I understand that the Schrodinger equation is actually a principle that cannot be proven. But can someone give a plausible foundation for it and give it some physical meaning/interpretation. I guess I'm searching for some intuitive solace here.

  • $\begingroup$ [Feynman's derivation PDF] (drchristiansalas.org.uk/MathsandPhysics/Research/…) $\endgroup$
    – Tony
    Commented Sep 17, 2014 at 3:04
  • $\begingroup$ For whatever it's worth, the true difficulty in understanding what the Schrodinger equation means is in understanding what state vectors mean. Or, if you prefer the Heisenberg picture, what operators mean. $\endgroup$
    – DanielSank
    Commented Sep 17, 2014 at 3:35
  • 6
    $\begingroup$ It's a statement of conservation of energy. $\endgroup$
    – user4552
    Commented Sep 17, 2014 at 5:38
  • 1
    $\begingroup$ One should be able to "derive" the Schroedinger equation as one special case of quantum field theory in the non-relativistic single particle effective potential limit, but the derivation may not be particularly useful. I have no idea what can be learned from it. $\endgroup$
    – CuriousOne
    Commented Sep 17, 2014 at 6:55
  • $\begingroup$ @CuriousOne I think you can show some interesting things in condensed matter physics with that sort of approach. $\endgroup$
    – innisfree
    Commented Sep 18, 2014 at 13:31

5 Answers 5


This is a fairly basic approach suitable for students who have finished at least one semester of introductory Newtonian mechanics, are familiar with waves (including the complex exponential representation) and have heard of the Hamiltonian at a level where $H = T + V$. As far as I understand it has no relationship to Schrödinger's historical approach.

Let's take up Debye's challenge to find the wave equation that goes with de Broglie waves (restricting ourselves to one dimension merely for clarity).

Because we're looking for a wave equation we will suppose that the solutions have the form $$ \Psi(x,t) = e^{i(kx - \omega t)} \;, \tag{1}$$ and because this is suppose to be for de Broglie waves we shall require that \begin{align} E &= hf = \hbar \omega \tag{2}\\ p &= h\lambda = \hbar k \;. \tag{3} \end{align}

Now it is a interesting observation that we can get the angular frequency $\omega$ from (1) with a time derivative and likewise wave number $k$ with a spacial derivative. If we simply define the operators1 \begin{align} \hat{E} = -\frac{\hbar}{i} \frac{\partial}{\partial t} \tag{4}\\ \hat{p} = \frac{\hbar}{i} \frac{\partial}{\partial x} \; \tag{5}\\ \end{align} so that $\hat{E} \Psi = E \Psi$ and $\hat{p} \Psi = p \Psi$.

Now, the Hamiltonian for a particle of mass $m$ moving in a fixed potential field $V(x)$ is $H = \frac{p^2}{2m} + V(x)$, and because this situation has no explicit dependence on time we can identify the Hamiltonian with the total energy of the system $H = E$. Expanding that identity in terms of the operators above (and applying it to the wave function, because operators have to act on something) we get \begin{align} \hat{H} \Psi(x,t) &= \hat{E} \Psi(x,t) \\ \left[ \frac{\hat{p}^2}{2m} + V(x) \right] \Psi(x,t) &= \hat{E} \Psi(x,t) \\ \left[ \frac{1}{2m} \left( \frac{\hbar}{i} \frac{\partial}{\partial x}\right)^2+ V(x) \right] \Psi(x,t) &= -\frac{\hbar}{i} \frac{\partial}{\partial t} \Psi(x,t) \\ \left[ -\frac{\hbar^2}{2m} \frac{\partial^2}{\partial x^2}+ V(x) \right] \Psi(x,t) &= i\hbar \frac{\partial}{\partial t} \Psi(x,t) \;. \tag{6}\\ \end{align} You will recognize (6) as the time-dependent Schrödinger equation in one dimension.

So the motivation here is

  1. Write a wave equation.
  2. Make the energy and momentum have the de Broglie forms, and
  3. Require energy conservation

but this is not anything like a proof because the pass from variable to operators is pulled out of a hat.

As an added bonus if you use the square of the relativistic Hamiltonian for a free particle $(pc)^2 - (mc^2)^2 = E^2$ this method leads naturally to the Klein-Gordon equation as well.

1 In very rough language an operator is a function-like mathematical object that takes a function as an argument and returns another function. Partial derivatives obviously qualify on this front, but so do simple multiplicative factors: because multiplying a function by some factor returns another function.

We follow a common notational convention in denoting objects that need to be understood as operators with a hat, but leaving the hat off of explcit forms.

  • $\begingroup$ This was the approach I got at Cornell -- I think it was David Muller teaching the course. $\endgroup$
    – CR Drost
    Commented Sep 5, 2017 at 1:22

I like @Simon 's answer, but my personal favorite method to "derive" the Schrodinger equation is this.

Think of the quantum state as encoding some information about your system. That is to say some quantum version of a probability distribution defined on a vector space (Hilbert space).

What do we want of a meaningful probability distribution? First it must be always normalized so that mutually exclusive outcomes add up to probability 1. Second, we want all the probabilities corresponding to these outcomes to always be positive or at least 0. The most general form of a time evolution operator - that is to say an operator acting on your state at time $t_0$ takes it to $t_1$ is a so-called completely positive trace preserving map -http://en.wikipedia.org/wiki/Quantum_operation

This means essentially that the map meets all the requirements I stated (there are some subtleties, but it will take longer to explain).

Now, we may ask what sort of dynamical equation corresponds to this map? We want the equation to be Markovian, that is local in time so that the system does not depend on what happened a long time ago because this would violate locality in some sense.

Lindblad has shown that the most general form of such an equation is,

$$ \dot\rho=-{i\over\hbar}[H,\rho]+\sum_{n,m = 1}^{N^2-1} h_{n,m}\left(L_n\rho L_m^\dagger-\frac{1}{2}\left(\rho L_m^\dagger L_n + L_m^\dagger L_n\rho\right)\right)$$

where $\rho$ is the state, $H$ is the Hamiltonian, $h_{m,n}$ are some rates and the $L_m$ are so called Lindblad operators which can be any operator.

However, as Banks, Susskind and Peskin have shown - http://adsabs.harvard.edu/abs/1984NuPhB.244..125B

this type of equation violates energy conservation or locality unless all $h_{m,n}$ are zero. If it violates energy conservation it can not describe a closed system which is invariant with respect to shifts in time. Therefore we set them to 0 and obtain just,

$$ \dot\rho=-{i\over\hbar}[H,\rho],$$

which is the von Neumann equation, which reduces to the Schrodinger equation for pure states, $\rho=|\psi \rangle \langle \psi|$

  • $\begingroup$ That's actually the von Neumann equation, isn't it? which can be derived from the Schrödinger equation (and vice-versa)? $\endgroup$
    – innisfree
    Commented Sep 18, 2014 at 13:09
  • $\begingroup$ @innisfree Yes, for pure states it reduces to Schrodinger equation. I've updated the answer to clarify. $\endgroup$
    – Bubble
    Commented Sep 18, 2014 at 15:25

It stands for dynamics for strange quantum particle that can be expressed as a wave $\psi$. Since quantum theory is fundamentally probabilistic we must write down classical Hamiltonian in expectation values:$$\langle H\rangle=\langle T\rangle+\langle V(x,t)\rangle$$

In quantum mechanics we use different operators that act on state $\psi$ $$\langle H\rangle=\int i\hbar \frac{d}{dt}\psi \centerdot \overline{\psi}dx$$ $$\langle T\rangle=\int - \frac{\hbar^2}{2m}\frac{d^2}{dx^2}\psi \centerdot \overline{\psi}dx$$ $$\langle V(x,t)\rangle=\int V(x,t)|\psi|^2dx$$

[NOTE: if you dare, you can actually derive previous expressions with Fourier methods and lesser assumptions, but genereally we take them granted since Schrödinger's equation is a fundamental postulate in physics]

After using variational lemma you get: $$ i\hbar \frac{d}{dt}\psi = - \frac{\hbar^2}{2m}\frac{d^2}{dx^2}\psi + V(x,t)\psi $$.

  • 1
    $\begingroup$ You are supposing that the solution to the Schrödinger equation is actually a function $\psi(x)$. But the Schrödinger equation does also hold in finite-dimensional Hilbert spaces (e.g. pure spin). $\endgroup$
    – ACuriousMind
    Commented Sep 21, 2014 at 13:29
  • 1
    $\begingroup$ Originally Schrödiner's equation is set upon energy conservation. If the resulting dynamics can offer meaningful results in "pure spin" cases... well, that's just great! :D I bet Mr. Schrödinger was not considering quantum-spin when he "derived" that equation. $\endgroup$
    – user59412
    Commented Sep 21, 2014 at 13:38
  • 2
    $\begingroup$ That's not how you calculate mean values; the correct formula is $\langle H \rangle = \int \psi^* (i\hbar \frac{d}{dt}) \psi\ dx$. $\endgroup$
    – Javier
    Commented Sep 21, 2014 at 15:26
  • $\begingroup$ @Javier. You are right, it's a typo. $\endgroup$
    – user59412
    Commented Sep 21, 2014 at 15:39

In order to understand the Schrodinger equation, you must know what the state vector is http://en.wikipedia.org/wiki/Quantum_state. Before we mesuare the system, its state could be any linear combinations of eigenvectors. The probability of obesrved value is square of the coefficient (also called wavefunction) of the corresponding eigenvector.

When state changes with time, we could apply an operator U to state vector, just as the way we get angular momentum operator by rotating state vector. According to information conservation, the eigenvectors can't mix up, so inner product is conserved and U is unitary vector. Assume U = I - iεH, I is identity operator, ε is a small number, then you can derive the Schordinger equation by Taylor expansion.

In a word, Schordinger equation is used to described how the state of system changes with time, and Hamilton operator is responsible for change.

I'm sorry for the unclear words, I was supposed to derive the equation for you , but , shamed ! I don't know how to use LaTex or MathJax. But I'm working on it.


The Schrodinger equation provides the definite energy space an electron can occupy inside a shell in the atom. The wave function, one of its variables, actually gives the probable location of an electron in space.

  • 1
    $\begingroup$ This is like saying that classical mechanics is for finding the periodicity of a pendulum where one of it's variables actually gives its probable location in space. $\endgroup$
    – user121330
    Commented Sep 5, 2017 at 1:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.