canonical commutator relation(CCR) can be derived by switching classical poisson bracket to quantum commutator and this process is called as canonical quantization.

$[x_i,p_j]_{PB}=\delta_{ij} →\frac{1}{ih}[x_i,p_j]=\delta_{ij}$

and also, the classical hamilton's equation of motion can be written by this quantum meachanical language,

$ \frac{dp}{dt}=[p,H]_{PB},\frac{dq}{dt}=[q,H]_{PB} →i\hbar\frac{dp}{dt}=[p,H],i\hbar\frac{dq}{dt}=[q,H]$

This is the heisenberg equation of motion as far as I know.

I want to know, how to derive schrodinger equation from this heisenberg equation of motion.

Maybe it looks like a stupid question. But please help me.


Take $\hat{O}$ to be operator of some observable.

In Heisenberg picture the states are time independent, i.e.

$$\partial_t |\psi_H\rangle=0,$$

but the operators depend on time

$$\partial_t \hat{O}_H=\frac{i}{\hbar}\left[\hat{H},\hat{O}_H\right]$$

A 'solution' to this equation is:

$$\hat{O}_H\left(t\right) = \exp\left(\frac{it}{\hbar}\hat{H}\right)\hat{O}\exp\left(-\frac{it}{\hbar}\hat{H}\right)$$

where $\partial_t \hat{O}=0$.

Any observable corresponding to the above operator will be:

$$O\left(t\right)=\langle \psi_H | \hat{O}_H \left(t\right) |\psi_H\rangle=\langle \psi_H |\exp\left(\frac{it}{\hbar}\hat{H}\right) \hat{O} \exp\left(-\frac{it}{\hbar}\hat{H}\right) |\psi_H\rangle$$

We now choose to define the Schrodinger picture as time-independent operators and time-dependent states:

$$|\psi\left(t\right)\rangle = \exp\left(-\frac{it}{\hbar}\hat{H}\right) |\psi_H\rangle$$

$$O\left(t\right)=\langle \psi \left(t\right) | \hat{O} |\psi \left(t\right)\rangle$$

It follows that: $\partial_t |\psi\left(t\right)\rangle= \frac{-i}{h}\hat{H}|\psi\left(t\right)\rangle $, usually written as:

$$i\hbar \partial_t |\psi\left(t\right)\rangle= \hat{H}|\psi\left(t\right)\rangle $$

  • $\begingroup$ thank you so much! $\endgroup$ – Einstein Jul 1 '19 at 9:07

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