Particle in a box: Difference between revisions
imported>Michael Underwood m (→The solutions: Changed caption to include negative integers) |
imported>Michael Underwood (→The 1D square well: Added alternate description, and fixed it so n must be positive) |
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One solution to this is of course <math>b=0</math>. However, this would mean that the wavefunction vanishes ''everywhere'' -- implying that there is no particle! The other way to satisfy this equality is to have the [[sine]] term vanish, which will happen if | One solution to this is of course <math>b=0</math>. However, this would mean that the wavefunction vanishes ''everywhere'' -- implying that there is no particle! The other way to satisfy this equality is to have the [[sine]] term vanish, which will happen if | ||
:<math>\sqrt{\frac{2mE}{\hbar^2}}L = 2n\pi\ ,</math> | :<math>\sqrt{\frac{2mE}{\hbar^2}}L = 2n\pi\ ,</math> | ||
where <math>n</math> | where mathematically <math>n</math> can be any [[integer]]. | ||
This problem is not purely mathematical though, and we know for physical reasons that <math>m</math>, <math>E</math>, <math>\hbar</math>, and <math>L</math> are all greater than zero. | |||
This means that <math>2n\pi</math> must also be greater than zero, so <math>n</math> is restricted to being a [[natural number]], <math>1,2,3,...</math>. | |||
We can now solve for the particle's energy, | |||
:<math>E_n=\frac{n^2\pi^2\hbar^2}{2mL^2}\ ,</math> | :<math>E_n=\frac{n^2\pi^2\hbar^2}{2mL^2}\ ,</math> | ||
where we have | where we have labelled the energy by the integer <math>n</math>. | ||
We have just derived energy quantization! Without the potential well, i.e. if the particle was [[free particle|free]], its energy would be allowed to take on any real number. Once inside the box though, only a specific [[discrete]] set of energy eigenvalues is permitted. | We have just derived energy quantization! Without the potential well, i.e. if the particle was [[free particle|free]], its energy would be allowed to take on any real number. Once inside the box though, only a specific [[discrete]] set of energy eigenvalues is permitted. | ||
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===Normalization=== | ===Normalization=== | ||
The probability of finding the particle ''somewhere'' must be | The probability of finding the particle ''somewhere'' must be <math>1</math>. Since the wavefunction vanishes everywhere outside <math>[0,L]</math>, | ||
the normalization condition is | |||
:<math>1 | :<math> 1 | ||
=\int_0^L\psi^*(x)\psi(x)\,\mathrm{d}x | =\int_0^L\psi^*(x)\psi(x)\,\mathrm{d}x | ||
=|b|^2\int_0^L\sin^2\left(\frac{n\pi x}{L}\right)\,\mathrm{d}x | =|b|^2\int_0^L\sin^2\left(\frac{n\pi x}{L}\right)\,\mathrm{d}x | ||
=|b|^2\frac{L}{2}\ . | =|b|^2\frac{L}{2}\ . | ||
</math> | </math> | ||
Solving | Solving for <math>b</math> tells us <math>b=e^{i\varphi}\sqrt{2/L}</math>, where <math>\varphi</math> is some overall [[quantum phase]] (a real number). | ||
Because overall phases do not affect measurable results we are free to choose any value we want for <math>\varphi</math> without | Because overall phases do not affect measurable results we are free to choose any value we want for <math>\varphi</math> without | ||
affecting the physics. For simplicity then, we set <math>\varphi=0</math>. | affecting the physics. For simplicity then, we set <math>\varphi=0</math>. | ||
===The solutions=== | ===The solutions=== | ||
[[Image:particle-in-a-box.png|thumb|320px|A plot of <math>\psi_n(x)</math> for <math>n=1</math> (darkest) to <math>n=7</math> (lightest) | [[Image:particle-in-a-box.png|thumb|320px|A plot of <math>\psi_n(x)</math> for the first 7 wavefunctions, <math>n=1</math> (darkest) to <math>n=7</math> (lightest).]] | ||
We have now solved the particle-in-a-box problem. According to the Schrödinger wave equation the particle confined to the (infinite) box can only take on certain energy [[eigenvalues]], namely | We have now solved the particle-in-a-box problem. According to the Schrödinger wave equation the particle confined to the (infinite) box can only take on certain energy [[eigenvalues]], namely | ||
:<math> E_n=\frac{n^2\pi^2\hbar^2}{2mL^2}\ ,</math> | :<math> E_n=\frac{n^2\pi^2\hbar^2}{2mL^2}\ ,</math> | ||
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:<math>\psi_n(x)=\sqrt{\frac{2}{L}}\sin\left(\frac{n\pi x}{L}\right)</math> | :<math>\psi_n(x)=\sqrt{\frac{2}{L}}\sin\left(\frac{n\pi x}{L}\right)</math> | ||
inside the box, and <math>\psi(x)=0</math> outside it. | inside the box, and <math>\psi(x)=0</math> outside it. | ||
===Alternate description=== | |||
Another common (equivalent) way to describe this problem is to shift the location of the box, so that it runs from <math>-L/2</math> to <math>L/2</math> instead of <math>0</math> to <math>L</math>. The derivation of the allowed wavefunctions is very similar to what we did here, or the results can be obtained via [[coordinate transformation]] (specifically, a [[translation (mathematics)|translation]]). With this different description of the square well the resulting wavefunctions are<ref>[[Albert Messiah|A. Messiah]] (year) ''Quantum Mechanics: Two volumes bound as one'' ISBN 0486409244</ref> | |||
:<math>\psi_n(x)=\sqrt{\frac{2}{L}}\cos\left(\frac{n\pi x}{L}\right),</math> | |||
if <math>n</math> is odd, and | |||
:<math>\psi_n(x)=\sqrt{\frac{2}{L}}\sin\left(\frac{n\pi x}{L}\right),</math> | |||
if <math>n</math> is even. | |||
==Properties / Discussion / Comments (under construction)== | ==Properties / Discussion / Comments (under construction)== | ||
Revision as of 13:22, 10 May 2007
The particle in a box or infinite square well problem is one of the simplest non-trivial solutions to Schrödinger's wave equation. As such it is often encountered in introductory quantum mechanics material as a demonstration of the quantization of energy. In its simplest form the problem is one-dimensional, and involves a single particle living in an infinite potential well.
The 1D square well
Setting up the problem
The potential can be given by
- Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle V(x)=0\,\!} for and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle V(x)=\infty} otherwise.
That is, in a region of length Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle L} (the box or well) the potential is zero; everywhere else it is infinite. An immediate consequence of this is that the particle must be located somewhere between Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x=0} and , since if it were anywhere else it would have to have infinite energy.
Since Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle V(x)} does not depend on time we can use the time-independent version of the Schrödinger equation,
- Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle -\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2}\psi(x)=E\psi(x)\ , }
where Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \hbar} is Planck's constant divided by 2π, and and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle E} are the mass and energy of the particle, respectively. This is nothing but the eigenvalue problem, and our task is to determine the energy eigenvalue(s) Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle E} and eigenstate(s) Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \psi} that solve it. The equation is a second-order linear ODE with constant coefficients, and the general solution can be written as
where Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle a} and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle b} are complex constants to be determined from conditions on the system. The first condition we can use is that the wavefunction must be continuous. We know that the particle cannot exist at positions Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x<0} , which tells us that the wavefunction must be identically zero there (in order for the probability of finding the particle there to also be zero). Continuity then implies that at the wavefunction is also zero, so we have
- Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 0=\psi(0)=a\cos(0)+b\sin(0)=a\ ,}
meaning that the wavefunction can now be written
- Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \psi(x)=b\sin\left(\sqrt{\frac{2mE}{\hbar^2}}x\right)\ .}
Energy quantization
Using the same continuity argument at Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x=L} tells us that
One solution to this is of course Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle b=0} . However, this would mean that the wavefunction vanishes everywhere -- implying that there is no particle! The other way to satisfy this equality is to have the sine term vanish, which will happen if
- Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sqrt{\frac{2mE}{\hbar^2}}L = 2n\pi\ ,}
where mathematically Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle n} can be any integer. This problem is not purely mathematical though, and we know for physical reasons that , Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle E} , Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \hbar} , and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle L} are all greater than zero. This means that must also be greater than zero, so Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle n} is restricted to being a natural number, Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 1,2,3,...} . We can now solve for the particle's energy,
- Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle E_n=\frac{n^2\pi^2\hbar^2}{2mL^2}\ ,}
where we have labelled the energy by the integer Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle n} . We have just derived energy quantization! Without the potential well, i.e. if the particle was free, its energy would be allowed to take on any real number. Once inside the box though, only a specific discrete set of energy eigenvalues is permitted.
Substituting Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle E_n} back into Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \psi} , we now have an infinite number of possible wavefunctions for the particle,
- Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \psi_n(x)=b\sin\left(\frac{n\pi x}{L}\right)\ .}
The final step is to determine the value of Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle b} . This requires another condition that we can impose upon the problem.
Normalization
The probability of finding the particle somewhere must be Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 1} . Since the wavefunction vanishes everywhere outside Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle [0,L]} , the normalization condition is
- Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 1 =\int_0^L\psi^*(x)\psi(x)\,\mathrm{d}x =|b|^2\int_0^L\sin^2\left(\frac{n\pi x}{L}\right)\,\mathrm{d}x =|b|^2\frac{L}{2}\ . }
Solving for Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle b} tells us Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle b=e^{i\varphi}\sqrt{2/L}} , where Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \varphi} is some overall quantum phase (a real number). Because overall phases do not affect measurable results we are free to choose any value we want for without affecting the physics. For simplicity then, we set Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \varphi=0} .
The solutions
We have now solved the particle-in-a-box problem. According to the Schrödinger wave equation the particle confined to the (infinite) box can only take on certain energy eigenvalues, namely
- Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle E_n=\frac{n^2\pi^2\hbar^2}{2mL^2}\ ,}
and the wavefunction for the particle when it is in the Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle n} th energy eigenstate is
- Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \psi_n(x)=\sqrt{\frac{2}{L}}\sin\left(\frac{n\pi x}{L}\right)}
inside the box, and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \psi(x)=0} outside it.
Alternate description
Another common (equivalent) way to describe this problem is to shift the location of the box, so that it runs from Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle -L/2} to Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle L/2} instead of Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 0} to Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle L} . The derivation of the allowed wavefunctions is very similar to what we did here, or the results can be obtained via coordinate transformation (specifically, a translation). With this different description of the square well the resulting wavefunctions are[1]
- Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \psi_n(x)=\sqrt{\frac{2}{L}}\cos\left(\frac{n\pi x}{L}\right),}
if Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle n} is odd, and
- Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \psi_n(x)=\sqrt{\frac{2}{L}}\sin\left(\frac{n\pi x}{L}\right),}
if Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle n} is even.
Properties / Discussion / Comments (under construction)
The Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \psi_n} are complete.
Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle E=0} is not allowed because Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle n=0} implies wavefunction vanishes everywhere; no particle. So lowest energy state is above zero.
Energy levels are degenerate; Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \psi_{-n}=-\psi_n} but Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle E_{-n}=E_n} .
As Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle L\to\infty} energy spacing goes to zero; recover the free space result.
Generalization to 3D (to be done)
- ↑ A. Messiah (year) Quantum Mechanics: Two volumes bound as one ISBN 0486409244