Question

When applying quantum mechanics, we often concentrate on states that qualify as “orthonormal”, The main point is this. If we evaluate a probability integral over all space of ${{\mathbf{\varphi }}_{\mathbf{1}}}^{\mathbf{*}}{\mathit{\varphi }}_{\mathbf{1}}$or of ${{\mathbf{\varphi }}_{\mathbf{2}}}^{\mathbf{*}}{\mathit{\varphi }}_{\mathbf{2}}$, we get 1 (unsurprisingly), but if we evaluate such an integral for${{\mathbf{\varphi }}_{\mathbf{1}}}^{\mathbf{*}}{\mathit{\varphi }}_{\mathbf{2}}\mathbf{}\mathit{o}\mathit{r}\mathbf{}\mathbf{}{{\mathbf{\varphi }}_{\mathbf{2}}}^{\mathbf{*}}{\mathit{\varphi }}_{\mathbf{1}}$ we get 0. This happens to be true for all systems where we have tabulated or actually derived sets of wave functions (e.g., the particle in a box, the harmonic oscillator, and the hydrogen atom). By integrating overall space, show that expression (7-44) is not normalized unless a factor of $\mathbf{1}\mathbf{/}\mathbf{2}$is included with the probability.

Short answer

By integrating overall space, you get that, a factor of ½ will be required to be included with the probability.

Step-by-step solution

Used formula:

The wave function transitional state can be represented by the following equation:

$$\begin{gathered} {\mathbf{\Psi }}_{\mathbf{t}\mathbf{r}\mathbf{a}\mathbf{n}\mathbf{s}\mathbf{i}\mathbf{t}\mathbf{i}\mathbf{o}\mathbf{n}}\left(r,t\right)\mathbf{=}{\mathbf{\Psi }}_{\mathbf{i}}\left(r,t\right)\mathbf{+}{\mathbf{\Psi }}_{\mathbf{f}}\left(r,t\right) \\ {\mathbf{\Psi }}_{\mathbf{transition}}\left(r,t\right)\mathbf{=}{\mathbf{\Psi }}_{\mathbf{i}}\mathbf{\left(}\mathbf{r}\mathbf{\right)}{\mathbf{e}}^{\mathbf{-}{\mathbf{iE}}_{\mathbf{i}}\mathbf{t}\mathbf{/}\mathbf{h}}\mathbf{+}{\mathbf{\Psi }}_{\mathbf{f}}\mathbf{\left(}\mathbf{r}\mathbf{\right)}{\mathbf{e}}^{\mathbf{-}{\mathbf{iE}}_{\mathbf{f}}\mathbf{t}\mathbf{/}\mathbf{h}}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{\dots }\mathbf{.}\mathbf{.}\mathbf{}\mathbf{(}\mathbf{7}\mathbf{-}\mathbf{44}\mathbf{)} \end{gathered}$$

Where, $\mathbf{\Psi }$ isWave function, ${\mathbf{\Psi }}_i$ is the Wave function of initial state, ${\mathbf{\Psi }}_f$ is the Wave function of final state, r is the radius, t is time, ${\mathit{E}}_{\mathbf{i}}$ is the initial energy, ${\mathit{E}}_{\mathbf{f}}$ is final energy, h is Planck’s constant.

Integrating overall space:

Consider the given data as below.

$\int \Psi .\Psi \hspace{0.17em}dV=1$

Now, if you introduce a factor A which can be adjusted to give probability equal to 1.

$$\begin{gathered} \int \left(\mathbf{A}.{{\mathbf{\Psi }}_i}^{*}\left(\mathbf{r}\right)e^{+i{E}_{i}t/h}+\mathbf{A}.{{\mathbf{\Psi }}_f}^{*}\left(r\right)e^{+e^{+i{E}_{i}t/h}}\right)\left(\mathbf{A}.{\mathbf{\Psi }}_i\left(\mathbf{r}\right)e^{-i{E}_{i}t/h}+\mathbf{A}.{\mathbf{\Psi }}_f\left(r\right)e^{+e^{-i{E}_{i}t/h}}\right)\hspace{0.17em}dV=1 \\ {\left|A\right|}^2\left[{\int \left|{\mathbf{\Psi }}_i\left(\mathbf{r}\right)\right|}^{2}dV+e^{+i(E_i+E_f)t/h}\int {{\mathbf{\Psi }}_i}^{*}\left(\mathbf{r}\right){\mathbf{\Psi }}_f\left(r\right)dV+e^{-i(E_i-E_f)t/h}\int {{\mathbf{\Psi }}_f}^{*}\left(\mathbf{r}\right){\mathbf{\Psi }}_i\left(r\right)dV+{\int \left|{\mathbf{\Psi }}_f\left(\mathbf{r}\right)\right|}^{2}dV\right]=1 \\ {\left|A\right|}^2\left(1+0+0+1\right)=1 \end{gathered}$$

$A=\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$\sqrt{2}$}\right.$

Hence, the factor$\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$\sqrt{2}$}\right.$must be included with each wave function.

As you know that probability density is the square of wave function.

Hence, a factor of $\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.$ will be required to be included with the probability.