sci.physics.research

"Igor Khavkine" < @ > wrote in message
news:slrnfbv8du.8dh. @ ...
> 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
>
I concur with all of the above, and they serve to get the ball rolling
by way of specific, well-defined examples.

Now, how do you go from these examples to S = int_{-oo to +oo} L (t,x)
d^4x? Is there some general rule here? Do +oo and -oo lend themselves
to imposing boundary conditions that can give us a general answer?

Also, perhaps as stepping stones to answer this, permit me please to ask
the following:

1) can you come up with a classical example that yields a discrete
spectrum?

2) can you come up with a quantum theory example that yields a discrete
spectrum?

3) can you articulate a general rule which will tell us when we will
obtain a continuous spectrum and when we will obtain a discrete
spectrum?

Jay.