**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: **

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.

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.

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.