direction of the electric field. (Observe that curvature in a spacetime plane is equivalent to acceleration or deceleration. In particular, curvature in a spacetime plane containing the   axis is equivalent to acceleration in a direction parallel to the   axis.) In this case the corresponding circulation is that of the 4-vector potential   around a spacetime loop.