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How is the Schrödinger equation

$$ i\hbar\frac {\partial }{\partial t}\psi=H{\psi }$$

derived?

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    $\begingroup$ I'm not aware of a way to actually derive the Schrodinger equation. However, it can be made plausible with some argument about the properties of the time evolution operator. Would you be interested in that? $\endgroup$ – Danu Mar 9 '14 at 16:08
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    $\begingroup$ You cannot cancel $\psi$ to get $H=i \hbar \frac{ \partial}{\partial t}$, since not any complex function obeys the Schrodinger equation. $\endgroup$ – user26143 Mar 9 '14 at 16:59
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    $\begingroup$ Related post physics.stackexchange.com/q/30537 $\endgroup$ – user26143 Mar 9 '14 at 17:46
  • $\begingroup$ Concerning the status of $\hat{H}=i \hbar \frac{ \partial}{\partial t}$, see this Phys.SE post. $\endgroup$ – Qmechanic Mar 9 '14 at 18:17
  • $\begingroup$ Take a look at my answer here: One can't derive the equation, but one can certainly argue strong grounds for its form with assumptions of linearity, conservation of probability and time shift invariance. $\endgroup$ – WetSavannaAnimal Mar 10 '14 at 7:00
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Schrodinger's equation cannot be derived. It was thought up using logical arguments and so far it has seemed to work experimentally.

The equations is essentially a re-write up for energy conservation:

$$E = T + V$$

Where $T$ is the Kinetic Energy and $V$ is the potential. However, to be more explicit we must work with operators (if you are unsure what operators are I suggest you look them up; this will give you a better understanding of what's going on).

The KE for a particle is given by the KE Operator: $$\hat{T} = \frac{-\hbar^2}{2m}\frac{\partial^2}{\partial x^2}$$.

This comes from the momentum operator of the particle/wave $\hat{p} = -ih\partial/\partial x$. You use this in the analogous classical mechanics equation for KE to obtain $\hat{T}$ (Try doing this as an exercise).

So now we are left with just putting it all together. The first equation turns into: $$\frac{-\hbar^2}{2m}\frac{\partial^2 \Psi}{\partial x^2} + V(x) \Psi = E \Psi$$ And we define the Hamiltonian $\hat{H}$ as: $$\hat{H} = \frac{-\hbar^2}{2m}\frac{\partial^2 }{\partial x^2} + V(x)$$

Thus:

$$\hat{H} \Psi = E \Psi$$

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You might be interested in this "elementary" derivation of the free particle Schroedinger equation from Maxwell's equations. It seems to be in the same spirit as Schroedinger's original reasoning. The niceness of this approach is that if you also include special relativity, it nets you both the free particle Schroedinger equation and its relativistic counterpart, the Klein Gordon equation.

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    $\begingroup$ Could you include an overview of the derivation, or at least some piece of the content of that paper so that the answer can stand on its own without a reader having to click the link? $\endgroup$ – David Z Mar 10 '14 at 1:22
  • $\begingroup$ For a connection between Schr. eq. and Klein-Gordon eq, see also e.g. A. Zee, QFT in a Nutshell, Chap. III.5, and this Phys.SE post plus links therein. $\endgroup$ – Qmechanic May 6 '16 at 19:57
  • $\begingroup$ That derivation is just heuristic nonsense. $\endgroup$ – Hulkster Oct 17 '18 at 0:31
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From Feynman's lectures :)

Where did we get that from? It's not possible to derive it from anything you know. It came out of the mind of Schrödinger.

ADDITION

Though a postulate such as Schrödinger equation cannot be proven, one can notice that in QM a state vector $\Psi$ is said to give the most complete description of a state of a system. So it is "natural" to assume that it also completely describes the evolution of the system with time - how to get the "next" state from the current state. Since in QM states form a linear space, this relation has to be linear too:

$$ d\Psi = \hat{A}\Psi dt$$

The norm of the state vector must be conserved $\left(\Psi+d\Psi, \Psi+d\Psi\right)=\left(\Psi, \Psi\right)=1$ thus

$$ \hat{A} = -i\hat{H} $$ where $\hat{H}=\hat{H}^\dagger$ is hermitian. This way you "derive" Schrödinger's equation.

By analogy, in classical physics a state is completely described by velocities and positions $(v_i,r_i)$ and you have Hamiltonian equations which are first order with respect to $v_i$ and $r_i$ (but nonlinear)

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    $\begingroup$ Postulates don't just fall out of the sky into our laps. Yes, you can't formally derive the Schrödinger equation, but you can persuasively motivate it, as in Alex R.'s answer. I think this and all the answers to the effect of, "It's a postulate, you can't derive it," are completely missing the point. $\endgroup$ – user27578 Mar 9 '14 at 22:51
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    $\begingroup$ By recognizing the Euclid's 5th postulate cannot be derived, one can open one's mind to general relativity. $\endgroup$ – DavePhD Mar 10 '14 at 1:34
  • $\begingroup$ @dgh, I've edited the post in case OP want's this kind of a "derivation" $\endgroup$ – xaxa Mar 10 '14 at 6:23
  • $\begingroup$ Take a look at my answer here: One can't derive the equation, but one can certainly argue strong grounds for its form with assumptions of linearity, conservation of probability and time shift invariance. I believe my argument I give is heavily influenced by the chapters following straight after the Feynman quote you give. $\endgroup$ – WetSavannaAnimal Mar 10 '14 at 7:04
  • $\begingroup$ @WetSavannaAnimalakaRodVance, your derivation is correct if time shift invariance is preserved. When the system is in external $t$-dependent field $\hat{H}$ does not correspond to energy (which is not conserved). However, the differential form of Schroedinger's equation still holds. $\endgroup$ – xaxa Mar 10 '14 at 7:16
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One can surely consistently derive the stationary state Schrodinger equation straight from the Dirac-von Neumann axioms. They are

The observables of a quantum system are defined to be the (possibly unbounded) self-adjoint operators $\mathcal{O}$ on the Hilbert space.

A state $\psi$ of the quantum system is a unit vector in Hilbert space.

The expectation value of an observable $\mathcal{O}$ for a system in a state $\psi$ is given by the inner product $\psi^{\dagger}\mathcal{O}\psi$.

On the physical grounds, we want the energy of the system to be minimum. The energy expectation value by the third axiom is $$\bar{\mathcal{H}}=\psi^{\dagger}\mathcal{H}\psi$$ Where $\mathcal{H}$ is the total energy operator (by the first axiom), i.e. the Hamiltonian. By the second axiom, we have $$\psi^{\dagger}\psi=1$$ Thus we are to minimise the following functional $\bar{\mathcal{H}}=\psi^{\dagger}\mathcal{H}\psi$ subject to the constraint $\psi^{\dagger}\psi=1$. We introduce the Lagrange multiplier $E$, and thus we are minimising the following functional $$F[\psi]=\psi^{\dagger}\mathcal{H}\psi-E\psi^{\dagger}\psi$$ let's first minimise with respect to $\psi$ $$\delta{F}[\psi]=\psi^{\dagger}\mathcal{H}\delta\psi-E\psi^{\dagger}\delta\psi=\psi^{\dagger}(\mathcal{H}-E)\delta\psi=0$$ this must hold for all variations $\delta\psi$, thus $$\psi^{\dagger}\mathcal{H}=\psi^{\dagger}E$$ then we minimise with respect to $\psi^{\dagger}$ $$\delta{F}[\psi]=\delta\psi^{\dagger}\mathcal{H}\psi-E\delta\psi^{\dagger}\psi=\delta\psi^{\dagger}(\mathcal{H}-E)\psi=0$$ this must hold for all variations $\delta\psi^{\dagger}$, thus $$\mathcal{H}\psi=E\psi$$ For the case of the time dependent Schrodinger equation you use the argument similar to that of @xaxa. Consider the transformation of the state on the Hilbert space $$\psi_{1}\rightarrow{U}\psi$$ The adjoint transforms as $$\psi^{\dagger}_{1}\rightarrow\psi^{\dagger}{U}^{\dagger}$$ But we must have $$\psi^{\dagger}_{1}\psi_{1}=\psi^{\dagger}{U}^{\dagger}U\psi=\psi^{\dagger}\psi=1$$ Thus $${U}^{\dagger}U=1$$ Hence, each transformation on the state vector must be unitary! From theory of Lie groups we know that any group element may be written as the exponential of the generator, i.e. $$U=e^{i\epsilon{J}}$$ Where $J$ is the generator and $\epsilon$ is the parameter of transformation. Hence $$U^{\dagger}U=(1-i\epsilon{J}^{\dagger}+...)(1+i\epsilon{J}+..)=1+i\epsilon(J-J^{\dagger})+...$$ Thus $$J=J^{\dagger}$$ Hence any generator of transformation of the quantum state is self adjoint. By the first axiom we conclude that generators of the transformations on the Hilbert space are the quantum observables. From time $0$ to time $t$ the state $\psi(0)$ transforms to the state $\psi(t)$, by our analysis there must exist a transformation $$\psi(t)=U(t)\psi(0)$$ And the only parameter of the transformation is $t$, thus we conclude that $U(t)=e^{itJ}$, hence $$\frac{d\psi(t)}{dt}=\frac{dU(t)}{dt}\psi(0)=iJU(t)\psi(0)=iJ\psi(t)$$ Thus $$i\frac{d\psi(t)}{dt}=-J\psi(t)$$ This is the only thing that can be concluded, the rate of change of the physical state of quantum system is given by the action on the state by some self adjoint operator. There exists no reasoning to explain why $J=-\mathcal{H}$, where $\mathcal{H}$ is the Hamiltoanian! Thus normally a third axiom is added, namely the Schrodinger equation of motion for the state!

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The time dependent Schrodinger equation is one of 5 (or 6) postulates of quantum mechanics. It is not proper to say that it is derived, unless you have a different set of postulates.

for example, in the references below, the time dependent Schrodinger equation is the 5th postulate.

http://vergil.chemistry.gatech.edu/notes/quantrev/node20.html

http://depts.washington.edu/chemcrs/bulkdisk/chem455A_aut10/notes_Lecture%20%206.pdf

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The original formulation of Schroedinger equation was given by Schroedinger himself based on Hamilton's studies on the optical mechanical analogy.

Hamilton noticed that in Hamilton-Jacobi theory, the momentum of the particle is given by $\vec\nabla S$, where the Hamilton's principal function $S$ is the action viewed as a function of the coordinates. By looking at level surfaces $S=\mathrm{const.}$ we see that particle's trajectory is orthogonal to the level surfaces. This is similar to light rays which travel perpendicularly to level surfaces corresponding to constant phase (wave fronts).

Schroedinger in 1926 conjectured that the action $S$ was indeed a phase of some wave process. Hence this wave should look like $$\psi=\psi_0\exp{\frac{iS}{\hbar}}=\psi_0\exp{\frac{i}{\hbar}\left[W(x)-Et\right]},$$ where $W(x)$ is the Hamilton's characteristic function. The constant $\hbar=h/2\pi$ is chosen so that this wave has frequency $\nu=E/h$, the Planck relation, which was known by that time. Plugging this wave into a wave equation one gets finally the Schroedinger equation $$-\frac{\hbar^2}{2m}\nabla^2\psi+V\psi=i\hbar\frac{\partial\psi}{\partial t}.$$ Classical mechanics can be understood as a limit case of Quantum Mechanics by plugging $\psi=\psi_0e^\frac{iS}{\hbar}$ into Schroedinger equation and taking the limit $\hbar\rightarrow 0$. The result is the Hamilton-Jacobi equation.

It is interesting to note that Hamilton was close to formulating wave mechanics a century before Schroedinger . He did not though, probably by lack of any experimental evidence.

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Schrödinger Equation is a postulate in traditional approaches to Quantum Mechanics.

I think you could postulate Klein-Gordon Equation or Dirac Equation instead, which appeared as relativistic generalizations of Schrödinger Equation, and derive the latter as a Classical Limit of the former(s).

Besides, there are other approaches with derivations of Schrödinger Equation from Statistical Physics principles. De la Peña derived it from the Theory of Markov processes, Olavo derived it from Liouville Equations and an infinitesimal Wigner-Moyal transformation. See below.

de la Peña, L. (1967). "A Simple Derivation of the Schrödinger Equation From the Theory of Markov processes". Physics Letters A. 24 (11): 603–604. http://www.sciencedirect.com/science/article/pii/0375960167906391?via%3Dihub

Olavo, L.S.F. (1999). "Foundations of quantum mechanics (II): equilibrium, Bohr–Sommerfeld rules and duality". Physica A: Statistical Mechanics and its Applications. Volume 271, Issues 3–4, 15 September 1999, Pages 260-302. http://www.sciencedirect.com/science/article/pii/S0378437199002162

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I have the answer to this question, but it is one that is controversial in nature. I wouldn't be surprised if my answer gets removed.

The Schrödinger Equation is a probabilistic Partial-Differential Equation. It is missing variables that would make it a deterministic Multivariate Equation. I used to not believe that it was a fundamental equation until I realized that I had derived it. It was a correction that I made to an equation of Quantum Mechanics. I saw a similarity in this equation after a few months had passed by. Time is mistreated as a spatial coordinate in physics leading to light-cone mapping and imaginary numbers so my correct treatment of time removed the imaginary part of the equation. Also the probabilistic Wave-Function is replaced by a deterministic Wave-Equation. This equation is the convergence of three equations that seem independent in their differentiated form, but are integrated on the Quantum Level. It integrates into a mathematical frame-work. I currently do not wish to disclose this final equation because my other work has not received academic acknowledgement. The Schrödinger Equation does model atoms, molecules, and the whole universe as what is stated about its possible role in physics. I believe that it is the Holy Grail of Physics and the base-knowledge of the Tree of Life.

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    $\begingroup$ Hi, @Stephen Tuck, can you please explain me your statement:_It is missing variables that would make it a deterministic Multivariate Equation_ ? I'll appreciate your explanation:) $\endgroup$ – user36790 Aug 14 '15 at 13:53
  • $\begingroup$ The inconvenience of your answer is that there is no arguments provided with it. The important thing is not the equation, but the considerations taken in its derivation. If you indeed have achieved what you claim, it will be very welcome, I am sure. If it is acknowledgement what worries you, I am pretty sure you can always claim ownership, since what you write here can be proven to be yours and the date is registered. $\endgroup$ – rmhleo Aug 14 '15 at 14:20
  • $\begingroup$ Partial-Differential Equations are not complete. The character ∂ denotes a Partial Derivative such as (∂z / ∂x). It is 'partial' because it isn't an exact solution. Multivariate Equations don't need approximations. A related topic is Difference Equations where the relationship involves values of the unknown function(s). Many methods to compute numerical solutions of Differential Equations involve approximation of the solution by a corresponding Difference Equation. There are underlying variables for which the probabilistic Schrödinger Equation approximates. It took me 6-7 years to find them. $\endgroup$ – Stephen Tuck Aug 14 '15 at 21:06
  • $\begingroup$ Physicists run into many singularities and infinities because of overlap or gaps in their approximate equations. You cannot successfully integrate these incomplete equation differentiations. You must find the underlying hidden variables that make the functions simple Multivariate Equations. Mathematical tricks like Perturbation or Renormalization only mask the true problem. I shall have to consider the consequences of full disclosure. The Multivariate Form of the Schrödinger Equation yields great knowledge of the Quantum Mechanics of the universe. $\endgroup$ – Stephen Tuck Aug 14 '15 at 21:22
  • $\begingroup$ I have decided that this isn't the correct place or time to disseminate the Multivariate Schrödinger Equation. It could be derived without the Unified Field Equation, but it is more difficult to realize without such mathematical framework. It is much easier to understand it's role by understanding gravity since it is the equation of Quantum Gravitation. It explains how gravitational fields arise from the quantum kinetic-energy of matter. Corrections to Special Relativity, General Relativity, and Quantum Mechanics are needed in order for anyone to truly understand this equation! $\endgroup$ – Stephen Tuck Aug 18 '15 at 15:41

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