## Problem 2.1

June 2020

Problem: Prove the following three theorems

(a) For normalizable solutions, the separation constant $$E$$ must be real. (b) The time-independent wave-function $$\psi(x)$$ can always be taken to be real (unlike $$\Psi (x, t)$$) (c) If $$V(x)$$ is an even function (that is, $$V(x)=V(-x)$$), then $$\psi(x)$$ can always be taken to be either even or odd.

Solution:

(a) We have the wave-function

$\Psi(x, t) = \psi(x)e^{-iEt/\hbar}$

Now, if we assume that $$E$$ is a complex number that takes on the form $$E_0 + \Gamma i$$, we have

$\Psi(x, t) = \psi(x)e^{-i(E_0 + \Gamma i)t/\hbar} = \psi(x)e^{(-iE_0t + \Gamma t)/\hbar}$

If $$\Psi$$ is normalizable, then

$A \int_{-\infty}^{\infty} \Psi^* \Psi dx = 1$

for some complex number $$A$$. Substituting in the wave-function, we have

$A \int_{-\infty}^{\infty} \Psi^* \Psi dx = A \int_{-\infty}^{\infty} \psi(x)e^{(iE_0t + \Gamma t)/\hbar} \psi(x)e^{(-iE_0t + \Gamma t)/\hbar} dx = A \int_{-\infty}^{\infty} \psi^2 (x) e^{2\Gamma t / \hbar}dx \ \ \ \ \ \ (1)$

The problem here is that if we were to integrate this, the normalization constant $$A$$ would become a function of $$t$$, meaning it wouldn't be constant, meaning the function doesn't remain normalized over time. The only way that $$A$$ could remain constant is if $$\Gamma = 0$$, implying that $$E$$ must be, practically speaking, a real number.

More generally, the same can easily be shown for non-separable wave-functions (which are simply linear combinations of separable ones). We have

$\Psi(x, t) = \sum\limits_{n=1}^{\infty} c_n \psi_n (x) e^{-i E_n t / \hbar} = \sum\limits_{n=1}^{\infty} c_n \psi_n (x) e^{(-iE_nt + \Gamma_n t)/\hbar}$

Thus, if $$\Psi$$ is normalizable,

$1 = A \int_{-\infty}^{\infty} \Psi^* \Psi dx = A \int_{-\infty}^{\infty} \Psi^* \Psi dx = A \int_{-\infty}^{\infty} \left(\sum\limits_{n=1}^{\infty} c_n \psi_n (x) e^{(iE_0t + \Gamma t)/\hbar}\right) \left( \sum\limits_{n=1}^{\infty} c_n \psi_n (x) e^{(-iE_0t + \Gamma t)/\hbar} \right) dx$

$= A \int_{-\infty}^{\infty} \left(c_1 \psi_1 (x)e^{(iE_1 t + \Gamma_1 t)/\hbar} + \cdots + c_{\infty} \psi_{\infty} (x)e^{(iE_{\infty} t + \Gamma_{\infty} t)/\hbar}\right) \left(c_1 \psi_1 (x)e^{(-iE_1 t + \Gamma_1 t)/\hbar} + \cdots + c_{\infty} \psi_{\infty} (x)e^{(-iE_{\infty} t + \Gamma_{\infty} t)/\hbar}\right)dx$

The product of these two sums will produce two kinds of terms. One will take on the form of the final expression in $$(1)$$, meaning $$E$$ must be real by the same logic used before. The second kind of term will contain the product of two or more time-independent wave-functions. I didn't know this as it wasn't yet explained in the textbook, but when consulting a friend I found out that the product of two time-independent wave-functions (also known as stationary states or energy eigenstates) must be zero because they are orthogonal.

Thus, the same condition ($$\Gamma = 0$$), must be true for any wave-function.

(b) Here, Griffiths specifies that while a time-independent solution can be complex, it can always be represented as a linear combination of solutions with the same energy that are real. So what he is actually asking us to prove is that the time-independent wave-function can always be represented as a linear combination of real time-independent wave-functions.

He points out that if $$\psi(x)$$ satisfies the time-independent Schrödinger equation,

$-\frac{h^2}{2m}\frac{d^2 \psi}{dx^2}+V\psi = E\psi$

for some given $$E$$, so must its complex conjugate, and subsequently so must the real linear combinations

$(\psi + \psi^*) \ \ \ \ \textrm{and} \ \ \ \ i(\psi - \psi^*)$

With this information in mind, we can construct a proof relatively easily. We can define $$\psi$$ as follows:

$\psi(x) = a(x) + ib(x)$

Thus,

$(\psi + \psi^*) = 2a(x) \ \ \ \ \ i(\psi - \psi^*) = -2b(x)$

Here, we have two real wave functions. Expressing $$\psi$$ as a linear combination of $$2a$$ and $$-2b$$, we have

$\psi(x) = \frac{1}{2} (2a + i 2b) = \frac{1}{2} ((\psi + \psi^*) - i^2(\psi - \psi^*))$

And so we've expresssed $$\psi$$ as a linear combination of two real wave-functions.

(c) Once again, it's important to note that what Griffiths means here by "taken to be..." is that $$\psi(x)$$ can be represented as a linear combination of even and odd solutions.

If $$\psi (x)$$ satisfies

$-\frac{h^2}{2m}\frac{d^2 \psi(x)}{dx^2}+V(x)\psi(x) = E\psi(x),$

then, the same must be true for $$-x$$, and so (assuming $$V(x)=V(-x)$$) we have,

$-\frac{h^2}{2m}\frac{d^2 \psi(-x)}{dx^2}+V(-x)\psi(-x) = -\frac{h^2}{2m}\frac{d^2 \psi(-x)}{dx^2}+V(x)\psi(-x) = E\psi(-x)$

This is important because if $$V(-x) \neq V(x)$$, then $$\psi(-x)$$ wouldn't satisfy the equation for the same $$E$$ as $$\psi(x)$$.

If both $$\psi(x)$$ and $$\psi(-x)$$ satisfy the time-independent Schrödinger equation, then so must $$\psi(x) \pm \psi(-x)$$. This gives us a valid, even solution to the time-independent Schrödinger equation,

$\psi(x)' = \psi(x) + \psi(-x)$

and a valid, odd solution

$\psi(x)'' = \psi(x) - \psi(-x)$

where the apostrophies have nothing to do with derivatives and are simply used to differentiate (pun unintended) between the different functions. Since

$\psi(x) = \frac{1}{2}(\psi(x)' + \psi(x)''),$

$$\psi$$ is a linear combination of even and odd solutions, and thus $$\psi$$ may as well be taken to be either even or odd.