sci.physics.research

On 2007-08-12, JohnMS wrote:
> On 11 Aug., 23:53, Igor Khavkine wrote:
>
>> There is the "action functional", S[x(t)], which is an integral
>> expression involging the Lagrangian, which associates a number to any
>> motion of the system, x(t). The variation of S[x(t)] with respect to
>> x(t) gives the dynamical equations of motion for ths system. I don't
>> know of any way to determine, empirically, the action functional for any
>> given system. In fact, there are often different but equivalent action
>> functionals that can correspond to the same physical system! So, the
>> "action functional" is not a likely candidate for a physical observable
>> (this statement can be made more technical and more forceful).
>
> This was what I meant. Now S=int L dt in classical physics
> can be determined (at least for the actual path) if L is
> specifically
> defined as Ekin-Epot. This quantity is surely measurable.

OK, this is doable. Consider a single particle system. Let x(t) denote
its position at time t and v(t) its velocity at time t. Both of these
are observable, without argument. Now, take a function of two arguments
L(x,v) and define S(T) = int_{0 to T} L(x(t),v(t)) dt, where the limits
of integration are chosen for convenience. Since S(T) is a function of
other observables, namely x(t) and v(t), then then S(T) itself is an
observable.

Incidentally, the quantity S(T) itself has a name: Hamilton's principle
function. When the dependence of S(T) on T as well as the initial
conditions of the motion are taken into account, it is a solution of the
Hamilton-Jacobi equation. This is another well studied topic in
analytical mechanics.

> If we use the same definition in quantum theory, S
> must be measurable as well !? (I add a question mark,
> because I agree that this topic is not often discussed,
> so I am somewhat in muddy waters.) Do you agree?

I don't know of any device that can directly measure S(T), as defined
above. But, int principle, taking measurements of a particle's position
and velocity at successive times and using the integral definition of
S(T), it can be obtained from observations.

> If there is agreement here, we can go on afterwards to see
> how to measure it in quantum theory.

While you may be trying to think of better ways of measuring S(T),
consider the following examples.

A. A simple harmonic oscillator. L = (m/2)*(v^2 - w^2*x^2). A straight
forward calculation gives S(T) = (m/2)*(V^2 - w^2*X^2)*sin(2*w*T)/(2*w),
where X and V are position and velocity at t = 0. Upon quantization, X
is promoted to the usual x operator in the Schroedinger picture, while V
is defined as p/m, where p is the Schroedinger momentum operator.

Now, there is a well posed question: what is the spectrum of the
quantized S(T) operator? Whenever T is an integral multiple of pi/(2*w),
S(T) is simply zero. But, for generic T, the answer is that the spectrum
is continuous, ranging from -infinity to +infinity. And its eigenstates
certainly do not look anything like energy eigenstates of the
oscillator. This conclusion can be drawn from noticing that S(T) is
merely a T-dependent scalar multiplying the Hamiltonian for the inverted
harmonic oscillator, where x^2 is replaced by -x^2 in the potential.
See, for instance, Pedrosa and Guedes arXiv:quant-ph/0307084.

B. A free particle. L = (m/2)*v^2. A similar calculation now gives
S(T) = (m/2)*V^2*T, with same notation as above. In this case we have
the identity S(T) = H*T, where H is the Hamiltonian of the free
particle.

So, since we know the spectrum of H, we also know the spectrum of S(T).
Namely, S(0) = 0, for T > 0, the spectrum is continuous from 0 to
+infinity, and for T < 0, the spectrum is continuous from 0 to
-infinity.

So, in both these cases, the spectrum of S(T) (a) depends on T and (b)
is not even discrete.

Igor