# Why do we pick trigonometric solutions, instead of complex exponentials, when solving the Schrödinger equation for a square well?

I was thinking about something that I never understood when I took my QM course. If I have an infinite square well, say $$V=0$$ for $$-a and $$V= \infty$$ otherwise. In the region of interest, the Schrödinger Equation reads: $$-\frac{\hbar}{2m}\frac{\partial^2\psi}{\partial x^2}=E\psi$$

so the differential equation to solve is:

$$\frac{\partial^2\psi}{\partial x^2}=-k^2\psi$$ with $$k^2=\frac{2mE}{\hbar^2}$$, as usual. Here, I remember that I was taught that we use the solutions $$\psi=A\sin(kx)+B\cos(kx),$$ which obviously satisfies the DE. However, I never quite understood why we don't pick $$\psi=Ae^{ikx}+Be^{-ikx},$$ since this function also would satisfy the DE.

During my QM course I always struggled choosing when to use an exponential solution versus a sinusoidal one. Could someone shed some light on this matter? There must be a clear reason to choose one over the other when facing a new problem.

• They’re equivalent, as the answer points out, but choosing the sin/cos form makes it more straightforward to satisfy the boundary conditions. In either case, you end up with just a sine. – G. Smith Nov 12 at 15:24

Expanding, we have:

$$Ae^{ikx}+Be^{-ikx}=(A+B)\sin(kx)+i(A-B)\cos(kx)$$

Setting $$A'=A+B$$ and $$B'=i(A-B)$$ we have that

$$Ae^{ikx}+Be^{-ikx}=A'\sin(kx)+B'\cos(kx)$$

so the two solutions are equivalent. It really doesn't matter in the end.

• Oh, for some reason I thought that the constants had to be real. I underdtand then, thanks! – Nick Heumann Nov 12 at 15:32

You're right, both are solutions to the Schrodinger equation for this potential. They are equivalent representations using Euler's Formula. The preference for complex exponentials vs sinusoids comes down to mathematical convenience. They are generally interchangeable.

As a rule of thumb, if the wave function is 0 outside of a bounded region of space, you probably want a sinusoid. In these scenarios, solutions are frequently standing waves, which are identical to traveling waves, waves of the form $$e^{ikx}$$ of the same magnitude moving at equal speeds and opposite directions. This is precisely the case for the solutions in the infinite square well. For the boundary conditions we have either $$B=A$$ or $$B=-A$$, but $$Ae^{ikx}+Ae^{-ikx}=2A\cos{kx}.$$ If $$B=-A$$, $$Ae^{ikx}-Ae^{-ikx}=2iA\sin{kx}.$$ Now, sines and cosines have known zeros. We can manipulate the argument to match the boundaries of our regions. So if the infinite square well is potential 0 from L and infinite elsewhere, we can readily choose a sine since all sines are zero at the origin and a simple formula for the other boundary point gives us our values for k. This technique would add extra work if you used just complex exponentials instead.

$$Ae^{ikx}$$ for any real values of k is a traveling wave, good for modeling scattering problems, say, particles incident on a finite potential well. Here the boundary conditions don't tend to allow the simple use of wave functions with equal and opposite wave numbers having the same intensity, or standing wave solutions. Consider the reflective wave function in the finite square well problem. The solution is a combination of wave functions with equal and opposite wave numbers, but the intensities are different. You can convert $$\psi=e^{ikx}+Re^{-ikx}=(1+R)\cos{kx}+i(1-R)\sin{kx}$$, but note the somewhat higher complexity in the representation. There's more terms to work with in the sinusoidal case.

It comes down to ease of mathematical manipulation which varies based on the problem. For bounded states, you often want to go with sinusoidals. For scattering states, you often want to go with complex exponentials instead.

$$\Psi(x)=Ae^{ikx}+Be^{-ikx}=Acos(kx)+Aisin(kx)+Bcos(kx)-Bisin(kx)=Ccos(kx)+Disin(kx)$$ You have to impose continuity of the solution at the border of the hole: $$\Psi(a)=Ccos(ka)+Disin(ka)=0$$ $$\Psi(-a)=Ccos(ka)-Disin(ka)=0$$ which are true if you put $$D=0$$ or $$C=0$$.
In the first case: $$0=cos(k^*a)$$ $$\Psi_{1}(x)=Ccos(k^*x)$$ In the second case: $$0=sin(k^/a)$$ $$\Psi_{2}(x)=iDsin(k^/x)$$ These two are good solutions, so by linearity you can sum the two and obtain another good solution. $$\Psi(x)=\Psi_{1}(x)+\Psi_{2}(x)=Ccos(k^*x)+iDsin(k^/x)$$ You have also to guarantee the continuity of the fist derivate in order to be able to get the second one: $$\frac{d}{dx}\Psi(x)=-k^*Csin(k^*x)+iDk^/cos(k^/x)$$ $$\frac{d}{dx}\Psi(x)|_{a}=-k^*Csin(k^*a)+iDk^/cos(k^/a)=0$$ $$\frac{d}{dx}\Psi(x)|_{-a}=k^*Csin(k^*a)+iDk^/cos(k^/a)=0$$

How you can see, the only possibility is $$C=0$$ and $$D=0$$ which means that you have the trivial solution. This is due to the fact that this is a nonphysical system and you can not find a good solution everywhere inside the hole.

There are many correct ways of answering this question, but I think the following is probably how an expert would actually think about it:

1. Schordinger's equation in the interior region is a second-order differential equation, so we expect it to have a two-dimensional vector space of solutions. We can choose any basis for a vector space. Omitting constants, both $$\{\sin x,\cos x\}$$ and $$\{e^{ix},e^{-ix}\}$$ are possible choices of basis. We're free to choose either.

2. In general, quantum mechanics involves complex-valued amplitudes, and there us no way to avoid this. However, in the special case of a standing wave we can usually get a real-valued wavefunction if we prefer it. Therefore it makes sense to pick a basis consisting of manifestly real functions, in hopes that we can also match the boundary conditions using real coefficients. This would just be nicer and easier.