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Line 15: − − ===Derivation=== − The following derivation, likely one that Schrodinger followed himself, is a completely non-rigorous method which takes a more intuitive approach. Whatever ambiguities arose in this derivation because of its questionable assumptions were wiped out by subsequent experiments verifying again and again the equation's ability to predict probabilities of particle location<ref>French, A.P. and Taylor, E.F.. ''An Introduction to Quantum Physics''. CRC Press, Boca Raton, FL. Copyright MIT 1978.</ref>. − − To start out, we assume that Planck and Einstein's quantized energy equation, <math>E = h\nu</math> where E is the energy, h is planck's constant, and nu is the frequency, is correct. We also assume that DeBroglie's wavelength of particles equation, <math>\lambda_{dB} = \frac{h}{p}</math> where lambda is the wavelength, h is planck's constant, and p is the momentum of the particle, is correct (these were indeed questionable assumptions during Schrodinger's time). Now we can rewrite the energy equation by multiplying and dividing the right hand side by <math>2 \pi</math> and turn it into − − − <math>E = \frac{h\omega}{2\pi}=\hbar\omega</math> − − − where omega is now the angular frequency. We can also rewrite DeBroglie's equation in the same way when divided by <math>2\pi</math>, turning it into − − − <math>\frac{\lambda}{2\pi} = \frac{\hbar}{p} or \frac{1}{k} = \frac{\hbar}{p}</math> − − − where k is the [[wavenumber]] of wavelength <math>\lambda_dB</math>. − − Now, we can derive energy equations using classical Newtonian mechanics and plug our results in from the new developments above. Energy in classical terms is − − − <math>E = K_E + V_E = \frac{p^2}{2m} + V</math> − − − where − E = energy, <math>K_E = \frac{1}{2}mv^2</math> (kinetic energy), <math>V_E = V</math> (potential energy) − − Note that we've rewritten kinetic energy in terms of momentum, <math>p = mv</math>. Subbing in energy from the quantized energy equation and momentum from DeBroglie's equation, we obtain − − − <math>\hbar\omega = \frac{\hbar^2k^2}{2m} + V</math> − − Notice this looks kind of like a quasi-wave equation. The one-dimensional [[wave equation]] from classical mechanics is given by the equation − − − <math>\frac{\partial^2y(x,t)}{\partial x^2} = \frac{1}{c^2}\frac{\partial^2y(x,t)}{\partial t^2}</math> − − − In classical wave mechanics, we assume the solution to be of the form <math>y = Ae^{i(kx-\omega t)}</math>. So the <math>\omega</math> in the left hand of the energy equation looks like a single partial derivative with respect to t, and the <math>k^2</math> on the right hand side looks like two partials with respect to x. If we assume a similar solution to the energy equation as in the classical wave equation but ''retain the imaginary parts'', we can set − − − <math>\Psi(x,t) = Ae^{i(kx-\omega t)}</math> − − − and each of its partials with respect to each corresponding side of the energy equation equal to each other. For now, we take ''V = 0'' for simplicity. For the partial with respect to x: − − − <math>\frac{\partial^2\Psi}{\partial x^2} = -k^2\Psi = \frac{-p^2}{\hbar^2}\Psi = \frac{-2mE}{\hbar^2}\Psi </math> − − − and with respect to t (stop at one partial derivative): − − − <math>\frac{\partial\Psi}{\partial t} = -i\omega\Psi = \frac{-iE}{\hbar}\Psi </math> − − − There are <math>E\Psi</math> terms in both equations, and we can equate those two together and get: − − − <math>\frac{-\hbar^2}{2m}\frac{\partial^2\Psi}{\partial x^2} = i\hbar\frac{\partial\Psi}{\partial t}</math> − − − Almost in its final form, we sub in the fact that <math> E = K_E + V_E </math> and realize this will just add an additional <math>V\Psi</math> term to the equation, we obtain the final form − − − <math>\frac{-\hbar^2}{2m}\frac{\partial^2\Psi}{\partial x^2} + V(x)\Psi = i\hbar\frac{\partial\Psi}{\partial t}</math> − Line 102:
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Line 45: − − − ==References== − <small><references/></small>
Schrodinger equation (view source)
Revision as of 14:15, September 29, 2009
, 14:15, September 29, 2009Reverted edits by Guardianofrice (Talk) to last version by BrianW
-\frac{\hbar^2}{2m}\frac{\partial^2\psi}{\partial x^2}+V(x)\psi=i\hbar\frac{\partial \psi}{\partial t}
-\frac{\hbar^2}{2m}\frac{\partial^2\psi}{\partial x^2}+V(x)\psi=i\hbar\frac{\partial \psi}{\partial t}
</math>
</math>
===Eigenvalue problems===
===Eigenvalue problems===
<math>E=\frac{\hbar^2 k^2}{2m}</math>
<math>E=\frac{\hbar^2 k^2}{2m}</math>
Physically, this corresponds to a wave traveling with a [[momentum]] given by <math>\hbar k</math>, where k can in principle take any value.
Physically, this corresponds to a wave travelling with a [[momentum]] given by <math>\hbar k</math>, where k can in principle take any value.
===Particle in a box===
===Particle in a box===
<math>E_n=\frac{\hbar^2 k^2}{2m}=\frac{\hbar^2n^2\pi^2}{2m}</math>
<math>E_n=\frac{\hbar^2 k^2}{2m}=\frac{\hbar^2n^2\pi^2}{2m}</math>
[[Category:Physics]]
[[Category:Physics]]