Prove the following three theorem;

a) For normalizable solutions the separation constant E must be real as ${\mathit{E}}_{\mathbf{0}}\mathbf{+}\mathit{i}\mathit{\tau }$and show that if equation 1.20 is to hold for all $\mathit{t}\mathbf{,}\mathit{\tau }$ must be zero.

b) The time - independent wave function localid="1658117146660" $\mathit{\psi }\mathbf{\left(}\mathbf{x}\mathbf{\right)}$ can always be taken to be real, This doesn’t mean that every solution to the time-independent Schrodinger equation is real; what it says is that if you’ve got one that is not, it can always be expressed as a linear combination of solutions that are . So, you might as well stick to$\mathit{\psi }$ ’s that are real

c) If is an even function then $\mathit{\psi }\mathbf{\left(}\mathbf{x}\mathbf{\right)}$can always be taken to be either even or odd

Schrödinger's wave equation, sometimes known as the Schrödinger equation, is a partial differential equation that uses the wave function to explain how quantum mechanical systems behave. The trajectory, position, and energy of these systems can be determined using the Schrödinger equation.

(a)

Consider$E=E_0+i\Gamma$, where$E_{°},\Gamma \in \mathrm{ℝ}$. Now, we calculate the probability density of the time-dependent wave function $\mathrm{\psi }$.

$$\begin{gathered} \left|\mathrm{\psi }{\left(\mathrm{x},\mathrm{t}\right)}^2\right|={\mathrm{\psi }}^{*}\mathrm{\psi } \\ =\left({\mathrm{\psi }}^{*}{\mathrm{e}}^{\left({\mathrm{E}}_0-\mathrm{i\Gamma }\right)\mathrm{t}/\mathrm{ħ}}\right)\left({\mathrm{\psi e}}^{-\left({\mathrm{E}}_0-\mathrm{i\Gamma }\right)\mathrm{t}/\mathrm{ħ}}\right) \\ ={\left|\mathrm{\psi }\right|}^2{\mathrm{e}}^{2\mathrm{\Gamma t}/\mathrm{ħ}} \end{gathered}$$

To check the normalizability of this wave-function, we integrate over all space.

$$\begin{gathered} {\int }_{-\infty }^{\infty }\left|\psi {\left(x,t\right)}^2\right|={\int }_{-\infty }^{\infty }{\left|\psi \right|}^2{e}^{2\Gamma t/ħ}dx \\ =e^{2\Gamma t/ħ}{\int }_{-\infty }^{\infty }{\left|\psi \right|}^{2}dx \\ =e^{2\Gamma t/ħ} \end{gathered}$$

So, it is not normalized (because the energy is not real), so to make it normalized we make $\mathrm{\Gamma }=0$, and in general,$E$ must be real for a normalized wave-function.

(b)

Eq.(2.5) is

$\frac{{ħ}^2}{2m}\frac{d^2\psi }{d{x}^2}+V\psi =E\psi$

Now, if $\mathrm{\psi }$satisfy this equation, then its conjugate ${\mathrm{\psi }}^{*}$also satisfy eq.(2.5), where $\mathrm{\psi }$and ${\mathrm{\psi }}^{*}$are complex.

$\frac{{ħ}^2}{2m}\frac{d^2\psi }{d{x}^2}+V{\psi }^{*}=E{\psi }^{*}$

So, any linear combination of these two function must satisfy eq.(2.5) too (i.g.,$\left(\mathrm{\psi }+{\mathrm{\psi }}^{*}\right)$and $i\left(\psi ={\psi }^{*}\right)$), in general

$\psi =a_1\omega +a_{2}x$

Where $\omega$and $x$are wave-functions, and $a_1$and $a_2$are complex constants.

$$\begin{gathered} -\frac{{ħ}^2}{2m}\frac{d^2\psi }{d{x}^2}+V\psi =-\frac{{ħ}^2}{2m}\frac{d^2\left(a_1\omega +a_{2}x\right)}{d{x}^2}+V\left(a_1\omega +a_{2}x\right) \\ =E\left(a_1\omega +a_{2}x\right) \end{gathered}$$

$$\begin{gathered} -\frac{{ħ}^2}{2m}\left(a_1\frac{d^2\omega }{d{x}^2}+a_2\frac{d^{2}x}{d{x}^2}\right)+a_{1}V\omega +a_2{V}_x=a_{1}E\omega +2Ex \\ -\frac{{ħ}^2}{2m}\left(a_1\frac{d^2\omega }{d{x}^2}+a_2\frac{d^{2}x}{d{x}^2}\right)+a_{1}V\omega +a_2{V}_x=a_{1}E\omega +2Ex \\ {a}_1\left(-\frac{{ħ}^2}{2m}\frac{d^2\omega }{d{x}^2}+V\omega \right)+a_2\left(-\frac{{ħ}^2}{2m}\frac{d^{2}x}{d{x}^2}+Vx\right)=a_{1}E\omega +a_{2}Ex \\ -\frac{{ħ}^2}{2m}\frac{d^2\psi }{d{x}^2}+V\psi =E\psi \end{gathered}$$

So from any complex solution, we can always construct two real solutions.

(c)

For an odd time-independent wave-function $\mathrm{\psi }\left(-\mathrm{x}\right)=-\mathrm{\psi }\left(\mathrm{x}\right)$, so

$-\frac{{\mathrm{ħ}}^2}{2\mathrm{m}}\frac{{\mathrm{d}}^2\mathrm{\psi }\left(\mathrm{x}\right)}{{\mathrm{dx}}^2}-\mathrm{V\psi }\left(\mathrm{x}\right)=-\mathrm{E\psi }\left(\mathrm{x}\right)$

Therefore,

$-\frac{{\mathrm{ħ}}^2}{2\mathrm{m}}\frac{{\mathrm{d}}^2\mathrm{\psi }\left(\mathrm{x}\right)}{{\mathrm{dx}}^2}+\mathrm{V\psi }\left(\mathrm{x}\right)=-\mathrm{E\psi }\left(\mathrm{x}\right)$

For an even time-independent wave-function$\psi \left(-x\right)=\psi \left(x\right)$, so

$$\begin{gathered} -\frac{{\mathrm{ħ}}^2}{2\mathrm{m}}\frac{{\mathrm{d}}^2\mathrm{\psi }\left(\mathrm{x}\right)}{{\mathrm{dx}}^2}+\mathrm{V\psi }\left(-\mathrm{x}\right)=\mathrm{E\psi }\left(-\mathrm{x}\right) \\ -\frac{{\mathrm{ħ}}^2}{2\mathrm{m}}\frac{{\mathrm{d}}^2\mathrm{\psi }\left(\mathrm{x}\right)}{{\mathrm{dx}}^2}+\mathrm{V\psi }\left(\mathrm{x}\right)=-\mathrm{E\psi }\left(\mathrm{x}\right) \end{gathered}$$

As we have see, both (the even and odd wave-function) satisfy eq.(2.5), so any

linear combination must satisfy this equation too, where${\psi }_{even}\left(x\right)=\frac{1}{2}\left[\psi \left(x\right)+\psi \left(-x\right)\right]$, and ${\psi }_{odd}\left(x\right)=\frac{1}{2}\left[\psi \left(x\right)-\psi \left(-x\right)\right]$.

The general solution can then be built from a linear combination of even and odd functions.