quantum mechanics - What does the Schrodinger Equation really mean?

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.

Answer

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.

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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.

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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.