Behind the Guesses: 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

[Click here for a PDF of this post with nicer formatting]
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.
³Kind of the reverse of how we got to this whole exponential notation in the first place...

5 comments:

Peeter JootMay 26, 2009 at 6:32 PM
That's an interesting approach, and very clearly written and understandable.

It seems to me however somewhat against the grain of "behind the guesses". In particular, equation 8, motivating the -i/h difference between the quantum and classical Hamiltonians, appears to be to have been pulled out of a magic hat. Although it obviously works to get the desired equation, a natural motivation for this factor or the quantum version of the poisson bracket would be desirable.

Also, it isn't clear to me that the generalization in (10) to "any" quantum operator from it's classical counterpart is justifyable or if it is the details of what that mean exactly are not clear (consider the one dimensional position operator with no i/hbar factors until you go to the momentum space representation).
Eli LanseyMay 26, 2009 at 6:51 PM
Peeter,
Thanks for your comment. I agree that Eq. (8) is a bit "out of the blue," but going through that derivation is really a bit more off topic. Maybe I'll cover it in another blog post :-) But see the Landau and Lifshitz book (ref [4]), section 9, pp. 27-27 and footnote for an excellent discussion (they differ by a factor of i).

Regarding your latter point, I think you are right -- I made a mistake (I'll correct it in a new post soon). Briefly, though, as I noted, any classical Poisson bracket can be transferred over to the QM commutator with the iℏ. This means that there's a difference of iℏ for only one of the operators. For the Hamiltonian equation (8) f can be anything, so my argument holds. But for the x and p operators there's a vagueness. So in one representation the p's get those factors, in another the x's do.
Peeter JootMay 27, 2009 at 12:30 PM
Perhaps offtopic, but what would you recommend as the best QM book(s), or online references, for self study. I've currently got Bohm's Quantum theory, French's Introductory QM, Pauli's Wave Mechanics, and Feynman's volume 3, but am lacking any text with a modern treatment.

I've been working my way through these kind of lock step. Bohm's text is very well laid out and the problems are helpful. Feynman's doesn't have problems which makes it fairly hard to actively use (will probably be better once I know the subject). Pauli's is fairly dense and takes a lot of puzzling out, but has helpful bits, and French's I plan to revisit in more detail later (having covered some of it eons ago when I did my engineering undergrad). I wasn't impressed with Liboff's book which I borrowed from the public library (too many magic hats used there).
LucasMay 27, 2009 at 6:43 PM
Hi,
You cited a couple books, but do you happen to have the pages/chapters in which the specific examples appear?
Thanks
Eli LanseyMay 27, 2009 at 8:06 PM
Lucas,
In Sakurai: pp. 68-72
Jackson: pp. 543-548
Landau Lifshitz CM: pp. 135-138
Landau Lifshitz QM: pp. 27-27
Goldstein: 407-408

Peeter,
Contact me via email: elansey@gmail.com
One of these days I'll make a recommended book section on the sidebar.

Behind the Guesses

Pages

Tuesday, May 26, 2009

The Schrödinger Equation

5 comments:

About Me

Subscription Options

Other sites to check out

Followers

Blog Archive