## Tuesday, May 26, 2009

### The Schrödinger Equation

Update: A corrected and improved version of this post is now up: http://behindtheguesses.blogspot.com/2009/06/schrodinger-equation-corrections.html

notElon asked me to discuss, and to try and derive the Schrödinger equation, so I'll give it a shot. This derivation is partially based on Sakurai,[1] with some differences.

A brief walk through classical mechanics
Say we have a function of $f(x)$ and we want to translate it in space to a point $(x+a)$. To do this, we'll find a ``space translation'' operator $\mathcal{S}_a$ which, when applied to $f(x)$, gives $f(x+a)$. That is,
$f(x+a)=\mathcal{S}_{a} f(x)$
(1)
We'll expand $f(x+a)$ in a Taylor series:
$f(x+a)&=f(x)+a\frac{df(x)}{dx}+\frac{a^2}{2!}\frac{d^2f(x)}{dx^2}+\ldots$
$=\left[1+a\frac{d}{dx}+\frac{a^2}{2!}\frac{d^2}{dx^2}+\ldots\right]f(x)$
(2)
which can be simplified using the series expansion of the exponential1 to
$e^{\left[a\frac{d}{dx}\right]}f(x)$
(3)
from which we can conclude that
$\mathcal{S}_a=e^{\left[a\frac{d}{dx}\right]}$
(4)
If you do a similar thing with rotations around the $z$-axis, you'll find that the rotation operator is
$\mathcal{R}_\theta=e^{\theta L_z},$
(5)
where $L_z$ is the $z$-component of the angular momentum.

Comparing (4) and (5), we see that both have an exponential with a parameter (distance or angle) multiplied by something ($\frac{d}{dx}$ or $L$). We'll call the something the ``generator of the transformation.'' So, the generator of space translation is $\frac{d}{dx}$ and the generator of rotation is $L$. So, we'll write an arbitrary transformation operator $\mathcal{O}$ through a parameter $\alpha$ as
$\mathcal{O}_a=e^{\alpha G}$
(6)
where $G$ is the generator of this particular transformation.2 See [2] for an example with Lorentz transformations.

From classical to quantum
In classical dynamics, the time derivative of a quantity $f$ is given by the Poisson bracket:
$\frac{df}{dt}=\{f,H\}$
(7)
where $H$ is the classical Hamiltonian of the system and $\{\,\,,\,\}$ is shorthand for a messy equation.[3] In quantum mechanics this equation is replaced with
$\frac{df}{dt}=i\hbar[f,\mathcal{H}]$
(8)
where the square brackets signify a commutation relation and $\mathcal{H}$ is the quantum mechanical Hamiltonian.[4] This holds true for any quantity $f$, and $i\hbar$ is a number which commutes with everything, so we can argue that the quantum mechanical Hamiltonian operator is related to the classical Hamiltonian by
$H=i\hbar\mathcal{H} \Rightarrow \mathcal{H}=-i H/\hbar$
(9)
specifically.

Additionally, we can extend from here that any quantum operator $\mathcal{G}$ is written in terms of its classical counterpart $G$ by
$\mathcal{G}=-i G/\hbar.$
(10)

So, using (4) the quantum mechanical space translation operator is given by
$\mathcal{S}_a=e^{\left[-i\frac{a}{\hbar}\frac{d}{dx}\right]}$
(11)
and, using (5), the rotation operator by
$\mathcal{R}_\theta=e^{-i\frac{\theta}{\hbar} L_z}$
(12)
or, from (6) any arbitrary (unitary) transformation, $\mathcal{U}$, can be written as
$\mathcal{U}=e^{-i \frac{\alpha}{\hbar} G},$
(13)
where $G$ is (an Hermitian operator and is) the classical generator of the transformation.

Time translation of a quantum state
Consider a quantum state at time $t$ described by the wavefunction $\psi(\vec{r},t)$. To see how the state changes with time, we want to find a ``time-translation'' operator $\mathcal{T}_{\Delta t}$ which, when applied to the state $\psi(\vec{r},t)$, will give $\psi(\vec{r},t+\Delta t)$. That is,
$\psi(\vec{r},t+\Delta t)=\mathcal{T}_{\Delta t}\psi(\vec{r},t).$
(14)
From our previous discussion we know that if we know the classical generator of time translation we can write $\mathcal{T}$ using (13). Well, classically, the generator of time translations is the Hamiltonian![5] So we can write
$\mathcal{T}_{\Delta t}=e^{-i \frac{\Delta t}{\hbar} H}$
(15)
and (14) becomes
$\psi(\vec{r},t+\Delta t)=e^{-i \frac{\Delta t}{\hbar} H}\,\psi(\vec{r},t).$
(16)

This holds true for any time translation, so we'll consider a small time translation and expand (16) using a Taylor expansion3 dropping all quadratic and higher terms:
$\psi(\vec{r},t+\Delta t)\approx\left[1-i\frac{\Delta t}{\hbar}H+\ldots \right]\psi(\vec{r},t)$
(17)
Moving things around gives
$H\psi(\vec{r},t)=i\hbar\left[\frac{\psi(\vec{r},t+\Delta t)-\psi(\vec{r},t)}{\Delta t}\right]$
(18)
In the limit $\Delta t\rightarrow 0$ the righthand side becomes a partial derivative giving the Schrödinger equation
$H\psi(\vec{r},t)=i\hbar\frac{\partial \psi(\vec{r},t)}{\partial t}$
(19)

For a system with conserved total energy, the classical Hamiltonian is the total energy
$H=\frac{\vec{p}\,^2}{2m}+V$
(20)
which, making the substitution for quantum mechanical momentum $\vec{p}=i\hbar\nabla$ and substituting into (19) gives the familiar differential equation form of the Schrödinger equation
$-\frac{\hbar^2}{2m}\nabla^2\psi(\vec{r},t)+V\psi(\vec{r},t)=i\hbar\frac{\partial \psi(\vec{r},t)}{\partial t}$
(21)

References
[1] J.J. Sakurai. Modern Quantum Mechanics. Addison-Wesley, San Francisco, CA, revised edition, 1993.
[2] J.D. Jackson. Classical Electrodynamics. John Wiley & Sons, Inc., 3rd edition, 1998.
[3] L.D. Landau and E.M. Lifshitz. Mechanics. Pergamon Press, Oxford, UK.
[4] L.D. Landau and E.M. Lifshitz. Quantum Mechanics. Butterworth-Heinemann, Oxford, UK.
[5] H. Goldstein, C. Poole, and J. Safko. Classical Mechanics. Cambridge University Press, San Francisco, CA, 3rd edition, 2002.

1 $e^x=\sum_{n=0}^{\infty} \frac{x^n}{n!}=1+x+\frac{x^2}{2!}+\ldots$
2 There are other ways to do this, differing by factors of $i$ in the definition of the generators and in the construction of the exponential, but I'm sticking with this one for now.
3 Kind of the reverse of how we got to this whole exponential notation in the first place...