Difference between revisions of "Schrodinger equation"
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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>. | 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>\ | + | 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 |
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<math>\hbar\omega = \frac{\hbar^2k^2}{2m} + V</math> | <math>\hbar\omega = \frac{\hbar^2k^2}{2m} + V</math> | ||
| − | Notice this looks kind of like | + | Notice this looks kind of like a quasi-wave equation. The one-dimensional [[wave equation]] from classical mechanics is given by the equation |
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| + | <math>\frac{\partial^2y(x,t)}{\partial x^2} = \frac{1}{c^2}\frac{\partial^2y(x,t)}{\partial t^2}</math> | ||
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| + | 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 | ||
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| + | <math>\Psi(x,t) = Ae^{i(kx-\omega t)}</math> | ||
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| + | 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: | ||
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| + | <math>\frac{\partial^2\Psi}{\partial x^2} = -k^2\Psi = \frac{-p^2}{\hbar^2}\Psi = \frac{-2mE}{\hbar^2}\Psi </math> | ||
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| + | and with respect to t (stop at one partial derivative): | ||
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| + | <math>\frac{\partial\Psi}{\partial t} = -i\omega\Psi = \frac{-iE}{\hbar}\Psi </math> | ||
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| + | There are <math>E\Psi</math> terms in both equations, and we can equate those two together and get: | ||
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| + | <math>\frac{-\hbar^2}{2m}\frac{\partial^2\Psi}{\partial x^2} = i\hbar\frac{\partial\Psi}{\partial t}</math> | ||
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| + | 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 | ||
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| + | <math>\frac{-\hbar^2}{2m}\frac{\partial^2\Psi}{\partial x^2} + V(x)\Psi = i\hbar\frac{\partial\Psi}{\partial t}</math> | ||
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===Eigenvalue problems=== | ===Eigenvalue problems=== | ||
Revision as of 23:08, September 28, 2009
The Schrodinger equation is a linear differential equation used in various fields of physics to describe the time evolution of quantum states. It is a fundamental aspect of quantum mechanics. The equation is named for its discoverer, Erwin Schrodinger.
Mathematical forms
General time-dependent form
The Schrodinger equation may generally be written
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle i\hbar\frac{\partial}{\partial t}|\Psi\rangle=\hat H|\Psi\rangle}
where Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle i}
is the imaginary unit,
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \hbar}
is Planck's constant divided by ,
is the quantum mechanical state or wavefunction (expressed here in Dirac notation), and
is the Hamiltonian operator.
The left side of the equation describes how the wavefunction changes with time; the right side is related to its energy. For the simplest case of a particle of mass m moving in a one-dimensional potential V(x), the Schrodinger equation can be written
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[1].
To start out, we assume that Planck and Einstein's quantized energy equation, 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, 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 and turn it into
where omega is now the angular frequency. We can also rewrite DeBroglie's equation in the same way when divided by Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 2\pi}
, turning it into
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{\lambda}{2\pi} = \frac{\hbar}{p} or \frac{1}{k} = \frac{\hbar}{p}}
where k is the wavenumber of wavelength Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \lambda_dB}
.
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
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle E = K_E + V_E = \frac{p^2}{2m} + V}
where
E = energy, Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle K_E = frac{1}{2}mv^2}
(kinetic energy), <mathV_E = V</math> (potential energy)
Note that we've rewritten kinetic energy in terms of momentum, Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle p = mv} . Subbing in energy from the quantized energy equation and momentum from DeBroglie's equation, we obtain
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \hbar\omega = \frac{\hbar^2k^2}{2m} + V}
Notice this looks kind of like a quasi-wave equation. The one-dimensional wave equation from classical mechanics is given by the equation
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{\partial^2y(x,t)}{\partial x^2} = \frac{1}{c^2}\frac{\partial^2y(x,t)}{\partial t^2}}
In classical wave mechanics, we assume the solution to be of the form Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle y = Ae^{i(kx-\omega t)}}
. So the Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \omega}
in the left hand of the energy equation looks like a single partial derivative with respect to t, and the Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle k^2}
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
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Psi(x,t) = Ae^{i(kx-\omega t)}}
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:
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{\partial^2\Psi}{\partial x^2} = -k^2\Psi = \frac{-p^2}{\hbar^2}\Psi = \frac{-2mE}{\hbar^2}\Psi }
and with respect to t (stop at one partial derivative):
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{\partial\Psi}{\partial t} = -i\omega\Psi = \frac{-iE}{\hbar}\Psi }
There are Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle E\Psi}
terms in both equations, and we can equate those two together and get:
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{-\hbar^2}{2m}\frac{\partial^2\Psi}{\partial x^2} = i\hbar\frac{\partial\Psi}{\partial t}}
Almost in its final form, we sub in the fact that Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle E = K_E + V_E }
and realize this will just add an additional Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle V\Psi}
term to the equation, we obtain the final form
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{-\hbar^2}{2m}\frac{\partial^2\Psi}{\partial x^2} + V(x)\Psi = i\hbar\frac{\partial\Psi}{\partial t}}
Eigenvalue problems
In many instances, steady-state solutions to the equation are of great interest. Physically, these solutions correspond to situations in which the wavefunction has a well-defined energy. The energy is then said to be an eigenvalue for the equation, and the wavefunction corresponding to that energy is called an eigenfunction or eigenstate. In such cases, the Schrodinger equation is time-independent and is often written
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle E\psi=\hat H\psi}
Here, E is energy, H is once again the Hamiltonian operator, and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \psi} is the energy eigenstate for E.
One example of this type of eigenvalue problem is an electrons bound inside an atom.
Examples for the time-independent equation
Free particle in one dimension
In this case, Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle V(x)=0} and so we see that the solution to the Schrodinger equation must be
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \psi=Ae^{-ikx}}
with energy given by
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle E=\frac{\hbar^2 k^2}{2m}}
Physically, this corresponds to a wave traveling with a momentum given by Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \hbar k} , where k can in principle take any value.
Particle in a box
Consider a one-dimensional box of width a, where the potential energy is 0 inside the box and infinite outside of it. This means that Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \psi} must be zero outside the box. One can verify (by substituting into the Schrodinger equation) that
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \psi=\sin(kx)}
is a solution if Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle k=n\pi} where n is any integer. Thus, rather than the continuum of solutions for the free particle, for the particle in a box there is a set of discrete solutions with energies given by
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle E_n=\frac{\hbar^2 k^2}{2m}=\frac{\hbar^2n^2\pi^2}{2m}}
References
- ↑ French, A.P. and Taylor, E.F.. An Introduction to Quantum Physics. CRC Press, Boca Raton, FL. Copyright MIT 1978.