Is it possible to derive Schrodinger equation in this way?

Question

Let's have wave-function $\lvert \psi \rangle$. The full probability is equal to one:

$$\langle \Psi\lvert\Psi \rangle = 1.\tag{1}$$

We need to introduce time evolution of $\Psi $; we know it in the initial moment of time. So it's naturally set

$$\lvert \Psi (t) \rangle = \hat {U}|\Psi (0) \rangle ,$$

and from $(1)$ it follows that

$$\hat {U}^{\dagger}\hat {U} = \hat {\mathbf E}.$$

So it may be represented as $U = e^{i \alpha \hat {H}t}$ (we suppose that $\hat {H}^{\dagger} = \hat {H}$ and $\hat {H} \neq \hat {H}(t)$ for simplifying derivation). Thus it is possible to write

$$\partial_{t}\lvert\Psi (t) \rangle = i\alpha \hat {H}| \Psi\rangle.$$

But how to get the physical interpretation for $\hat {H}$?

Answer 1

One can indeed motivate the Schrödinger equation along the lines you suggest. The crucial point you are missing is time shift invariance of your quantum system. It is this that lets you write down $U = \exp(i\alpha\,H\,t)$.

To explain further:

1. The fact of $\psi(t) = U(t) \psi(0)$ is simply a linearity assumption.

2. The evolution wrought by your state transition matrix $U(t)$ over time n units is simply the matrix product of the individual transition operations over 1 unit, so $U(n) = U(1)^n$ for integer $n$. This is simply saying the evolution operator for any fixed time interval is the same, no matter when it is imparted to the quantum state (i.e time shift invariance). The evolution in the same experiment doesn't care whether I do it now or in ten minutes time after I have my cup of tea (as long as I don't spill any tea on the experiment, say). It's a Copernican notion. Arguing like this, and slicing the time interval $t$ in different ways, you can quickly prove things like $U(p) = U(1)^p$ for all rational $p$ and $U(t+s) = U(s)U(t) = U(t) U(s);\,\forall s,t\in\mathbb{R}$. The only continuous matrix function with all these properties is $U = \exp(K t)$, for some constant $K$.

3. Now the assumption of probability conservation ("full possibilities") is brought to bear. This means that $U$ must be unitary - your $U^\dagger U = U U^\dagger = I$. This means $\exp(K t) \exp(K^\dagger t) = \exp(K^\dagger t) \exp(K t) = I$. So $K$ and $K^\dagger$ must commute and $K + K^\dagger = 0$, whence $K = -K^\dagger$. $K$ is thus skew Hermitian. Now every skew Hermitian $K$ can be written as $K = i H$, for some Hermitian $H$. And we can pull out any nonzero, real scalar factor we like to get $U(t) = \exp(i\alpha H)$

The rest of your argument follows. How do we get the physical interpretation for $H$? Its simply a hunch. With a bit of work (converting from the Schrödinger to Heisenberg picture) you can show that any observable that commutes with $H$ is constant in time - its eigenvalues do not evolve. $H$'s eigenvalues are thus. So, if we postulate that $H$ is indeed an observable $\hat{H}$ with its full blown recipe for measurement interpretation instead of simply a plain boring old operator, then all the statistics of its measurements are constant in time. It represents a time-shift-invariant symmetry. So, by analogy with the classical energy and Noether's theorem for the classical energy, we simply postulate that $H = \hat{H} = {\rm energy\, observable}$.

Richard Feynman uses exactly this approach in the early chapters of Volume 3 of his Lectures on Physics. There is a chapter 7 called "The Dependence of Amplitude on Time" wherein he does this much better than I do.