Schrodinger's Equation

**Hyperreality** · Dec27-04, 12:30 AM

Can Schrodinger's Equation be derived without a boundary condition?

Particles according to quantum physics are only "partly localised", so does it mean that Schrodinger's equation can only be applied in a confined region of space?

Also, from what I read from my text book, Schrodinger's Equation is applied to wave packets, because it has an "estimated" boundary of $LaTeX Code: \\Delta x$ of large magnitude. If so, how can a simple harmonic quantum oscillator exist? An ideal simple harmonic motion is represented by pure sine or cosine waves, where $LaTeX Code: \\Delta x = \\infty$ .

**dextercioby** · Jan2-05, 09:03 PM

Originally Posted by Hyperreality

Can Schrodinger's Equation be derived without a boundary condition?

Schroedinger's equation cannot be derived.It is accepted as a postulate.U mean 'solved'.In that case,it depends on the typical problem it is applied.Generally we take into account wavefunctions defined on all R^{3n}.

Originally Posted by Hyperreality

Particles according to quantum physics are only "partly localised", so does it mean that Schrodinger's equation can only be applied in a confined region of space?

No,Schroedinger's equation can be applied for all space,for all conditions,for all possible cases.However,not every solution to this equation describes a possible quantum state of a system.

Originally Posted by Hyperreality

Also, from what I read from my text book, Schrodinger's Equation is applied to wave packets, because it has an "estimated" boundary of $LaTeX Code: \\Delta x$ of large magnitude.

Yes,it's applied to to wave packets,simply because de Broglie's plane momentum waves do not represent phyiscal states of a quantum particle as they cannot be normalized.

Originally Posted by Hyperreality

If so, how can a simple harmonic quantum oscillator exist?An ideal simple harmonic motion is represented by pure sine or cosine waves, where $LaTeX Code: \\Delta x = \\infty$ .

It can exist,as all solution of the Schroedinger's equation can be normalized and hence describe possible quantum states.Pick one state of the QSHO and compute $LaTeX Code: \\Delta \\hat{x}$ and see whether it is infinite or not...

Daniel.

**wizzart** · Jan3-05, 01:45 PM

@dextercioby:

I've seen you state a few times that Schrodinger's equation cna't be derived, but needs to be accepted as a postulate or axiom. I think that would be a bit strange, since what then would bring schrodinger to formulate it? Also, I've read a derivation of the equation following from Feynman's path integral formulation of QM, that, as far as I could tell, didn't take the Schrodinger equation as a known fact. But I'm quite the newbie on path integral QM, so I could be wrong.

**dextercioby** · Jan3-05, 02:06 PM

Traditional nonrelativistic QM has 3 versions/formulations:
1.Formulation with vectors and operators,due to mostly to Paul Adrien Maurice Dirac and founded mathematically by John von Neumann.
2.Formulation with operators due to John von Neumann.
3.Formulation with path-integrals due to Richard P.Feynman.

At conceptual level all these 3 formulations are equivalent.That means one implies the other and viceversa.In particular,the Schroedinger equation can be obtained from the treatment by Feynman and naturally viceversa.Actually Feynman took Schroedinger's eq.for granted and proved its complete equivalence with equations containing path-integrals.

So,in the Schroedinger picture of Dirac/traditional formulation of nonrelativistic QM,the IV-th postulate contains the Schroedinger's eq.But in the Feynman formulation,this eq.can be derived through a more tricky procedure.This formulation due to Fynman is most commonly made in the Heisenberg picture,so one has to use the equivalence between the Heisenberg picture and the Schroedinger's one to get the equation.

Anyway,the key-word is "equivalence".

Daniel.

**jtbell** · Jan3-05, 05:39 PM

Originally Posted by wizzart

I've seen you [dextercioby] state a few times that Schrodinger's equation cna't be derived, but needs to be accepted as a postulate or axiom. I think that would be a bit strange, since what then would bring schrodinger to formulate it?

Schrödinger's inspiration (not a derivation in the rigorous mathematical sense) was to make an analogy between mechanics and optics. I have a PDF copy of Schrödinger's first English-language paper on the subject [Phys. Rev. 28, 1049 (1926)] following on his original German-language papers. He starts with the Hamiltonian action integral of classical mechanics:

$LaTeX Code: W=\\int{(T-V)dt}$

and considers the family of surfaces in space on which W is constant. It turns out that the possible trajectories of a particle in that system are perpendicular (normal) to those constant-W surfaces. He comments:

Originally Posted by Schrödinger

The above-mentioned construction of normals dn is obviously equivalent to Huygens' principle. The orthogonal curves of our system of W-surfaces form a system of rays in our optical picture; they are possible orbits of the material points in the mechanical problem.

A bit further on:

Originally Posted by Schrödinger

The well-known mechanical principle due to and named after Hamilton can very easily be shown to correspond to the equally well-known optical principle of Fermat.

He discusses the relationship between geometrical (ray) optics and wave optics, and then:

Originally Posted by Schrödinger

Now compare with these considerations the very striking fact, of which we have today irrefutable knowledge, that ordinary mechanics is really not applicable to mechanical systems of very small, viz. of atomic dimensions.

[...] is one not greatly tempted to investigate whether the non-applicability of ordinary mechanics to micro-mechanical problems is perhaps of exactly the same kind as the non-applicability of geometrical optics to the phenonema of diffraction or interference and may, perhaps, be overcome in an exactly similar way?

[...] At any rate the equations of ordinary mechanics will be of no more use for the study of these micro-mechanical wave-phenomena than the rules of geometrical optics are for the study of diffraction phenomena. Well known methods of wave theory, somewhat generalized, lend themselves readily. The conceptions, roughly sketched in the preceding are fully justified by the success which has attended their development.

Schrödinger associates Hamilton's W-function with the phase of his $LaTeX Code: \\psi$ -function:

$LaTeX Code: \\psi=A(x,y,z) \\sin(W/\\hbar)$

He assumes that this has to satisfy the usual differential wave equation, with velocity

$LaTeX Code: u=E/\\sqrt{2m(E-V)}$

and on substituting these into the differential wave equation, out pops what we now know as the time-independent Schrödinger equation!

**dextercioby** · Jan3-05, 05:57 PM

To carry on from where Jtbell left off,i'll post Schroedinger's view on the (Hydrogen) atom.

Originally Posted by Erwin Schroedinger

So the whole of the wave-phenomenon,though mathematically spreading throughout all space,is essentially restricted to a small sphere of a few Angstroms diameter which may be called "the atom" according to ondulatory mechanics.

However,though it has been obtained from classical phyiscs,this equation is still the basis of quantum physics and it cannot be applied to macroscopic level/classical systems.

Daniel.

**jvicens** · Mar5-05, 10:58 PM

I never had a formal QM course but I'm jumping into it by myself. For the last four days I've tried to understand the "derivation" of Schrodinger's equation. As a matter of fact, as I now type this reply I have 5 books on my desk open in Schrodinger chapter. Of course I have not been able to fully understand why the "derivation" process was jumping from one integral to another. Now I see.
I guess I should not feel bad if I don't understand its derivation as there is no real rigorous mathematical derivation but it is in part based on considerations. Right?

**dextercioby** · Mar6-05, 04:41 AM

Well,there are formulations of QM in which SE is not a postulate.Even in Dirac'ss formulation,if you postulate Heisenberg equation,you'll be able to prove SE...

Daniel.

**thinker** · Mar6-05, 07:25 AM

Originally Posted by Hyperreality

Can Schrodinger's Equation be derived without a boundary condition?

Particles according to quantum physics are only "partly localised", so does it mean that Schrodinger's equation can only be applied in a confined region of space?

Also, from what I read from my text book, Schrodinger's Equation is applied to wave packets, because it has an "estimated" boundary of $LaTeX Code: \\Delta x$ of large magnitude. If so, how can a simple harmonic quantum oscillator exist? An ideal simple harmonic motion is represented by pure sine or cosine waves, where $LaTeX Code: \\Delta x = \\infty$ .

I have the same question.:-)

**ZapperZ** · Mar6-05, 08:27 AM

Originally Posted by Hyperreality

Can Schrodinger's Equation be derived without a boundary condition?

Particles according to quantum physics are only "partly localised", so does it mean that Schrodinger's equation can only be applied in a confined region of space?

Also, from what I read from my text book, Schrodinger's Equation is applied to wave packets, because it has an "estimated" boundary of $LaTeX Code: \\Delta x$ of large magnitude. If so, how can a simple harmonic quantum oscillator exist? An ideal simple harmonic motion is represented by pure sine or cosine waves, where $LaTeX Code: \\Delta x = \\infty$ .

You have a considerable misunderstanding of what a "Schrodinger Equation" is, and, I suspect, a lack of knowledge on the mathematics involved. The Schrodinger equation itself is INDEPENDENT of any boundary condition. It is, after all, simply a 2nd order differential equation. It is the SOLUTIONS to the Schrodinger equation that is dependent on the boundary conditions. The solutions are the ones tailored to whatever boundaries that exist.

Does this, outright, answer your question?

Zz.

**marlon** · Mar8-05, 01:00 PM

Originally Posted by Hyperreality

Can Schrodinger's Equation be derived without a boundary condition?

The SE is independent of any boundary condition. There is a simple way to "prove" the SE. Let's assume we have no problem with wavefunctions (why should we, right ?). Given the equation for a wavefunction (ie the exponential structure) we now that taking the second derivative with respect to position yields the momentum p squared : p² (with some constants). The first derivative with respect to time yields the energy E. Now we know that E = p²/2m so if you substitute p² and E with the derivatives of the wavefunctions, you get the SE

regards
marlon

~~Crosson~~ · Mar8-05, 05:05 PM

marlon, your derivation does nothing more than show:

$LaTeX Code: \\psi(\\vec {r}, t) = e^i^(^\\vec {p} \\cdot \\vec{r} + Et)$

Is a solution to a particular equation, a game we could play all night (it is like asking "for what equations is 2 a solution?").

That is really a backwards derivation, and I don't see how it applies in any case other than that of a plane wave (at most, maybe you only meant one dimensional).

If you want to see the quick and dirty derivation of SE:

1) Assume there is a wave function which contains all of the observable information it is possible to know about a particle.

Cosideration: In order to find this function we must construct it by giving it properties. That is, it should be constructed as to give us information about the system when we apply a relatively simple operator.

2) Whatever the wave function is, I want to get momentum with a space derivative and energy with a time derivative, that would be simple.

Since energy and momentum are related as they are, we have the shrodinger equation! Fortunately the uniqueness of solutions to a partial differential equation gaurantees our success.

**dextercioby** · Mar8-05, 05:13 PM

Originally Posted by Crosson

marlon, your derivation does nothing more than show:

$LaTeX Code: \\psi(\\vec {r}, t) = e^{i\$\\vec {p} \\cdot \\vec{r} + Et\$}$

Is a solution to a particular equation, a game we could play all night (it is like asking "for what equations is 2 a solution?").

That is really a backwards derivation, and I don't see how it applies in any case other than that of a plane wave (at most, maybe you only meant one dimensional).

If you want to see the quick and dirty derivation of SE:

1) Assume there is a wave function which contains all of the observable information it is possible to know about a particle.

Cosideration: In order to find this function we must construct it by giving it properties. That is, it should be constructed as to give us information about the system when we apply a relatively simple operator.

2) Whatever the wave function is, I want to get momentum with a space derivative and energy with a time derivative, that would be simple.

That's both fishy and logically incorrect,because you know where you to get (the SE) and you know that the result you're expecting (SE) is correct...

Daniel.

**marlon** · Mar8-05, 05:30 PM

Originally Posted by Crosson

marlon, your derivation does nothing more than show:

$LaTeX Code: \\psi(\\vec {r}, t) = e^i^(^\\vec {p} \\cdot \\vec{r} + Et)$

I don't really think you got the point. However, since i don't want to play with you all night long, let us just drop it...ok ?

marlon

EL · Mar9-05, 04:39 AM

well, it's not the first time this is discussed here...

**dextercioby** · Mar9-05, 07:38 AM

True,these subjects in Quantum Physics forum are kinda circular.The EPR gedankenexperiment is still the most debated.

Daniel.