Schrodinger's Equation
kushal
What are the characteristics of the Schrodinger's Equation? Are they
the single particle trajectories that we get from Newton's Force laws?
If I remember correctly, Schrodinger had derived this equation
starting from classical action. So, there must be a connection between
the single particle trajectories and the solutions and characteristics
of the Schrodinger's equation.
Thanks,
Kushal.
robert bristow-johnson
>
> What are the characteristics of the Schrodinger's Equation? Are they
> the single particle trajectories that we get from Newton's Force laws?
it doesn't describe a trajectory of a particle in the same sense as
classical mechanics does. it describes the probablity of existence of
the particle at a particular place in space and time.
> If I remember correctly, Schrodinger had derived this equation
> starting from classical action.
i always thought he hypothesized it as one of several differential
equations that have the DeBroglie wave equation as a solution.
> So, there must be a connection between
> the single particle trajectories and the solutions and characteristics
> of the Schrodinger's equation.
i think the connection between what classical physics predicts (an
equation for a trajectory) and what wave mechanics predicts is in the
Expectation Value. long ago when i was a student, i tried to apply
this "correspondance principle" in a naive way. i tried to solve the
time-dependant Schrodingers equation for a simple particle in a box
(or maybe it was the hamonic oscillator) and then, leaving t
variable, solve for the expection value of x (as a function of t). i
was hoping to see as the mass of the particle got large, that the
result of the expection would approach what classical mechanics would
say. but i was unsuccessful and never really understood my prof's
explanation for why.
r b-j
one thing that i tried to do
kushal
<r...@audioimagination.com> wrote:
> On Aug 8, 3:39 pm, kushal <atmabo...@gmail.com> wrote:
>
>
>
> > What are the characteristics of the Schrodinger's Equation? Are they
> > the single particle trajectories that we get from Newton's Force laws?
>
> it doesn't describe a trajectory of a particle in the same sense as
> the particle at a particular place in space and time.
>
> > If I remember correctly, Schrodinger had derived this equation
> > starting from classical action.
>
> i always thought he hypothesized it as one of several differential
> equations that have the DeBroglie wave equation as a solution.
>
> > So, there must be a connection between
> > the single particle trajectories and the solutions and characteristics
> > of the Schrodinger's equation.
>
> i think the connection between what classical physics predicts (an
> equation for a trajectory) and what wave mechanics predicts is in the
> this "correspondance principle" in a naive way. =A0i tried to solve the
> time-dependant Schrodingers equation for a simple particle in a box
> variable, solve for the expection value of x (as a function of t). =A0i
> was hoping to see as the mass of the particle got large, that the
> result of the expection would approach what classical mechanics would
> explanation for why.
>
> r b-j
>
> one thing that i tried to do
Yeah, its true that QM does not describe particle trajectories in the
sense of CM, but there should be some connection between a wave
function and the trajectories, since Schrodinger's equation [SE] was
derived from classical action using the ideas of wave optics.
Although its a wide belief that SE is just a hypothesis, but its not
really so. Though one cannot rigorously prove SE, but this equation
has been developed in a very elegant way. A reading of the original
paper by S can be very helpful in this regard.
Thanks,
Kushal.
Rich L.
viewpoints. In classical mechanics we are representing the problem in
position space, but in quantum mechanics the representation is in
energy or momentum space. To make this clearer, consider a vibrating
string. In classical mechanics we talk of the individual atoms of the
string and how they are connected elastically to their neighbors. We
understand the motion of the string in terms of the motions of the
individual atoms. An alternative representation is as the collective
motion of the string as waves, where the motions of the individual
atoms are not explicitly represented. This wave can be described in
terms of its wavelength (i.e. momentum) and/or energy (amplitude).
Both representations are correct. (Whether each is complete is an
other question...)
The collective motion of the vibrating string makes as much sense as
the collective motion of a pendulum, even though the individual atoms
are not making the same movements at the same time, they are still
part of the same motion and are all interconnected. Anything that
alters the vibration of the string alters the collective motion
exactly analogous to a pendulum being disturbed. An just as a
pendulum can be abstracted to a point mass at the end of a massless
rod, the wave on string can be abstracted as a single pendulum.
I recall reading several years ago about a calculation for electrons
in highly excited states in an atom. If the wave functions for many
excited energy levels are combined in the appropriate way, they got a
wave function that showed a localized electron orbiting the atom. I
think this is a key point. If you study only one energy level you
only see the wave nature of the particles. If you want to see the
particle nature you need to superimpose many wave funtions
corresponding to different energy/momentum states.
Again back to the analogy with the vibrating string: A localized pulse
on the string (similar to a localized electron) can be represented by
a superposition of many vibration modes. When you do this, the pulse
will propagate back and forth along the string (assuming it has fixed
ends) just like you'd expect an electron to in a closed box. A
quantum mechanical calculation with many superimposed modes will show
the same behavior. It is the superposition of many modes (eigen
states) that gives the classical particle-like behavior.
Rich L.
robert bristow-johnson
>
> Yeah, its true that QM does not describe particle trajectories in the
> sense of CM, but there should be some connection between a wave
> function and the trajectories,
i guess, in my simple-minded understanding, that connection would be
the Expectation Value (applied to the solution of the time-dependent
Shrodinger Equations) where the expected position of the particle
(x,y,z) is solved for as a function of t. but i'm just an electrical
engineer whose physics was limited to what they make EEs take (and i
took an elective in astrophysics in the 70s, which might be completely
superceded now) where i had one semester of QM followed by a semester
of semiconductor physics. so, really, i don't know diddley.
when i review my third-semester physics book, i see that they did this
for the harmonic oscillator and the time-INdependent Schrodinger
equation for 1 dimension. so the result of the expectation was not a
function of time, but simply the probability (over all time) of
particle to be at position x. it looks like if you were to low-pass
filter out the oscillations, that probability function would look like
what classical mechanics would predict. but it still is not a
trajectory, there is no position as a function of time. "t" was
eliminated from the problem at the outset.
>
> Although its a wide belief that SE is just a hypothesis,
i hadn't said that it's *presently* just a hypothesis. but, before it
stood the test of physical experiment, it was. that's true for any
other physical law, ain't it?
r b-j
kushal
Yeah, I agree that the idea of a trajectory in CM is not valid in QM
where only the expectation values are important. But the Schrodinger's
equation is a PDE, 1st order in time and 2nd order in space. And,
usually, for solving PDEs, we use the method of characteristics. So, I
was wondering if this method of characteristics applies to the
Schrodinger's equation also. If it does, then what are the
characteristics? If it does not apply, then why so? Both these cases
would be interesting.
The Schrodinger's Equation is not provable in the strict sense, but
this equation has emerged out of a very smart way of looking at wave
motion in optics and coupling that to Hamiltonian systems. The
Schrodinger's equation has a very deep connection with Classical
Hamiltonian Mechanics. A reading of Schrodinger's original paper can
be very enlightening in this regard.
Thanks,
Kushal.
On Aug 10, 12:08 am, robert bristow-johnson
kushal
@Rich: I guess I would agree with what you have said. A superposition
of many eigen states can be taken to represent a "localized" wave
packet and hence a particle. But the term "localized" has to be
carefully defined. The mean motion of this wave packet should
represent the classical trajectory. A single eigen state of SE does
not have direct correspondence with CM. But since a collection of
eigen states leads to a mean classical trajectory kind of thing,
should not the individual eigen states also carry the classical
information in some form. Because if the individual states did not
carry the classical information, how would a collection of them be
able to display this classical property.
@Robert: Yes, its true that any physical law has to be proved by
experiment, but we can't say that its a hypothesis till being proved.
A law can be called a hypothesis if it is based on come conjecture
based on pure intuition without having any solid rational ground for
its proposal. But the SE was founded upon wave optics and Hamiltonian
mechanics. So, in my view, it won't be correct to call it a
hypothesis.
Thanks,
Kushal.
Materion
In my view, you may use Schrodinger's equation to characterize the
spinning motion of a arrow-like object (needle, rod, wheelspoke,
nail...). If the orientation of a needle changes, the differential
between the vector representing that arrow at two subsequent instants
may be written as:
d|arrow> = i omega |arrow> dt
which is the time-dependent form of the SE. The imaginary 'i'
indicates that the vector difference d|arrow> is perpendicular to the
arrow.
So the SE describes the single arrow-like particles spinning motion
that are subject to the fluctuation of the surrounding field.
I've tried to explain it on my blog
http://commonsensequantum.blogspot.com/2008/07/second-video-sequence-schrdinger.html
Best regards,
Arjen Dijksman
Rock Brentwood
> What are the characteristics of the Schrodinger's Equation? Are they
> the single particle trajectories that we get from Newton's Force laws?
It's a parabolic differential equation, same form as the heat
equation. Equivalently, it's a hyperbolic differential equation in one
higher dimension with a first-order differential constraint.
Rewrite it in the following form:
(del^2 + 2 d^2/dtdu - z d^2/dt^2) psi = 0
and
(i h-bar d/du - m) psi = 0.
In momentum-energy space, p = -i h-bar del, and H = i h-bar d/dt, and
you get the following
p^2 - 2mH - z H^2 = 0.
The case z = 0 gives you the non-relativistic Schroedinger equation.
The case z = (1/c)^2 gives you the relativistic form (i.e. the Klein-
Gordon equation), up to a phase shift (the phase shift, itself,
amounts to projecting out the 5th coordinate, u). The phase change is
equivalent to replacing the kinetic energy H by the total energy E = H
+ mc^2 (in which case, the quadratic invariant reduces to p^2 - (E/
c)^2 + (mc)^2 = 0).
The z > 0 case projects onto a hyperbolic differential equation (which
is what the Klein-Gordon equation is), while z < 0 projects onto an
elliptical differential equation (the Euclidean form of the Klein-
Gordon equation).
> If I remember correctly, Schrodinger had derived this equation
> starting from classical action.
The equation is, more properly, derived from the transformation theory
underlying the space-time symmetry group: the Galilean group for non-
relativistic physics, the Poincare' group for relativistic physics.
So, in the Galilean case, one *starts* with the generators J, K for
rotations and boosts; P, H for space translations and time
translations, and the invariant m (for "translation" along the u
"coordinate"). One then ends up proving that P^2 - 2mH is an
invariant. For the family characterized by m != 0, one then defines r
= K/m, and S = J - r x P and can then *prove* that (r,P) forms 3 sets
of Heisenberg conjugates, while S satisfies the algebra for spin.
The Stone von Neumann theorem may then be used to show that a state-
space representation for (r,P) reduces to the Schroedinger
representation (up to unitary equivalence)
P = -i h-bar del, r = multiplicative operator.
Then -- in this representation -- the invariant P^2 - 2mH = constant
becomes
-h-bar^2 del^2 - 2m i h-bar d/dt = constant.
The constant on the right can be reabsorbed by redefining the phase,
so one can write
i h-bar d/dt = -h-bar^2/2m del^2
which is the Schroedinger equation for the free particle.
The interpretation of H = i h-bar d/dt, in contrast to p = -i h-bar
del, does NOT come out of the Stone von-Neumann theorem or any notion
of (H,t) being Heisenberg conjugates. They're not. Instead, it comes
out of the starting point above, where H was defined as the generator
of time translations.
There are only 3 Heisenberg conjugates instead of 4, because a
prospective 4th one is projected out by the quadratic invariant (P^2 -
2mH). As a consequence, there is also a 1-parameter's worth of
ambiguity in the definition of the position vector: r = K/m - (P/m) s;
where s is the extra parameter. Consequently, the definitions for J
and K reduce to
J = r x P = -i h-bar r x del, K = mr - tP = mr + i h-bar s P.
The extra parameter, s, of course then goes on to play the role of the
particle's clock time; and one has, in effect, the equivalence s = t.
The same process works in the relativistic case (for slower-than-light
particles), but the definition of the position operator is MUCH more
complicated. The "Schroedinger equation" as per the above discussion
will reduce to
p^2 - 2mH - z H^2 = constant;
so one ends up getting
-h-bar^2 del^2 - 2m i h-bar d/dt + (1/c)^2 h-bar^2 d^2/dt^2 =
constant
Adding a suitable constant on the left, this becomes
h-bar^2 (del^2 - 1/c^2 (d/dt - imc^2/h-bar)) = constant
which is the Klein-Gordon equation, up to a phase change. Again, this
is equlvalent to replacing the kinetic energy H by the total energy E.
In operator form, the transformation is
H = i h-bar d/dt going to H + mc^2 = i h-bar d/dt
or
d/dt --> d/dt - imc^2/h-bar.
Analogous to before, the particle's clock time s plays the role of the
extra parameter and one has a kind of "soldiering relation" s = t + u/
c^2, relating it to the extra u coordinate.
For the non-relativistic family characterized by m = 0, there is
nothing in the literature I'm aware of discussing it. It doesn't have
a Schroedinger equation, per se -- at least not in the usual form.
Instead, one has a Heisenberg conjugate pair (H, t), where this time
one can actually DEFINE the time operator t = -K.P/P^2. It has no
position operator r (at least not one whose components x,y,z commute
amongst themselves), so the constraint P^2 - 2mH = P^2 = constant does
NOT reduce to a Poisson equation representation (-h-bar del^2 =
constant).
I'm not sure what kind of representation the m = 0 sector has, nor do
I know exactly what its differential equation would look like. It's,
essentially, the non-relativistic form of both the photon and tachyon;
"going" at infinite speed, transmitting an impulse j = |P| directly
between two points in space at a single time, t.
For the relativistic case, neither photons nor tachyons have position
operators either (at least, not in the usual sense, again, where the
components x,y,z commute with one another). So, the usual route to
Schroedinger equation from representation theory is blocked here too.
In energy-momentum representation, however, everything is simple.
There are Dirac equations, for all the cases above that comes about
simply by taking the "square root" of the quadratic invariant P^2 -
2mH - z H^2 into (alpha.P + delta H + epsilon m), where
{alpha_i, alpha_j} = delta_{ij}; {alpha_i, delta} = 0 = {alpha_i,
epsilon}
delta^2 = -z, epsilon^2 = 0, {delta, epsilon} = -1.
The resulting algebra is equivalent to the complex 4x4 matrix algebra,
regardless of what z is. The non-relativistic case (z = 0, m != 0)
gives you the Schroedinger equation. The relativistic case (z > 0, P^2
- 2mH - zH^2 = 0) gives you the Klein-Gordon equation.
Arnold Neumaier
> On Aug 8, 2:39 pm, kushal <atmabo...@gmail.com> wrote:
>> What are the characteristics of the Schrodinger's Equation? Are they
>> the single particle trajectories that we get from Newton's Force laws?
>
> It's a parabolic differential equation, same form as the heat
> equation.
The Schoedinger equation is _not_ a parabolic equation, but
a hyperbolic equation.
Parabolic equations cannot have plane wave solutions - all their
solutions decay!
> Equivalently, it's a hyperbolic differential equation in one
> higher dimension with a first-order differential constraint.
parabolic and hyperbolic is never equivalent.
Arnold Neumaier
Lou Pecora
<36b63aa9-914d-40e5...@v39g2000pro.googlegroups.com>,
Rock Brentwood <mark...@yahoo.com> wrote:
> On Aug 8, 2:39 pm, kushal <atmabo...@gmail.com> wrote:
> > What are the characteristics of the Schrodinger's Equation? Are they
> > the single particle trajectories that we get from Newton's Force laws?
>
> It's a parabolic differential equation, same form as the heat
> equation. Equivalently, it's a hyperbolic differential equation in one
> higher dimension with a first-order differential constraint.
>
> Rewrite it in the following form:
> (del^2 + 2 d^2/dtdu - z d^2/dt^2) psi = 0
> and
> (i h-bar d/du - m) psi = 0.
[cut]
Interesting. Got a reference for all that? Thanks.
--
-- Lou Pecora
kushal
> In article
> <36b63aa9-914d-40e5-b1a3-90469c6a0...@v39g2000pro.googlegroups.com>,
> Rock Brentwood <markw...@yahoo.com> wrote:
>
> > On Aug 8, 2:39 pm, kushal <atmabo...@gmail.com> wrote:
> > > What are the characteristics of the Schrodinger's Equation? Are they
> > > the single particle trajectories that we get from Newton's Force laws?
>
> > It's a parabolic differential equation, same form as the heat
> > equation. Equivalently, it's a hyperbolic differential equation in one
> > higher dimension with a first-order differential constraint.
>
> > Rewrite it in the following form:
> > (del^2 + 2 d^2/dtdu - z d^2/dt^2) psi = 0
> > and
> > (i h-bar d/du - m) psi = 0.
>
> [cut]
>
> Interesting. Got a reference for all that? Thanks.
>
> --
> -- Lou Pecora
If the Schrodinger's equation is a hyperbolic equation, then the
method of characteristics should work for it.
Quoting from wiki (http://en.wikipedia.org/wiki/Method_of_characteristics) :
"In mathematics, the method of characteristics is a technique for
solving partial differential equations. Typically, it applies to first-
order equations, although more generally the method of characteristics
is valid for any hyperbolic partial differential equation. "
So, are the characteristics of SE, the particle trajectories of
Classical Mechanics?
Kushal.