$\sin$ and $\cos$ components in symmetric infinite potential well problem - Physics Stack Exchange most recent 30 from physics.stackexchange.com 2019-08-25T15:31:54Z https://physics.stackexchange.com/feeds/question/375975 http://www.creativecommons.org/licenses/by-sa/3.0/rdf https://physics.stackexchange.com/q/375975 3 $\sin$ and $\cos$ components in symmetric infinite potential well problem ahmed https://physics.stackexchange.com/users/179492 2017-12-23T17:35:38Z 2017-12-24T15:48:46Z <p>Consider an infinite potential well in one dimension with boundaries at $\pm a/2$. Can $\psi(x) = A \sin(kx) + B \cos(kx)$ for this system?</p> <p>The way it was answered was "mathematically acceptable but physically unacceptable" using the boundary conditions, but I want to understand more, like can't a superposition of a sine wave and a cosine wave meet the boundary conditions?</p> https://physics.stackexchange.com/questions/375975/-/375984#375984 2 Answer by Bill N for $\sin$ and $\cos$ components in symmetric infinite potential well problem Bill N https://physics.stackexchange.com/users/63690 2017-12-23T18:13:50Z 2017-12-24T09:06:46Z <p>You can't have a free particle in an infinite square well. It's a bound particle for which the potential function is finite in a certain region. For example, if the problem is for a 1-dimensional system, $V=V_0$ for $a&lt;x&lt;b$, and $V=\infty$ everywhere else. The particle energy, $E$, is greater than $V_0$. Often, $V_0$ is set to zero, so let's do that.</p> <p>When you solve the time-independent Schrodinger equation for the region $a&lt;x&lt;b$ you get the solution you propose: $$\psi (x) = A \sin (kx) + B \cos (kx).$$ where $$k=\frac{2mE}{\hbar^2}$$</p> <p>For the other regions, consider a very large potential, $G$, where $G&gt;E$, and take the limit as $G\to \infty$. The SWE becomes $$\frac{\mathrm{d^2}\psi(x)}{\mathrm{d}x^2}=\frac{2m}{\hbar^2}(G-E)\psi(x).$$</p> <p>The solution for this is $$\psi(x)=Ce^{\kappa x}+De^{-\kappa x},$$ where $$\kappa =\frac{2m}{\hbar^2}\left(G-E\right).$$</p> <p>This solution must be bounded for both $x\to +\infty$ and $x\to -\infty$, as well as $G\to \infty$. The only way for this to happen is for $\psi(x\le a)=0$ and $\psi(x\ge b)=0$. That establishes the boundary conditions for the sinusoidal type solution in the $a&lt;x&lt;b$ region because the solutions must be continuous at the boundaries of the well.</p> <p>So, for your system, because the potential is symmetric about zero, your solutions must have definite parity about zero which means that the set of solutions will have $A=0$ (positive parity) for some and $B=0$ (negative parity) for other solutions.</p> https://physics.stackexchange.com/questions/375975/-/376004#376004 0 Answer by Zack Hutchens for $\sin$ and $\cos$ components in symmetric infinite potential well problem Zack Hutchens https://physics.stackexchange.com/users/96636 2017-12-23T20:38:21Z 2017-12-23T20:38:21Z <p>Solving from the Schrodinger equation, you have, step by step:</p> <p>$$\hat{H}\Psi = E\Psi$$ where $$\hat{H} = K + V = \frac{-\hbar^2}{2m}\frac{\partial^2}{\partial x^2}$$ because $V_0 = 0$. Therefore, we have the differential equation $$\frac{-\hbar^2}{2m}\frac{\partial^2\Psi}{\partial x^2} = E\Psi$$ $$\rightarrow \frac{\partial^2\Psi}{\partial x^2} = -\frac{2mE}{\hbar^2}\Psi .$$ The superposition of sine and cosine you have suggested, $\Psi(x) = A\sin(kx) + B\cos(kx)$, is mathematically acceptable here, where $k = \frac{\sqrt{2mE}}{\hbar}$.</p> <p>Let us apply the value at boundary $x = a$ and $x=-a$. Then $$\Psi(a) = A\sin(\frac{\sqrt{2mE}}{\hbar}a) + B\cos(\frac{\sqrt{2mE}}{\hbar}a) = 0.$$ and $$\Psi(-a) = A\sin(-\frac{\sqrt{2mE}}{\hbar}a) + B\cos(-\frac{\sqrt{2mE}}{\hbar}a) = 0.$$ Invoking the odd/even natures of sines and cosines, this second equation is $$\Psi(-a) = -A\sin(\frac{\sqrt{2mE}}{\hbar}a) + B\cos(\frac{\sqrt{2mE}}{\hbar}a) = 0.$$ Now compute $\Psi(a) + \Psi(-a) = 0$. You get that $B = 0$. Now compute $\Psi(a) - \Psi(-a)$. You get that $A = 0$. The problem here is that the well is symmetric about zero, if you force your initial values to make either $A$ or $B$ zero, then you will get only half of the solutions.</p> https://physics.stackexchange.com/questions/375975/-/376018#376018 0 Answer by freecharly for $\sin$ and $\cos$ components in symmetric infinite potential well problem freecharly https://physics.stackexchange.com/users/129209 2017-12-23T22:43:23Z 2017-12-24T15:48:46Z <p>This is a purely mathematical question. It signifies whether non-trivial wave functions $$\psi(x) = A \sin(kx) + B \cos(kx)$$ exist with $$\psi(x=\pm a/2)=0$$ This results in a homogenous system of two linear equations for $A$ and $B$ $$A \sin(ka/2) + B \cos(ka/2)=0$$ $$-A \sin(ka/2) + B \cos(ka/2)=0$$ which has nonzero solutions for $A$ and $B$ only when the coefficient determinant is zero $$\sin(ka/2)\cos(ka/2)+\sin(ka/2)\cos(ka/2)=2\sin(ka/2)\cos(ka/2)=0$$ Thus an infinite number of wavefunctions of the above form with nonzero $A$ and $B$ exists for $$ka/2=\pm n\pi$$ or $$ka/2=\pm (n+1/2)\pi$$ The first condition gives the sine functions ($B=0$) $$\psi(x) = A \sin(kx)$$ with arbitrary $A$, the second the cosine functions ($A=0$) $$\psi(x) = B \cos(kx)$$ with arbitrary $B$. No other solutions are allowed except for the trivial mathematical solution $\psi(x) = 0$.</p>