# Full time-derivative of a function and Schrodinger equation

From Hamiltonian formalism there is well known equation, $$\frac{d F}{dt} = \frac{\partial F}{\partial t} + \{F, H\}_{PB},$$ where $\{H, F\}_{PB}$ is the Poisson bracket.

After using Hamiltonian formalism in quantum mechanics it transforms into $$\frac{d \hat {F}}{dt} = \frac{\partial \hat {F}}{\partial t} + \frac{i}{\hbar }[\hat {H}, \hat {F}]$$ in the Heisenberg picture, where $[\hat {H}, \hat {F}]$ is the commutator.

How can I get Schroedinger equation from this expression?

• In the second equation, the subscript $P$ should be omitted because it's a commutator, not a Poisson bracket. The second equation is the Heisenberg equation of motion for the Heisenberg picture where operators change but the wave function doesn't. It is equivalent to Schrödinger's equation where the operators are fixed (up to their explicit time derivative) and the wave function changes. You may verify that the (truly observable) matrix elements $\langle \psi |F_1 F_2\dots |\psi\rangle$ will have the same time derivative in both pictures for all products of operators $F$. – Luboš Motl Jul 9 '13 at 16:22

The link you're looking for is hidden by the fact that Heisenberg kets $| \psi \rangle_H$ are not Schroedinger kets $|\psi \rangle _S$. The tranlation from the Heisenberg to the Schroedinger pictures is done via the time-evolution operator $U(t - t_0) = \exp\{-i H * (t - t_0)/\hbar\}$ (I do not explicity put hats on operators, since it should be pretty obrvious what's an operator and what's not).

In the Heisenberg picture, kets $|\psi \rangle_H \equiv |\psi (t_0) \rangle_H$ are not time-dependent. Heisenbergs idea was that the state of a system is not time-dependent, but the observables are. So an operator in the Heisenberg picture at a time $t$ is $$F(t) = U^\dagger(t - t_0) F(t_0) U(t-t_0)$$ The Heiseberg equation follows from this: $$\frac{d}{dt} F(t) = U^\dagger(t-t_0) (\partial_t F(t_0)) U(t-t_0) + (\partial_t U^\dagger (t-t_0)) F(t_0) U(t-t_0) + + U^\dagger(t-t_0) F(t_0) (\partial_t U(t-t_0))$$ The Leibniz terms give exacly the commutator from the Heisenberg equation.

In the Schroedinger picture, the operators are constant $F(t) \equiv F(t_0)$, but the state changes $| \psi \rangle_S = | \psi(t) \rangle_S$. As all physics needs to be equal in both pictures, we can look at expectation values: \begin{aligned}\langle \psi | F | \psi \rangle &= {}_H\langle \psi | F(t) | \psi \rangle_H &&= \langle \psi | U^\dagger(t-t_0) F(t_0) U(t-t_0) | \psi \rangle \\ & = {}_S \langle \psi(t) | F(t_0) | \psi(t) \rangle_S &&= \langle \psi(t_0) | U^\dagger(t-t_0) F(t_0) U(t-t_0) | \psi(t_0) \rangle \end{aligned} where we take the basis of our Hilbert space to conincide for both pictures at $t_0$.

Thus, assuming the Heisenberg euqation is correct, the equation for the time-evolution of a state in the Schroedinger picture is $$| \psi(t) \rangle_S = U(t-t_0) | \psi(t_0) \rangle_S$$ The time derviative then gives you the Schroedinger equation $$\frac{d}{dt} |\psi(t)\rangle = \Big(\frac{d}{dt} U(t-t_0)\Big) |\psi(t_0)\rangle = -\frac{i}{\hbar} H | \psi(t_0) \rangle$$

In the Heisenberg representation, the "equivalent" of the Schrodinger equation is :

$$\hat H(t) = \frac{\hat P^2(t)}{2m} + V(\hat X(t))$$ with $[\hat P(t), \hat X(t)] = i\hbar$

If you are looking at eigenstates and eigenvalues of the hamiltonian, you will look for a constant Hamiltonian.

For instance, for the harmonic oscillator, you will have :

$$Constant ~ operator ~ \hat H(t) = \frac{\hat P^2(t)}{2m} + \frac{m \omega^2}{2} \hat X^2(t)$$

For the harmonic oscillator, solutions are :

$\hat X(t) = \sqrt{\frac{\hbar}{m \omega}} (a e^{i \omega t} + a^+e^{-i \omega t})$, with $Constant ~ operator ~ \hat H(t) = (a^+a + 1/2) \hbar \omega$

Here $a$ and $a^+$ are constant operators such as $[a,a^+] = 1$, explicitely the only non-null members of the operator $a$ are $a_{n+1,n} = \sqrt{n}$

Explicitely, $\hat H(t)$ is a diagonal operator, with $H_{nn} = (n+1/2) \hbar \omega$

In this representation, the "wavefunction" is simply a vector in the vector basis upon which these operators act.