CPL.Luke Posted September 17, 2005 Share Posted September 17, 2005 is anyone here capable of deriving the schorindger equation, or at least know why it is the way it is? I tried googleing briefly and couldn't find anything really. Link to comment Share on other sites More sharing options...

5614 Posted September 17, 2005 Share Posted September 17, 2005 I don't think it was derived as such, Schrödinger wanted a wave equation and this worked. Link to comment Share on other sites More sharing options...

Severian Posted September 17, 2005 Share Posted September 17, 2005 A plane wave of momentum [math]\vec{p}[/math] and energy [math]E[/math]is [math]\psi = N e^{-i(Et-\vec{p}\cdot\vec{x})}[/math]. You can extract the momentum and energy with operators [math]\hat \vec{p}[/math] and [math]\hat E[/math] such that [math]\hat \vec{p}\psi = \vec{p} \psi[/math] and [math]\hat E \psi = E \psi[/math], if you make the choice of operators: [math]\hat \vec{p} \equiv - i \hbar \vec{\nabla}[/math] and [math]\hat E \equiv i \hbar \frac{\partial}{\partial t}[/math]. Now, we know from classical mechanics that the total energy is the sum of the kinetic and potential energies. Also, the kinetic energy (in a non-relativistic system) is [math]E_K = \frac{1}{2} m \vec{v}^2 = \frac{1}{2m} \vec{p}^2[/math]. In other words, the total energy is: [math]E = \frac{1}{2m} \vec{p}^2 + V[/math]. Simply replacing the energy and momentum by the right operators for a plane wave gives the Schrodinger Equation: [math]i \frac{\partial \psi}{\partial t} = \left( - \frac{\hbar^2}{2m} \vec{\nabla}^2 + V \right) \psi [/math]. Using the same principals, can you derive the relativistic version (the Klien-Gordon Equation)? Link to comment Share on other sites More sharing options...

ydoaPs Posted September 17, 2005 Share Posted September 17, 2005 i can't see your [math]\LaTeX[/math] Link to comment Share on other sites More sharing options...

Severian Posted September 17, 2005 Share Posted September 17, 2005 Hmm... it was working a minute ago. Edit: Weird - I just edited it but didn't change anything and now it is working! Edit 2: or not... seems to be a site problem rather than me. Link to comment Share on other sites More sharing options...

Locrian Posted September 18, 2005 Share Posted September 18, 2005 But why use a single derivative of phi with respect to time? I mean, this is a wave we're discussing here... Link to comment Share on other sites More sharing options...

Severian Posted September 19, 2005 Share Posted September 19, 2005 But why use a single derivative of phi with respect to time? I mean, this is a wave we're discussing here... Good question. Historically this was to avoid solutions which had negative energy. If you can derive the relativistic form (as I suggested) you will see what I mean.... Link to comment Share on other sites More sharing options...

DQW Posted September 19, 2005 Share Posted September 19, 2005 A plane wave of momentum [math]\vec{p}[/math] and energy [math]E[/math]is [math]\psi = N e^{-i(Et-\vec{p}\cdot\vec{x})}[/math]... A nice way of arriving the the SE is by looking at time evolution of a quantum state' date=' and modeling this through a unitary operator. Let the state [imath']| \alpha \rangle [/imath] evolve with time as : [math]| \alpha \rangle ~ \xrightarrow{time~evolution} | \alpha,t \rangle [/math] We create an opertor [imath]{\cal U} (t,0) [/imath] that provides this time-evolution, so that : [math] | \alpha ,t \rangle = {\cal U} (t,0) ~| \alpha \rangle [/math] Now, this time evolution operator is required to satisfy a whole bunch of properties. For instance, one must have [math] \lim _{dt \rightarrow 0} {\cal U} (dt,0) = \mathbf {1} [/math] All these requirements are satisfied by writing [imath] {\cal U} (dt,0) = \mathbf {1} - i \mathbf{\Omega} dt[/imath], where [imath]\mathbf{\Omega} [/imath] is a Hermitian operator. Now borrowing from classical mechanics, the idea that the Hamiltonian is the generator of time-translation (just as momentum is the generator of spatial translations), we choose [imath]\mathbf{\Omega} [/imath] to be [imath] {\cal H} / \hbar [/imath]. And the Schrodinger Equation follows automatically from playing with the composition property of the time-evolution operator. This property requires [math] {\cal U} (dt_1 + dt_2, 0) = {\cal U} (dt_2,dt_1) ~{\cal U} (dt_1, 0) [/math] The need for this property is obvious : if you evolve a state through time [imath]dt_1[/imath] and then through [imath]dt_2[/imath], you expect the final state to be the same as that caused by a time-evolution through [imath]dt_1 + dt_2[/imath]. Now, we can use this composition property to write [math] {\cal U} (t+dt, 0) = {\cal U} (t+dt,t) ~ {\cal U} (t,0) = \left( \mathbf{1} - \frac{i {\cal H} dt}{\hbar} \right) {\cal U} (t,0) [/math] Multiplying out, and rearranging terms, this can be written in the differential form : [math]i \hbar \frac {\partial}{\partial t} {\cal U} (t,0) = {\cal H} {\cal U} (t,0) [/math] Oprating this on the initial state ket [imath] | \alpha \rangle = |\alpha, t=0 \rangle [/imath] gives [math]i \hbar \frac {\partial}{\partial t} {\cal U} (t,0)~| \alpha \rangle = {\cal H} {\cal U} (t,0)~| \alpha \rangle [/math] But from the definition of the time-evolution operator, this gives [math]i \hbar \frac {\partial}{\partial t} | \alpha , t \rangle = {\cal H} | \alpha ,t \rangle [/math] ta dah ! Link to comment Share on other sites More sharing options...

Severian Posted September 19, 2005 Share Posted September 19, 2005 That isn't the Schrodinger Equation though. That is showing (rather nicely) that the Hamiltonian is the time-evolution operator. You still need to find the form of the Hamiltonian (although you might consider that as trivial). Link to comment Share on other sites More sharing options...

DQW Posted September 19, 2005 Share Posted September 19, 2005 It's not trivial. You must arrive at the general result that the momentum operator (in position representation) is given by the gradient. This requires going over a similar process as above with spatial translations... but I decided to skip that. Or alternatively, I could say "But you already did that for me !" Link to comment Share on other sites More sharing options...

Locrian Posted September 19, 2005 Share Posted September 19, 2005 Good question. Historically this was to avoid solutions which had negative energy. If you can derive the relativistic form (as I suggested) you will see what I mean.... I admit I haven't done that yet. That is certainly a mark against me. However, I believe your answer affirms a point I'd like to make: while the SE may be considered derivable in retrospect (knowing what we know now), it wasn't at the time. In fact, I suggest it should really be considered an assumption that has just been repeatedly validated by theoretical and experimental tests. Link to comment Share on other sites More sharing options...

DQW Posted September 19, 2005 Share Posted September 19, 2005 Severian's approach postulates the wavefunction of a spinless free particle to be such-and-such, and derives the SE from there (but only for a free-particle). My (incomplete) approach starts with the classical mechanics and quantizes the various elements of it. On the other hand, the SE can itself be postulated (as in Dirac's formulation of NRQM) and things calculated from there. Link to comment Share on other sites More sharing options...

Severian Posted September 19, 2005 Share Posted September 19, 2005 It's not trivial. You must arrive at the general result that the momentum operator (in position representation) is given by the gradient. This requires going over a similar process as above with spatial translations... but I decided to skip that. So then you would have shown that [math]{\cal H} = i \hbar \frac{\partial}{\partial t}[/math] and [math]p = -i \hbar \nabla[/math] But this is still not the Schrodinger equation - for that you need to show how [math]{\cal H}[/math] and [math]p[/math] are related. Actually, to be more correct, you should derive the SE by writing down the most general Lagranian for a scalar field that you can think of, and use the Euler Lagrange equations. Link to comment Share on other sites More sharing options...

DQW Posted September 20, 2005 Share Posted September 20, 2005 But this is still not the Schrodinger equation - for that you need to show how [math]{\cal H}[/math] and [math]p[/math'] are related.Right. And you "did that" by writing out H = p^2/2m + V = total energy (at least when the Lagrangian is not explicitly time-dependent). The idea was to essentially "modify" the Classical Hamiltonian by using the momentum operator (as "derived") instead of the classical canonical momentum. Link to comment Share on other sites More sharing options...

## Recommended Posts

## Create an account or sign in to comment

You need to be a member in order to leave a comment

## Create an account

Sign up for a new account in our community. It's easy!

Register a new account## Sign in

Already have an account? Sign in here.

Sign In Now