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The Schrödinger equation is the basis to understanding quantum mechanics, but how can one derive it? I asked my instructor but he told me that it came from the experience of Schrödinger and his experiments. My question is, can one derive the Schrödinger equation mathematically?

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Be aware that a "mathematical derivation" of a physical principle is, in general, not possible. Mathematics does not concern the real world, we always need empirical input to decide which mathematical frameworks correspond to the real world.

However, the Schrödinger equation can be seen arising naturally from classical mechanics through the process of quantization. More precisely, we can motivate quantum mechanics from classical mechanics purely through Lie theory, as is discussed here, yielding the quantization prescription

$$ \{\dot{},\dot{}\} \mapsto \frac{1}{\mathrm{i}\hbar}[\dot{},\dot{}]$$

for the classical Poisson bracket. Now, the classical evolution of observables on the phase space is

$$ \frac{\mathrm{d}}{\mathrm{d}t} f = \{f,H\} + \partial_t f$$

and so its quantization is the operator equation

$$ \frac{\mathrm{d}}{\mathrm{d}t} f = \frac{\mathrm{i}}{\hbar}[H,f] + \partial_t f$$

which is the equation of motion in the Heisenberg picture. Since the Heisenberg and Schrödinger picture are unitarily equivalent, this is a "derivation" of the Schrödinger equation from classical phase space mechanics.

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    $\begingroup$ What about the "derivation" via path integrals? $\endgroup$ Oct 19, 2014 at 22:57
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    $\begingroup$ @LoveLearning: It all depends on where you want to start. In my view, the most mysterious element of both the Schrödinger equation and the path integral is the appearance of $\mathrm{i}$. You can indeed derive the SE from the path integral (and vice versa), but then you have to explain why the heck you are integrating over $e^{iS/\hbar}$ in the first place. The procedure of geometric quantization at least gives a mathematical motivation for that, starting from classical mechanics. Of course, if you believe that we should not start from classical mechanics, then you'll not find this convincing. $\endgroup$
    – ACuriousMind
    Oct 19, 2014 at 23:08
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    $\begingroup$ Derive Schrödinger equation via path integrals can be at most a "physical" derivation, but never a mathematical deivation, since path integrals in the sense of Feynman do not have a mathematical meaning. $\endgroup$ Oct 20, 2014 at 0:40
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    $\begingroup$ @LoveLearning See my newly added answer for more clarifications. $\endgroup$
    – Ellie
    Oct 20, 2014 at 10:32
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    $\begingroup$ @ACuriousMind +1: I think I'm going to make a T-shirt that reads "It all depends on where you want to start" and start selling it. That way, you can just point to your chest next time. I will also benefit immensely from wearing it when I walk around campus. Can I mark you down for one order? $\endgroup$ Oct 31, 2014 at 4:58
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Small addition to ACuriousMind's great answer, in reply to some of the comments asking for a derivation of Schrödinger wave equation, using the results of Feynman's path integral formalism:

(Note: not all steps can be included here, it would be too long to remain in the context of a forum-discussion-answer.)

In the path integral formalism, each path is attributed a wavefunction $\Phi[x(t)]$, that contributes to the total amplitude, of let's say, to go from $a$ to $b.$ The $\Phi$'s have the same magnitude but have differing phases, which is just given by the classical action $S$ as was defined in the Lagrangian formalism of classical mechanics. So far we have: $$ S[x(t)]= \int_{t_a}^{t_b} L(\dot{x},x,t) dt $$ and $$\Phi[x(t)]=e^{(i/\hbar) S[x(t)]}$$

Denoting the total amplitude $K(a,b)$, given by: $$K(a,b) = \sum_{paths-a-to-b}\Phi[x(t)]$$

The idea to approach the wave equation, describing the wavefunctions as a function of time, we should start by dividing the time interval between $a$-$b$ into $N$ small intervals of length $\epsilon$, and for a better notation, let's use $x_k$ for a given path between $a$-$b$, and denote the full amplitude, including its time dependance as $\psi(x_k,t)$ ($x_k$ taken over a region $R$):

$$\psi(x_k,t)=\lim_{\epsilon \to 0} \int_{R} \exp\left[\frac{i}{\hbar}\sum_{i=-\infty}^{+\infty}S(x_{i+1},x_i)\right]\frac{dx_{k-1}}{A} \frac{dx_{k-2}}{A}... \frac{dx_{k+1}}{A} \frac{dx_{k+2}}{A}... $$

Now consider the above equation if we want to know the amplitude at the next instant in time $t+\epsilon$:

$$\psi(x_{k+1},t+\epsilon)=\int_{R} \exp\left[\frac{i}{\hbar}\sum_{i=-\infty}^{k}S(x_{i+1},x_i)\right]\frac{dx_{k}}{A} \frac{dx_{k-1}}{A}... $$

The above is similar to the equation preceding it, the difference relying on the hint that, the added factor with $\exp(i/\hbar)S(x_{k+1},x_k)$ does not involve any of the terms $x_i$ before $i<k$, so the integration can be preformed with all such terms factored out. All this reduces the last equation to:

$$\psi(x_{k+1},t+\epsilon)=\int_{R} \exp\left[\frac{i}{\hbar}\sum_{i=-\infty}^{k}S(x_{i+1},x_i)\right]\psi(x_k,t)\frac{dx_{k}}{A}$$

Now a quote from Feynman's original paper, regarding the above result:

This relation giving the development of $\psi$ with time will be shown, for simple examples, with suitable choice of $A$, to be equivalent to Schroedinger's equation. Actually, the above equation is not exact, but is only true in the limit $\epsilon \to 0$ and we shall derive the Schroedinger equation by assuming this equation is valid to first order in $\epsilon$. The above need only be true for small $\epsilon$ to the first order in $\epsilon.$

In his original paper, following up the calculations for 2 more pages, from where we left things, he then shows that:

Canceling $\psi(x,t)$ from both sides, and comparing terms to first order in $\epsilon$ and multiplying by $-\hbar/i$ one obtains

$$-\frac{\hbar}{i}\frac{\partial \psi}{\partial t}=\frac{1}{2m}\left(\frac{\hbar}{i}\frac{\partial}{\partial x}\right)^2 \psi + V(x) \psi$$ which is Schroedinger's equation.

I would strongly encourage you to read his original paper, don't worry it is really well written and readable.


References: Space-Time Approach to Non-Relativistic Quantum Mechanics by R. P. Feynman, April 1948.

Feynman Path Integrals in Quantum Mechanics, by Christian Egli

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  • $\begingroup$ The Schroedinger Equation is simply the Hamiltonian ie. Kinetic + Potential energy as a function of momenta and coordinates alone, written with Quantum operators for momentum replacing the classical definition of momentum. Hamilton's equation is well known from Classical Physics, has been tested for ~2 Centuries, and is easy to use. The only 'new' idea is the Quantum operator for momentum, which isn't intuitive or obvious, but is used because it gives the correct answer. $\endgroup$ Mar 5, 2016 at 17:41
  • $\begingroup$ Do you, per chance, have a link to Schroedinger's papers in English? $\endgroup$ Apr 22, 2017 at 21:12
  • $\begingroup$ @MadPhysicist unfortunately I cannot find the very early ones in English, but at least there's his paper on "An Undulatory Theory of the Mechanics of Atoms and Molecules". Among the very first ones by Heisenberg and afterwards Schrödinger were "Über quantentheoretische Umdeutung kinematischer und mechanischer Beziehungen." and "Quantisierung als Eigenwertproblem" respectively. Try to look for the English translation of these. What are you exactly interested in? Maybe I can recommend more modern material to you. $\endgroup$
    – Ellie
    Apr 23, 2017 at 15:59
  • $\begingroup$ I was particularly interested in his papers pertaining to expanding on the work of de Broglie and producing the Schroedinger equation. $\endgroup$ Apr 24, 2017 at 9:59
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According to Richard Feynman in his lectures on Physics, volume 3, and paraphrased "The Schrodinger Equation Cannot be Derived". According to Feynman it was imagined by Schrodinger, and it just happens to provide the predictions of quantum behavior.

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    $\begingroup$ See also this $\endgroup$
    – HDE 226868
    Oct 19, 2014 at 21:56
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    $\begingroup$ That was in the 1960's. But I also found this 2006 paper in the American Journal of Physics: arxiv.org/abs/physics/0610121 which claims a derivation. $\endgroup$
    – docscience
    Oct 19, 2014 at 21:59
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Fundamental laws of physics cannot be derived (turtles all the way down and all that).

However, they can be motivated in various ways. Direct experimental evidence aside, you can argue by analogy - in case of the Schrödinger equation, comparisons to Hamiltonian mechanics and the Hamilton-Jacobi equation, fluid dynamics, Brownian motion and optics have been made.

Another approach is arguing by mathematical 'beauty' or necessity: You can look at various ways to model the system and go with the most elegant approach consistent with constraints you imposed (ie reasoning in the vein of 'quantum mechanics is the only way to do X' for 'natural' or experimentally necessary values of X).

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While it is in general impossible to derive the laws of physics in the mathematical sense of the word, a strong motivation or rationale can be given most of the time. Such impossibility arises from the very nature of physical sciences which attempt to stretch the all-to-imperfect logic of the human mind onto the natural phenomena around us. In doing so, we often make connections or intuitive hunches which happen to be successful at explaining phenomena in question. However, if one had to point out which logical sequence was used in producing the hunch, he would be at a loss - more often than not such logical sequence simply does not exist.

"Derivation" of the Schroedinger equation and its successful performance at explaining various quantum phenomena is one of the best (read audacious, mind-boggling and successful) examples of the intuitive thinking and hypothesizing which lead to great success. What many people miss is that Schroedinger simply took the ideas of Luis de Broglie further to their bold conclusion.

In 1924 de Broglie suggested that every moving particle could have a wave phenomenon associated with it. Note that he didn't say that every particle was a wave or vice versa. Instead, he was simply trying to wrap his mind around the weird experimental results which were produced at the time. In many of these experiments, things which were typically expected to behave like particles also exhibited a wave behavior. It is this conundrum which lead de Broglie to produce his famous hypothesis of $\lambda = \frac{h}{p}$. In turn, Schroedinger used this hypothesis as well as the result from Planck and Einstein ($E = h\nu$) to produce his eponymous equation.

It is my understanding that Schroedinger originally worked using Hamilton-Jacobi formalism of classical mechanics to get his equation. In this, he followed de Broglie himself who also used this formalism to produce some of his results. If one knows this formalism, he can truly follow the steps of the original thinking. However, there is a simpler, more direct way to produce the equation.

Namely, consider a basic harmonic phenomenon:

$ y = A sin (wt - \delta)$

for a particle moving along the $x$-axis,

$ y = A sin \frac{2\pi v}{\lambda} (t - \frac{x}{v}) $

Suppose we have a particle moving along the $x$-axis. Let's call the wave function (similar to the electric field of a photon) associated with it $\psi (x,t)$. We know nothing about this function at the moment. We simply gave a name to the phenomenon which experimentalists were observing and are following de Broglie's hypothesis.

The most basic wave function has the following form: $\psi = A e^{-i\omega(t - \frac{x}{v})}$, where $v$ is the velocity of the particle associated with this wave phenomenon.

This function can be re-written as

$\psi = A e^{-i 2 \pi \nu (t - \frac{x}{\nu\lambda})} = A e^{-i 2 \pi (\nu t - \frac{x}{\lambda})}$, where $\nu$ - the frequency of oscillations and $E = h \nu$. We see that $\nu = \frac{E}{2 \pi \hbar}$ The latter is, of course, the result from Einstein and Planck.

Let's bring the de Broglie's result into this thought explicitly:

$\lambda = \frac{h}{p} = \frac{2\pi \hbar}{p}$

Let's substitute the values from de Broglie's and Einstein's results into the wave function formula.

$\psi = A e^{-i 2 \pi (\frac{E t}{2 \pi \hbar} - \frac{x p}{2 \pi \hbar})} = A e^{- \frac{i}{\hbar}(Et - xp)} (*)$

this is a wave function associated with the motion of an unrestricted particle of total energy $E$, momentum $p$ and moving along the positive $x$-direction.

We know from classical mechanics that the energy is the sum of kinetic and potential energies.

$E = K.E. + P.E. = \frac{m v^2}{2} + V = \frac{p^2}{2 m} + V$

Multiply the energy by the wave function to obtain the following:

$E\psi = \frac{p^2}{2m} \psi + V\psi$

Next, rationale is to obtain something resembling the wave equation from electrodynamics. Namely we need a combination of space and time derivatives which can be tied back into the expression for the energy.

Let's now differentiate $(*)$ with respect to $x$.

$\frac{\partial \psi}{\partial x} = A (\frac{ip}{\hbar}) e^{\frac{-i}{\hbar}(Et - xp)}$

$\frac{\partial^2 \psi}{\partial x^2} = -A (\frac{p^2}{\hbar^2}) e^{\frac{-i}{\hbar}(Et - xp)} = \frac{p^2}{\hbar^2} \psi$

Hence, $p^2 \psi = -\hbar^2 \frac{\partial^2 \psi}{\partial x^2}$

The time derivative is as follows:

$\frac{\partial \psi}{\partial t} = - A \frac{iE}{\hbar} e^{\frac{-i}{\hbar}(Et - xp)} = \frac{-iE}{\hbar}\psi$

Hence, $E \psi = \frac{-\hbar}{i} \frac{\partial \psi}{\partial t}$

The expression for energy we obtained above was $E\psi = \frac{p^2}{2m} \psi + V\psi$

Substituting the results involving time and space derivatives into the energy expression, we obtain

$\frac{-i}{\hbar} \frac{\partial \psi}{\partial t} = \frac{- \hbar ^2}{2m} \frac{\partial ^2 \psi}{\partial x^2} + V\psi$

This, of course, became better known as the Schroedinger equation.

There are several interesting things in this "derivation." One is that both the Einstein's quantization and de Broglie's wave-matter hypothesis were used explicitly. Without them, it would be very tough to come to this equation intuitively in the manner of Schroedinger. What's more, the resulting equation differs in form from the standard wave equation so well-known from classical electrodynamics. It does because the orders of partial differentiation with respect to space and time variables are reversed. Had Schrodinger been trying to match the form of the classical wave equation, he would have probably gotten nowhere.

However, since he looked for something containing $p^2\psi$ and $E\psi$, the correct order of derivatives was essentially pre-determined for him.

Note: I am not claiming that this derivation follows Schroedinger's work. However, the spirit, thinking and the intuition of the times are more or less preserved.

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There are already a lot of good answers to this question. Of course, as already pointed out by many others, there is no "mathematical derivation" of Schrödinger's equation from first principles, since it is one of the axioms of quantum mechanics. However, there a many different ways to motivate this equation, from which many were already described by others here. One of them, which I like personally very much, is still missing, and hence I would like to add it here at this point.

Lets take the other axioms of quantum mechanics for granted, like the description of states using unit rays in a seperable Hilbert space and observables in terms of hermitian operators. Then there is very nice way to motivate the Schrödinger equation from physical ideas and some very deep and beautiful mathematical theorems.

The starting point is the following: A physical state is represented by a vector $\vert\psi_{0}\rangle\in\mathcal{H} $ in some Hilbert space $\mathcal{H}$. Now, we would like to analyse the time-evolution of this state. For this, we take a mapping $U:\mathbb{R}_{\geq 0}\to\mathcal{B}(\mathcal{H})$, where $\mathcal{B}(\mathcal{H})$ is the set of bounded operators in $\mathcal{H}$, such that $$\mid\psi(t)\rangle=U(t)\vert\psi_{0}\rangle.$$ Now, by the probabilistic description of quantum mechanics, the state $\mid\psi_{0}\rangle$ should be normalized and similarely, also $\mid\psi(t)\rangle$ should be normalized. As a consequence, we see that the operator $U(t)$ has to be unitary for all $t$. Furthermore, it is easy to see that we must have that $$U(t+s)=U(t)U(s).$$ To sum up, we see that the time-evolution operator $\{U(t)\}_{t\in\mathbb{R}_{\geq 0}}$ has to have the following properties:

  1. $U(t)$ is unitary for all $t$ and $U(0)=1_{\mathcal{H}}$.
  2. The mapping $U:\mathbb{R}_{\geq 0}\to\mathcal{B}(\mathcal{H})$ is a group homomorphism, i.e. $U(t+s)=U(t)U(s)$ for all $t,s$.
  3. From the mathematical side, we may also assume that the family of operators is "strongly-continuous", i.e. $\lim_{t\to s}\Vert U(t)-U(s)\Vert=0$.

A family of such operators $\{U(t)\}_{t\in\mathbb{R}_{\geq 0}}$ is called a "strongly-continuous unitary one-parameter group" in mathematics. This type of objects is a special case of "strongly continuous one-parameter semigroups", which are well-studied in the mathematics literature (i.e. theorem of Hille-Yosida, etc.).

Now, how is this related to the Schrödinger equation? It turns out that the statement "the time-evolution operator defines a strongly-continuous unitary one-parameter group" is equivalent to say "the state satisfies the Schrödinger equation" for a suitable operator $H$! Let me discuss this in the following: First of all, it is a general fact, which is not too hard to prove, that for every (possibly unbounded operator) $A:\mathcal{D}(A)\to\mathcal{H}$, where $\mathcal{D}(A)\subset\mathcal{H}$ is some domain, the family $$U(t):=e^{-itA}$$ defines a strongly-continous one-parameter group. Let me also stress that this expression is totally well-defined: In physics, one often just writes the exponential of an operator without discussing how to make sense out of these expressions, but this can be done in total rigour. For bounded operators, one can just take the usual exponential series, which always converges. For unbounded operators (like differential operators) one can use the "spectral theorem", which provides a rigours way to write expressions like $U(t):=e^{itA}$ (and in fact reduces to the standard exponential series for bounded operators). Using all the properties of the functional calculus coming from the spectral theorem, one can also prove that the group $U(t)=e^{itA}$ has the following properties:

  1. The domain $\mathcal{D}(A)$ can be characterised as $$\mathcal{D}(A)=\bigg\{\vert\psi\rangle\in\mathcal{H}\,\bigg\vert\,\frac{\mathrm{d}}{\mathrm{d}t}\bigg\vert_{t=0} U(t)\vert\psi\rangle:=\lim_{t\to 0}\frac{U(t)\vert\psi\rangle -\vert\psi\rangle}{t}\, \text{exists} \bigg\}.$$
  2. Every $\vert\psi\rangle\in\mathcal{D}(A)$ satisfies the following equation $$A\vert\psi\rangle=i\frac{\mathrm{d}}{\mathrm{d}t}\bigg\vert_{t=0} U(t)\vert\psi\rangle.$$

Now, the main point is that also the reverse is true: For every strongly-continuous unitary one-parameter group $\{U(t)\}_{t\in\mathbb{R}_{\geq 0}}$, there is a self-adjoint operator $A:\mathcal{D}(A)\to\mathcal{H}$, such that $U(t)=e^{-itA}$. This is the so-called Stone's theorem.

To sum up, we have seen that the fact that a time-evolution should preserve the norm of a state, implies that the time-evolution is described by a strongly-continuous one-parameter group $\{U(t)\}_{t\in\mathbb{R}_{\geq 0}}$. Furthermore, we have seen that (by Stone's theorem) there exists a self-adjoint operator $H:\mathcal{D}(H)\to\mathcal{H}$ such that $U(t)=e^{-itH}$. Defining a time-dependent state via $\vert\psi(t)\rangle=U(t)\vert\psi_{0}\rangle$, we also have seen that this satisfies the equation $$H\vert\psi(t)\rangle=i\frac{\mathrm{d}}{\mathrm{d}s}\bigg\vert_{s=0} U(s)\vert\psi(t)\rangle=i\frac{\mathrm{d}}{\mathrm{d}s}\bigg\vert_{s=0} \vert\psi(t+s)\rangle=i\frac{\mathrm{d}}{\mathrm{d}s}\bigg\vert_{s=t} \vert\psi(s)\rangle=i\frac{\mathrm{d}}{\mathrm{d}t} \vert\psi(t)\rangle,$$ but this is nothing else then Schrödinger's equation! Furthemore, also the reverse is true. Let us start from Schrödinger's equation $$H\vert\psi(t)\rangle=i\frac{\mathrm{d}}{\mathrm{d}t} \vert\psi(t)\rangle.$$ Choosing an initial value $\vert\psi(0)\rangle=\vert\psi_{0}\rangle$, then one can show that this equation has a unique solution, which is given by $$\vert\psi(t)\rangle=e^{-itH}\vert\psi_{0}\rangle$$ and as we have seen above, the family $U(t):=e^{-itH}$ is in fact a strongly-continuous unitary one-parameter group.

To sum up, the existence of a unitary time-evolution (which physically comes from the fact that we want to preserve the probability) is equivalent to the statement of Schrödinger's equation!


Let me stress that the above arguments are only valid if one assumes that the Hamilton operator itself is time-independent. However, one can generalize many of the argument above using "two-parameter unitary groups" $U(t,s)$. In this case, the story is mathematically more involved, but still one can proof a lot of nice results using this. For more details, I would recomment the second volume of Reed and Simons book "Methods of Modern Mathematical Physics".

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To my mind there are two senses in which we can "derive" a result in physics. New theories try to address the shortcomings of older ones by upgrading what we already have, giving new results. They also recover old results. I suppose we can call both derivations.

For example, the TISE and TDSE were first obtained because quantum mechanics said that, where classical mechanics would imply $f=0$, we should have $\hat{f}\left|\psi\right\rangle = 0$, with $\hat{f}$ the operator promotion of $f$, which in this case is $f=E-\frac{p^2}{2m}-V$ with operators $E=i\hbar\partial_t,\,\mathbf{p}=-i\hbar\boldsymbol{\nabla}$. (Some results become the weaker $\left\langle\psi\right|\hat{f}\left|\psi\right\rangle = 0$, e.g. with $f=\frac{d\mathbf{p}}{dt}+\boldsymbol{\nabla}V$, so I'm not being entirely honest here. But we expect $\hat{E}$-eigenstates are important because the probability distribution of $E$ is conserved.)

Note that the above paragraph summarises how Schrödinger was derived in the first sense, and its ending parenthesis hints at how Newton's second law was "derived" in my second sense. And everyone talking about path integrals is hinting at a type-2 derivation for both results (path integrals obtain a transition amplitude in terms of $e^{iS/\hbar}$ with $S$ the classical action now miraculously coming out of a hat, so technically our direct recovery is of Lagrangian mechanics rather than the equivalent Newtonian formulation).

I'll leave people to fight over which, if either, type of derivation is "valid" or "better", but physical insight requires frequent doses of both. I think it's worth distinguishing them in a discussion like this.

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It is common in books on path integrals to see the claim that we can derive the Schrodinger equation (indeed Ref. [2] below claims it can indeed be derived via path integrals), so it is surprising to see people say that it can't be derived.

I will say that of course there is a way to directly derive the Schrodinger equation simply (i.e. without the formal machinery of one-parameter groups and Stone's Theorem etc...) without path integrals and within the logic of (Copenhagen) quantum mechanics. It is given in (the absolutely canonical textbook of) reference [1]. Another derivation is given by Dirac in ([3], Sec. 27).

Assuming the existence of a wave function $$\Psi(q,t)$$ (where $q$ are generalized coordinates), and assuming the wave function completely describes the state of a physical quantum system (i.e. assuming we can obtain 'complete information' about the system and do not need density matrices, to which a slight modification of what follows still applies, see [1]), this implies that the state of the system at all future instants is determined (to within the obvious limits of quantum mechanics). This means that the time derivative $\frac{\partial \Psi}{\partial t}$ of the wave function must be determined by the value of $\Psi(q,t)$ at that instant, and the superposition principle implies this relationship has to be a linear relationship, so that $$\frac{\partial \Psi(q,t)}{\partial t} = \tilde{\hat{H}} \Psi(q,t)$$ should hold, for some linear operator $\tilde{\hat{H}}$ (notice the hat and the tilde on $\tilde{\hat{H}}$).

Further assuming the existence of classical mechanics as a limit of the theory (a fundamental unavoidable assumption at least in Copenhagen quantum mechanics), we have to describe how a quantum wave function reduces to give a classical picture (and in fact disappears in the classical limit).

The perspective taken in [1] is that this limiting procedure is done by requiring that the wave function $\Psi(q,t)$, which is a complex number, written in complex exponential form $$\Psi(q,t) = |\Psi(q,t)|e^{i \phi},$$ behaves in the same way that for example an electromagnetic field, which can also be written in complex exponential form (of course there we must still take the real part), behaves when we take the limit of geometric optics, where the notion of definite 'paths' can be defined, despite the wave-like nature of electromagnetism.

Since quantum mechanics says that paths don't exist, but that they should exist more and more accurately until they completely exist in a to-be-defined 'classical limit', this is a very natural assumption to make about the behavior of a wave function as we approach the 'classical limit'. What this means is that we should ignore the amplitude of $\Psi(q,t)$ like we ignore the amplitude in geometric optics approximation, and we should treat the argument of the exponential (the eikonal) like a function which determines 'classical paths'.

But there is obviously some function we know of that does precisely this, the 'classical action $S$'. Thus we assume in a 'quasi-classical' limit, where the wave function more and more accurately describes a classical system, should reduce to the form $$\Psi(q,t) = |\psi(q,t)| e^{i S/\hbar}$$ with $|\psi(q,t)|$ being a slowly varying amplitude that we can neglect the more classical we get, and at this stage $\hbar$ is literally just a constant with the dimensions of action $[S] = [\hbar] = [E][T]$ used to ensure the argument of the exponential is dimensionless.

We now see the classical limit is taken by sending $$\hbar \to 0$$ which means the argument of the exponential of the quasi-classical wave function will become infinite and make no sense unless the action $S$ is minimized in the $\hbar \to 0$ limit, which is exactly the condition needed for a classical action to completely describe the dynamics of a classsical system.

Therefore in the quasi-classical limit, if we take a time derivative of $\Psi(q,t)$, we can neglect contributions from the amplitude, to find $$\frac{\partial \Psi(q,t)}{\partial t} = \frac{i}{\hbar} \frac{\partial S}{\partial t} \Psi(q,t).$$ From classical analytical mechanics we know that $$\frac{\partial S}{\partial t} = - H$$ so we now approximately have $$\frac{\partial \Psi(q,t)}{\partial t} = - \frac{i}{\hbar} H \Psi(q,t).$$ ignoring contributions from the amplitude, or rather $$i \hbar \frac{\partial \Psi(q,t)}{\partial t} = H \Psi(q,t).$$

We now see that the linear operator $\tilde{\hat{H}}$ must reduce to the classical Hamiltonian $H$ (multiplied by $- \frac{i}{\hbar}$) in the classical limit, and we refer to $$i \hbar \frac{\partial \Psi(q,t)}{\partial t} = \hat{H} \Psi(q,t)$$ (with $\hat{H}$ reducing to the classical Hamiltonian $H$ in the classical limit) as the Schrodinger equation.

The form of the operator $\hat{H}$ is still left arbitrary, indeed we have not yet invoked non-relativistic Galilean or Relativistic Poincare symmetry principles which are required to fix the form of $\hat{H}$.

We can similarly determine the momentum operator from the quasi-classical limit, noting from $$\nabla \Psi = \frac{i}{\hbar} (\nabla S) \Psi$$ and the classical analytical mechanics definition of momentum as $$\mathbf{p} = \nabla S$$ that the $\hat{\mathbf{p}}$ in $$\hat{\mathbf{p}} \Psi = - i \hbar \hat{\nabla} \Psi$$ is the operator reducing to the classical momentum in the classical limit.

One immediate consequence of this perspective is by differentiating the expected value of some operator $\hat{f}$ $$ \overline{\hat{f}} = \int dq \Psi^* \hat{f} \Psi$$ we can easily derive that the time derivative of $\hat{f}$ is \begin{align*} \frac{d \hat{f}}{dt} &= \frac{\partial \hat{f}}{\partial t} + \frac{i}{\hbar}(\hat{H} \hat{f} - \hat{f}\hat{H}) \\ &= \frac{\partial \hat{f}}{\partial t} + \frac{i}{\hbar}[\hat{H},\hat{f}]. \end{align*} But we also have the classical result that some function $f$ of the canonical variables satisfies \begin{align*} \frac{d f}{dt} &= \frac{\partial f}{\partial t} + (\frac{\partial f}{\partial q_i} \frac{d q_i}{dt} + \frac{\partial f}{\partial p_i} \frac{d p_i}{dt}) \\ &= \frac{\partial f}{\partial t} + (\frac{\partial f}{\partial q_i} \frac{\partial H}{\partial p_i} - \frac{\partial f}{\partial p_i} \frac{\partial H}{\partial q_i}) \\ &= \frac{\partial f}{\partial t} - \{H,f \}_{P.B.} \end{align*} Thus the time derivative of the quantum version of a classical function $f$ should reduce to these equations of motion in the classical limit, showing that in the quasi-classical limit the commutator $$[\hat{H},\hat{f}] = \hat{H} \hat{f} - \hat{f}\hat{H}$$ reduces to $$[\hat{H},\hat{f}] = i \hbar \{H,f\}_{P.B.} + \mathcal{O}(\hbar^2),$$ justifying the fundamental quantization rule $$\{\dot{},\dot{}\}_{P.B.} \mapsto \frac{1}{\mathrm{i}\hbar}[\dot{},\dot{}]$$ for operators. In other words, to set up the basic principles of the theory, at a fundamental level it ultimately reduces to invoking this classical limit, which is why people take so seriously the existence of a classical limit and stress that (at least Copenhagen) quantum mechanics does not exist without this limit.

To fix the form of $\hat{H}$ we note, e.g. if we assume that we are dealing with a non-relativistic quantum system, then on taking the quasi-classical limit of $\Psi(q,t)$ we see the action $S$ in the quasi-classical limit should reduce to the non-relativistic classical Hamiltonian.

This is the fundamental idea behind why an approach such as the Schrodinger functional picture will apply in quantum field theory, we instead just assume the Hamiltonian obeys relativistic invariance principles (and something else important which I'll leave as a mystery) and depending on the fields involved in the action/Hamiltonian we then just apply the above thinking to get the Schrodinger functional equation and solve it (easier said than done).

If one thinks about it, they will notice some of this looks similar to the path integral approach (e.g. the quasi-classical limit). Indeed the above is constructed on the basis of assuming 'no paths' existing. What do we do when setting up path integrals? We instead assume all possible paths affect our results. Reference [2] makes explicit how assuming the existence of $e^{iS/\hbar}$ (with $S$ evaluated on all possible paths between two points) is a postulate in setting up quantum mechanics from a path integral perspective.

This is the logically consistent (within the principles of quantum mechanics) way to justify those rules of sending classical quantities to their operator analogues. It's completely inconsistent to just take a classical equation and just insert the operator forms without at least going through something equivalent to this dance, but with the above proviso's it's as consistent as one can expect. It's of course why books claim the Schrodinger equation can be derived from a path integral perspective, i.e. within that logical framework books like [2] say it can be derived, so of course it can also be derived within the canonical Copenhagen framework also.

References:

  1. Landau and Lifshitz, "Quantum Mechanics", 3rd Ed.
  2. Kaku, "Quantum Field Theory: A Modern Introduction", Ch. 8.
  3. Dirac, "Principles of Quantum Mechanics", 4th Ed.
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In Mathematics you derive theorems from axioms and the existing theorems.

In Physics you derive laws and models from existing laws, models and observations.

In this case we can start from the observations in photoelectric effect to get the relation between photon energy and frequency. Then continue with the special relativity where we observed the speed of the light is constant in all reference frames. From this when generalizing the kinetic energy we can get the mass energy equivalence. Combining the two we can assign mass to the photon, consequently we can get the momentum of a photon as function of the wavenumber.

Generalizing the energy-frequency and the momentum-wavenumber relation when have the De-Broglie relations. Which is applicable to any particles.

Assuming that a particle have 0 energy when it stands still (you can do it), although it doesn't cause too much trouble if you leave the constant term there, in the later phases you can simply put it into the left side of the equation. We can deal with the kinetic energy. Substituting the non-relativistic kinetic into the relation and reordering we can have the following dispersion relation:

$$\omega = \frac{\hbar k^2}{2m}$$

The wave equation can be derived from the dispersion relation of the matter waves using the way I mentioned in that answer.

In this case we will need the laplacian and first time derivative:

$$\nabla^2 \Psi + \partial_t \Psi = -k^2\Psi - \frac{i \hbar k^2}{2m}\Psi$$

Multiplying the time derivative with $-\frac{2m}{i\hbar}$, we can zero the right side:

$$\nabla^2 \Psi - \frac{2m}{i\hbar} \partial_t \Psi = -k^2\Psi + k^2\Psi = 0$$

We can reorder it to obtain the time dependent schrödinger equation of a free particle:

$$ \partial_t \Psi = \frac{i\hbar}{2m} \nabla^2 \Psi$$

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A new physical theory is based on a new physical principle that allows to formulate useful models. In particular, differential equations, like the Schrödinger equation, are based on first-principles. In quantum mechanics the founding principle is the uncertainty principle that mixed with probabilistic classical mechanics allows to derive the quantum dynamics that gives the practical theory. For a first-principle derivation of the Schrödinger equation, see e.g. Another look through Heisenberg's microscope.

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    $\begingroup$ You're confusing "revolutionary" with "wrong". This answer is being downvoted because it is incorrect, plain and simple. $\endgroup$ Dec 9, 2021 at 9:51

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