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How do you derive the Schrödinger equation (wave mechanics, time dependent state) from Heisenberg's Matrix Mechanics (matrix based, time dependent operators)

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The basic idea is simple enough: one seeks a unitary transformation on the Hilbert space that preserves all physical measurements, which is equivalent to keeping entities of the form $\langle \psi,\,\hat{A}\,\psi\rangle$ unchanged, where $\psi$ is a quantum state and $\hat{A}$ an observable (or any of its powers). Furthermore, we want to "freeze" the states. So suppose we begin with a Schrödinger picture state $\psi$ evolving by:

$$i\,\hbar\,\partial_t\,\psi = \hat{H}\,\psi$$

or, equivalently, $\psi(t) = \exp\left(-\frac{i}{\hbar}\,\hat{H}\,t\right)\,\psi(0)$. We can easily justify the the existence of some self-adjoint $\hat{H}$ from any unitary one parameter group of evolutions in the finite dimensional case; in the infinite dimensional case, one appeals to the Stone Theorem on One Parameter Unitary Groups.

So now, how do our physical measurements vary? They vary like:

$$M(t) = \langle \psi(t),\,\hat{A}\,\psi(t)\rangle$$

when we write them down in the Schrödinger picture. But now:

$$\begin{array}{lcl}M(t) &=& \langle \psi(t),\,\hat{A}\,\psi(t)\rangle\\ &=& \langle \exp\left(-\frac{i}{\hbar}\,\hat{H}\,t\right)\,\psi(0),\,\hat{A}\,\exp\left(-\frac{i}{\hbar}\,\hat{H}\,t\right)\,\psi(0)\rangle\\ &=&\langle \psi(0),\,\left(\exp\left(-\frac{i}{\hbar}\,\hat{H}\,t\right)\right)^\ast\,\hat{A}\,\exp\left(-\frac{i}{\hbar}\,\hat{H}\,t\right)\,\psi(0)\rangle\\ &=&\langle \psi(0),\,\exp\left(+\frac{i}{\hbar}\,\hat{H}\,t\right)\,\hat{A}\,\exp\left(-\frac{i}{\hbar}\,\hat{H}\,t\right)\,\psi(0)\rangle \end{array} $$

so that we'll keep all our measurements unchanged with frozen states if and only if our observables evolve following:

$$\hat{A}(t) = \exp\left(+\frac{i}{\hbar}\,\hat{H}\,t\right)\,\hat{A}\,\exp\left(-\frac{i}{\hbar}\,\hat{H}\,t\right)$$

so this is the unitary transformation of observables we seek. Let's work out the differential form of this one: we get:

$$\partial_t\hat{A}(t) = \frac{i}{\hbar}\,\left(\hat{H}\,\hat{A}(t)-\hat{A}(t)\,\hat{H}\right) = \frac{i}{\hbar}[\hat{H},\,\hat{A}]$$

which of course is the Heisenberg equation, the quantum analogue of the Liouville equation from Hamiltonian mechanics.

Going the other way is straightforward; one uses the inverse transformation . Or one can derive it from scratch using the same principle as above: we must keep our measurements fixed.

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  • $\begingroup$ But what if I wanted to say, derive wave mechanics completely from Heisenberg's matrices, including the form of the operators? $\endgroup$ – user140323 Mar 20 '18 at 13:06
  • $\begingroup$ Perhaps you could use the Stone-Von Neumann theorem, which says that up to unitary equivalence, there is only one choice for the position and momentum operators that satisfy the commutation relation. $\endgroup$ – user1379857 Mar 20 '18 at 13:34
  • $\begingroup$ @user1379857 Yes that would certainly do it if the OP wanted to take the CCR as fundamental. But I'm not really sure what he/she wants to begin from. $\endgroup$ – WetSavannaAnimal Mar 20 '18 at 13:47
  • $\begingroup$ @user140323 See my comment above on Stone-von Neumann: I'm unsure exactly what you want to begin from and perhaps you should rewrite your question to state it. $\endgroup$ – WetSavannaAnimal Mar 20 '18 at 13:48

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