2 improved formatting (improper V and spacing around "=") edit approved Dec 24 '17 at 9:06 Przemysław Scherwentke 31511 gold badge44 silver badges88 bronze badges 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 $$a, and V=$$\infty$$$$V=\infty$$ everywhere else. The particle energy, $$E$$, is greater than $$V_0$$. Often, $$V_0$$ is set to zero, so let's do that. When you solve the time-independent Schrodinger equation for the region $$a you you get the solution you propose: $$\psi (x) = A \sin (kx) + B \cos (kx).$$ where $$k=\frac{2mE}{\hbar^2}$$ For the other regions, consider a very large potential, $$G$$, where $$G>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}\left(G-E\right)\psi(x).$$$$\frac{\mathrm{d^2}\psi(x)}{\mathrm{d}x^2}=\frac{2m}{\hbar^2}(G-E)\psi(x).$$ The solution for this is $$\psi(x)=Ce^{\kappa x}+De^{-\kappa x},$$ where $$\kappa =\frac{2m}{\hbar^2}\left(G-E\right).$$ 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 $$a region because the solutions must be continuous at the boundaries of the well. So, for your system, because the potential is symmetric about zero, your solutions must have definite parity about zero which means that the set of solutions will have $$A=0$$ (positive parity) for some and $$B=0$$ (negative parity) for other solutions. You can't have a free particle in an infinite square well. It's a bound particle for which the potential function is finite in a certain region. For example, if the problem is for a 1-dimensional system, $$V=V_0$$ for $$a, 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. When you solve the time-independent Schrodinger equation for the region $$a you get the solution you propose: $$\psi (x) = A \sin (kx) + B \cos (kx).$$ where $$k=\frac{2mE}{\hbar^2}$$ For the other regions, consider a very large potential, $$G$$, where $$G>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}\left(G-E\right)\psi(x).$$ The solution for this is $$\psi(x)=Ce^{\kappa x}+De^{-\kappa x},$$ where $$\kappa =\frac{2m}{\hbar^2}\left(G-E\right).$$ This solution must be bounded for both $$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 region because the solutions must be continuous at the boundaries of the well. 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. 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 $$a, and $$V=\infty$$ everywhere else. The particle energy, $$E$$, is greater than $$V_0$$. Often, $$V_0$$ is set to zero, so let's do that. When you solve the time-independent Schrodinger equation for the region $$a you get the solution you propose: $$\psi (x) = A \sin (kx) + B \cos (kx).$$ where $$k=\frac{2mE}{\hbar^2}$$ For the other regions, consider a very large potential, $$G$$, where $$G>E$$, and take the limit as $$G\to \infty$$. The SWE becomes $$\frac{\mathrm{d^2}\psi(x)}{\mathrm{d}x^2}=\frac{2m}{\hbar^2}(G-E)\psi(x).$$ The solution for this is $$\psi(x)=Ce^{\kappa x}+De^{-\kappa x},$$ where $$\kappa =\frac{2m}{\hbar^2}\left(G-E\right).$$ 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 $$a region because the solutions must be continuous at the boundaries of the well. So, for your system, because the potential is symmetric about zero, your solutions must have definite parity about zero which means that the set of solutions will have $$A=0$$ (positive parity) for some and $$B=0$$ (negative parity) for other solutions. 1 answered Dec 23 '17 at 18:13 Bill N 10.5k11 gold badge2424 silver badges4343 bronze badges You can't have a free particle in an infinite square well. It's a bound particle for which the potential function is finite in a certain region. For example, if the problem is for a 1-dimensional system, $$V=V_0$$ for $$a, 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. When you solve the time-independent Schrodinger equation for the region $$a you get the solution you propose: $$\psi (x) = A \sin (kx) + B \cos (kx).$$ where $$k=\frac{2mE}{\hbar^2}$$ For the other regions, consider a very large potential, $$G$$, where $$G>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}\left(G-E\right)\psi(x).$$ The solution for this is $$\psi(x)=Ce^{\kappa x}+De^{-\kappa x},$$ where $$\kappa =\frac{2m}{\hbar^2}\left(G-E\right).$$ This solution must be bounded for both $$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 region because the solutions must be continuous at the boundaries of the well. 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.