Question 2.1: Prove the following three theorems: (a) For normalizable sol...

(a)

$\Psi(x, t)=\psi(x) e^{-i\left(E_{0}+i \Gamma\right) t / \hbar}=\psi(x) e^{\Gamma t / \hbar} e^{-i E_{0} t / \hbar} \Longrightarrow|\Psi|^{2}=|\psi|^{2} e^{2 \Gamma t / \hbar}$.

$\int_{-\infty}^{\infty}|\Psi(x, t)|^{2} d x=e^{2 \Gamma t / \hbar} \int_{-\infty}^{\infty}|\psi|^{2} d x$.

The second term is independent of t, so if the product is to be 1 for all time, the first term $(e^{2Γt/\hbar})$ must also be constant, and hence Γ = 0.     QED.

(b) If ψ satisfies Eq. 2.5

$-\frac{h^{2}}{2 m} \frac{d^{2} \psi}{d x^{2}}+V \psi=E \psi$      (2.5).

then (taking the complex conjugate and noting that V and E are real): $-\frac{h^{2}}{2 m} \frac{d^{2} \psi^*}{d x^{2}}+V \psi^*=E \psi^*$ , so $\psi^*$ also satisfies Eq. 2.5. Now, if $ψ_1$ satisfy Eq. 2.5, so too does any linear combination of them ($ψ_3 ≡ c_1 ψ_1 + c_2 ψ_2$ ):

$-\frac{h^{2}}{2 m} \frac{d^{2} \psi}{d x^{2}}+V \psi=E \psi$      (2.5).

$=c_{1}\left(E \psi_{1}\right)+c_{2}\left(E \psi_{2}\right)=E\left(c_{1} \psi_{1}+c_{2} \psi_{2}\right)=E \psi_{3}$.

Thus, $(ψ + ψ^*)$ and $i (ψ – ψ^*)$ – both of which are real – satisfy Eq. 2.5. Conclusion: From any complex solution, we can always construct two real solutions (of course, if ψ is already real, the second one will be zero). In particular, since $\psi=\frac{1}{2}\left[\left(\psi+\psi^{*}\right)-i\left(i\left(\psi-\psi^{*}\right)\right)\right]$ , ψ can be expressed as a linear combination of two real solutions.     QED.

(c) If ψ(x) satisfies Eq. 2.5, then, changing variables x → -x and noting that d²/d(-x)² = d² /dx²,

$-\frac{h^{2}}{2 m} \frac{d^{2} \psi}{d x^{2}}+V \psi=E \psi$      (2.5).

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

so if V (-x) = V (x) then ψ(-x) also satisfies Eq. 2.5. It follows that $ψ_+(x) ≡ ψ(x) + ψ(-x)$ (which is even: $ψ_+(-x) = ψ_+(x)$) and $ψ_-(x) ≡ ψ(x) – ψ(-x)$ (which is odd: $ψ_-(x) = -ψ_-(x)$ ) both satisfy Eq.2.5. But $\psi(x)=\frac{1}{2}\left(\psi_{+}(x)+\psi_{-}(x)\right)$, so any solution can be expressed as a linear combination of even and odd solutions.    QED.