Derivation of Schrodinger equation (chicken and egg problem?)

[Hamiltonian]

The classical wave equation in 1-D reads:
$$\frac{\partial^2 u}{\partial x^2}(x,t) = \frac{1}{v^2}\frac{\partial^2 u}{\partial t^2}(x,t)$$

The D'alembert solution to the wave equation is:
$$u(x,t) = f(x+vt) + g(x-vt)$$
so a allowed wave function solution to the 1-Dimensional classical wave equation is:
$$u(x,t) = Ae^{(kx \pm \omega t)}$$

the Schrodinger equation is in some sense derived from the De Broglie plane wave
$$\psi(x,t) = Ae^{i(kx-\omega t)} = Ae^{\frac{i}{\hbar}(px - Et)}$$
$$\frac{\partial \psi}{\partial x}(x,t) = \frac{i}{\hbar}pAe^{\frac{i}{\hbar}(px - \omega t)} = \frac{i}{\hbar}p\psi(x,t)$$
and from that we get the momentum operator:
$$\hat p = -i\hbar \partial_x$$
we can again find the energy operator in a similar manner
$$\frac{\partial \psi}{\partial t}(x,t) = -\frac{i}{\hbar}EAe^{\frac{i}{\hbar}(px - Et)} = -\frac{i}{\hbar}E\psi(x,t)$$
$$\hat E = i\hbar \partial_t$$
and we know that the total energy of a system is just the KE+PE
$$\hat E = \frac{\hat p^2}{2m} + V(x) = i\hbar\frac{\partial}{\partial t}$$
hence we have the time-dependent Schrodinger equation:
$$-\frac{\hbar^2}{2m}\frac{\partial \psi}{\partial x^2}(x,t) + V(x)\psi(x,t) = i\hbar\frac{\partial \psi}{\partial t}(x,t)$$
a solution to this equation would be of the form:$$\psi(x,t) = Ae^{i(kx - \omega t)}$$

My question is if a wave function must satisfy the Schrodinger equation then how can the Schrodinger equation itslef be derived from ##Ae^{\frac{i}{\hbar}(px - \omega t)}##. This form of a wave function is the solution to the classical wave function (except there is no ##i## in that) so how could it directly be carried on to QM? This kind of feels like the chicken and egg problem did the Schrodinger equation come first or did the above wave function solution come first?
am I missing something very obvious

 

[dextercioby]

I do not think there are somewhere Schroedinger's 1924-1925 notes/notebooks which would give us the real starting point of his work culminating in the 4 articles from 1926. Anyways, this is just a historical issue. What is really relevant is that in QM (traditional, so-called Copenhagen - university textbook interpretation) Schroedinger's equation is an axiom.
It becomes a theorem, for example, in the Weyl-Wigner symmetry formulation.

Rephrased, it is pointless to ask: which came first, the particular form of the WF, or Schroedinger's equation? Actually, one needs to go to L dB PhD thesis and notebooks in its preparation to see LdB used a form of wavefunction when he came up with his "standing and traveling waves" theory.

 

[Frabjous]

https://www.physicsforums.com/threads/how-to-derive-schrodingers-equation.1004975/#post-6514138

Check out the excerpts in post 6.

 

[Demystifier]

$$-\frac{\hbar^2}{2m}\frac{\partial \psi}{\partial x^2}(x,t) + V(x)\psi(x,t) = i\hbar\frac{\partial \psi}{\partial t}(x,t)$$
a solution to this equation would be of the form:
$$\psi(x,t) = Ae^{i(kx - \omega t)}$$

My question is if a wave function must satisfy the Schrodinger equation then how can the Schrodinger equation itslef be derived from ##Ae^{\frac{i}{\hbar}(px - \omega t)}##.

When ##V(x)## is not a constant, then ##Ae^{\frac{i}{\hbar}(px - \omega t)}## is not a solution of the Schrodinger equation. More to the point, for non-constant ##V(x)## we cannot easily guess ##\psi(x,t)##. To find it we must really solve the Schrodinger equation, which is a very non-trivial job.

 

[Hamiltonian]

I have come to accept that the Schrodinger equation can only be motivated and that one must accept it as a model.

Is it correct to say that the bellow equation is the Schrodinger equation when we treat Energy in the classical sense and as it doesn't account for SR the bellow equation isn't very fundamental and is only an approximation of how a "real" quantum system evolves.
$$i\hbar\frac{\partial \psi}{\partial t}(\vec r,t) = [-\frac{\hbar^2}{2m} \nabla^2 + V(\vec r)]\psi(\vec r, t)$$
and that the actual Schrodinger equation is the one bellow, in which we must substitute a more accurate Hamiltonian to describe more accurately a real quantum system:
$$\hat H \psi(\vec r, t) = i\hbar \partial_t \psi(\vec r, t)$$
what I am trying to ask in short is if someone asks me what the Schrodinger equation is, will the most accurate answer that I can give them be:
$$\hat H | \psi \rangle = i\hbar\partial_t| \psi \rangle$$

 

[Demystifier]

treat Energy in the classical sense

I'm not sure what do you mean by that, but otherwise you are right about everything in the post above.

 

[Philip Koeck]

the Schrodinger equation is in some sense derived from the De Broglie plane wave
$$\psi(x,t) = Ae^{i(kx-\omega t)} = Ae^{\frac{i}{\hbar}(px - Et)}$$
$$\frac{\partial \psi}{\partial x}(x,t) = \frac{i}{\hbar}pAe^{\frac{i}{\hbar}(px - \omega t)} = \frac{i}{\hbar}p\psi(x,t)$$
and from that we get the momentum operator:
$$\hat p = -i\hbar \partial_x$$
we can again find the energy operator in a similar manner
$$\frac{\partial \psi}{\partial t}(x,t) = -\frac{i}{\hbar}EAe^{\frac{i}{\hbar}(px - Et)} = -\frac{i}{\hbar}E\psi(x,t)$$
$$\hat E = i\hbar \partial_t$$
and we know that the total energy of a system is just the KE+PE
$$\hat E = \frac{\hat p^2}{2m} + V(x) = i\hbar\frac{\partial}{\partial t}$$
hence we have the time-dependent Schrodinger equation:
$$-\frac{\hbar^2}{2m}\frac{\partial \psi}{\partial x^2}(x,t) + V(x)\psi(x,t) = i\hbar\frac{\partial \psi}{\partial t}(x,t)$$
a solution to this equation would be of the form:$$\psi(x,t) = Ae^{i(kx - \omega t)}$$

The deBroglie plane wave is only a solution of the SE for constant (zero) potential and the energy E in the exponent is only the KE.
See also Demystifiers post.
The generalization happens when you insert the total energy, KE + PE, in the place of E.
That's why it's not a derivation.
I've attached a little text that should explain everything.

 

[Mark Hadley]

Schrödinger's equation is derived neatly in Ballentine's book on quantum theory. While Weinberg derives the relativistic equations in volume 1.

They start from the probability rules and then look at the effects of continuous symmetry operations.

Why the probabilities are given by a quadratic expression is a more subtle issue that is a requirement for quantum probabilities to be context dependent.

 

[vharihar]

The deBroglie plane wave is only a solution of the SE for constant (zero) potential and the energy E in the exponent is only the KE.
See also Demystifiers post.
The generalization happens when you insert the total energy, KE + PE, in the place of E.
That's why it's not a derivation.
I've attached a little text that should explain everything.

I like how you put it (quoting you):
>>
Generalizations like this lead to a new theory which cannot be derived from any other theory. It is not possible to derive a theory, one can only develop it. One theory never follows from another. If it did it wouldn’t be a theory
>>

The SE seems to be motivated by other simpler theories, but cannot be derived from something else. It is motivated in the sense that as you suggest, an approach was perhaps started out from the simpler theories, and then on top of that a generalization/abstraction was added in after some trial and error the intent being to explain phenomena observed at that time hitherto unexplainable by simpler theories, in such a manner that it collapses to the simpler theories asymptotically under limiting simplifications.

That's the way I have rationalized it to myself.

 

[Fra]

My question is if a wave function must satisfy the Schrodinger equation then how can the Schrodinger equation itslef be derived from ##Ae^{\frac{i}{\hbar}(px - \omega t)}##. This form of a wave function is the solution to the classical wave function (except there is no ##i## in that) so how could it directly be carried on to QM?

Another angle: If one considers the Fourier transform of ##P(x,t)^{1/2}## instead of debroglie waves, and defines E and p from the conjugate variables, even general solutions can be considered in the form but one gets instead then expectation values for the potential ##<V(x)>## instead of the Classical potential ##V(<x>)##, except for trivial cases.

/Fredrik

 

[PeterDonis]

I've attached a little text

Please do not post attachments. Text and equations should be posted directly, using the PF LaTeX feature for equations. Otherwise it is impossible for anyone else to quote what you post in replying or asking questions.

 

[Philip Koeck]

The classical wave equation in 1-D reads:
$$\frac{\partial^2 u}{\partial x^2}(x,t) = \frac{1}{v^2}\frac{\partial^2 u}{\partial t^2}(x,t)$$
The D'alembert solution to the wave equation is:
$$u(x,t) = f(x+vt) + g(x-vt)$$
so a allowed wave function solution to the 1-Dimensional classical wave equation is:
$$u(x,t) = Ae^{(kx \pm \omega t)}$$
The Schrodinger equation is in some sense derived from the De Broglie plane wave
$$\psi(x,t) = Ae^{i(kx-\omega t)} = Ae^{\frac{i}{\hbar}(px - Et)}$$
My question is if a wave function must satisfy the Schrodinger equation then how can the Schrodinger equation itslef be derived from ##Ae^{\frac{i}{\hbar}(px - \omega t)}##. This form of a wave function is the solution to the classical wave function (except there is no ##i## in that) so how could it directly be carried on to QM?

A small comment: As you correctly state, every function of (kx ±ωt) is a solution of the classical wave equation.

In particular the harmonic function is also one, so you can have an i in the exponent of the exponential function, even for a classical wave, and if you want to describe a harmonic wave you have to have an i.

The function you give as a solution to the classical equation doesn't describe a harmonic wave.

 

[Philip Koeck]

The Schrodinger equation is in some sense derived from the De Broglie plane wave
$$\psi(x,t) = Ae^{i(kx-\omega t)} = Ae^{\frac{i}{\hbar}(px - Et)}$$
$$\frac{\partial \psi}{\partial x}(x,t) = \frac{i}{\hbar}pAe^{\frac{i}{\hbar}(px - \omega t)} = \frac{i}{\hbar}p\psi(x,t)$$
and from that we get the momentum operator:
$$\hat p = -i\hbar \partial_x$$
we can again find the energy operator in a similar manner
$$\frac{\partial \psi}{\partial t}(x,t) = -\frac{i}{\hbar}EAe^{\frac{i}{\hbar}(px - Et)} = -\frac{i}{\hbar}E\psi(x,t)$$
$$\hat E = i\hbar \partial_t$$
and we know that the total energy of a system is just the KE+PE
$$\hat E = \frac{\hat p^2}{2m} + V(x) = i\hbar\frac{\partial}{\partial t}$$
hence we have the time-dependent Schrodinger equation:
$$-\frac{\hbar^2}{2m}\frac{\partial \psi}{\partial x^2}(x,t) + V(x)\psi(x,t) = i\hbar\frac{\partial \psi}{\partial t}(x,t)$$
a solution to this equation would be of the form:$$\psi(x,t) = Ae^{i(kx - \omega t)}$$

Another question: Is there a reason why one chooses the first time derivative and the second spatial derivative in this approach to the SE?
What would happen if one decided for something else?
Is it simply in order to match the energy equation?

 

[Hamiltonian]

Another angle: If one considers the Fourier transform of ##P(x,t)^{1/2}## instead of debroglie waves, and defines E and p from the conjugate variables, even general solutions can be considered in the form but one gets instead then expectation values for the potential ##<V(x)>## instead of the Classical potential ##V(<x>)##, except for trivial cases.

is ##P(x,t)^{1/2}## the Probability density function? and what do you mean by "defines E and p from the conjugate variables".

A small comment: As you correctly state, every function of (kx ±ωt) is a solution of the classical wave equation.

In particular the harmonic function is also one, so you can have an i in the exponent of the exponential function, even for a classical wave, and if you want to describe a harmonic wave you have to have an i.

The function you give as a solution to the classical equation doesn't describe a harmonic wave.'

for a classical harmonic wave shouldn't that wave function solution be $$u(x,t) = Re(e^{i(kx \pm \omega t)})$$ and not just ##e^{i(kx \pm \omega t)}## as a solution to the classical wave equation must be strictly real.

Another question: Is there a reason why one chooses the first time derivative and the second spatial derivative in this approach to the SE?
What would happen if one decided for something else?
Is it simply in order to match the energy equation?

the second spatial derivative appears in the Schrodinger equation because of the ##\hat p = -i\hbar \nabla## in the KE part of the Hamiltonian. I read somewhere because the SE contains a second derivative wrt to space and a single derivative wrt to time(hence does not treat spac-time in the same way), it's hence not lorentz invariant so I don't think one chooses the equation to be that way.

 

[Philip Koeck]

the second spatial derivative appears in the Schrodinger equation because of the ##\hat p = -i\hbar \nabla## in the KE part of the Hamiltonian. I read somewhere because the SE contains a second derivative wrt to space and a single derivative wrt to time(hence does not treat spac-time in the same way), it's hence not lorentz invariant so I don't think one chooses the equation to be that way.

If you use the second derivative wrt time you match it to the relativistic energy equation and you end up with the Dirac equation.

 

[vanhees71]

The question is, what is meant by "derive" some fundamental equation of physics. To derive it you need of course some assumptions. From a modern (or rather 20th-century) point of view the most fundamental arguments underlying to a large extent all of the physical theories (Newtonian mechanics, classical electrodynamics, and quantum theory) are the symmetry principles.

Everything starts with the exploitation of the symmetries of the spacetime model, which is a prerequisite to start formulating any physical theory. Indeed, that's where in 1905 Einstein initiated the thinking in terms of symmetries in his famous paper on the electrodynamics of moving bodies, i.e., his groundbreaking solution of the tension between Maxwell's equations and the Newtonian spacetime structure including the experimental evidence that there seems to be no possibility to define a preferred inertial frame of reference (aka an aether rest frame) by any physical observations (though Einstein doesn't quote the corresponding Michelson Morley experiment but an example concerning a magnet and a wire loop, considering the relative motion of each other as the physically adequate description rather than any preferred frame of reference like either the rest frame of the magnet or that of the wire loop).

After the historical dust had settled in 1926, it is clear how a general QT looks like in terms of the Hilbert space formalism (Dirac's naive formulation is almost always sufficient for the physicist), it is clear that a continuous symmetry like the spacetime symmetries underlying Newtonian mechanics (Galilei group) or special relativity (proper orthochronous Poincare group) dictates to a large extent the fundamental equations of motion for time evolution in terms of Hamilton's principle in Hamilton's canonical form.

The result are either the extensions of Schrödinger's equation (valid for massive particles with spin 0) to corresponding equations for particles with arbitrary spin. Mass occurs as a central charge of the underlying Poincare algebra, and the rotation group SO(3) has to be substituted with its covering group SU(2), leading also to half-integer spins in addition to integer spins.

In the case of the Poincare group one is inevitably lead to the existence of anti-particles and (for interacting particles) necessarily to a QFT description, because of the inevitability of particle creation and annihilation processes. Here one has massive and massless representations with mass as a Casimir operator of the Poincare group rather than a central charge (which implies that there is no mass superselection rule as in Newtonian QT). The resulting equations are the Klein-Gordon (spin 0), Dirac (spin 1/2), the Maxwell equations for massless spin-1 particles (which are necessarily gauge fields), and all the other named and unnamed equations of "higher spin". These, however, apply to quantized fields, i.e., field operators rather than "wave functions" as in the non-relativistic case (although of course the non-relativistic QT can be equivalently formulated as a QFT (dubbed unfortunately often as "2nd quantization"), which is of great advantage to treat many-body problems compared to the "1st-quantization formlulation").

 

[Philip Koeck]

The second spatial derivative appears in the Schrodinger equation because of the ##\hat p = -i\hbar \nabla## in the KE part of the Hamiltonian. I read somewhere because the SE contains a second derivative wrt to space and a single derivative wrt to time(hence does not treat spac-time in the same way), it's hence not lorentz invariant so I don't think one chooses the equation to be that way.

I imagine it should be possible to show that this choice of 1. derivative wrt time and 2. wrt to space ensures that non-relativistic energy and momentum are conserved in all solutions of the time dependent SE.
I can't remember seeing that anywhere, but maybe somebody here knows.

 

[vanhees71]

The symmetry under Galilei transformations determines the usual form of the Hamilton function in classical and thus also the Hamilton operator in quantum mechanics for a closed system of point particles to be
$$H=\sum_{j=1}^N \frac{1}{2 m_j} \vec{p}_j^2 + \frac{1}{2} \sum_{j \neq k} V(|\vec{x}_j-\vec{x}_k|),$$
where I have restricted myself to the usual pair-forces (higher-order ##n##-particle interactions occur only very rarely, one example being interactions between nucleons, i.e., protons and neutrons).

This Hamiltonian commutes with the usual Noether-conserved quantities of the Galilei symmetry, i.e., total momentum, total angular momentum, and center-of-mass velocity.

In the Schrödinger picture of time evolution the corresponding wave function obeys the Schrödinger equation,
$$-\mathrm{i} \hbar \partial_t \psi(t,\vec{x}_1,\ldots,\vec{x}_2)=\hat{H} \psi(t,\vec{x}_1,\cdots,\vec{x}_N).$$
From this the usual parabolic form of the equation follows via the fundamental principles of QT: 1st order in the time derivative and 2nd order in spatial derivatives.