Question

The general form for symmetric and antisymmetric wave functions is${\mathbf{\psi }}_{\mathbf{n}}\mathbf{\left(}{\mathbf{x}}_{\mathbf{2}}\mathbf{\right)}{\mathbf{\psi }}_{\mathbf{n}}\mathbf{\left(}{\mathbf{x}}_{\mathbf{2}}\mathbf{\right)}\mathbf{\pm }{\mathbf{\psi }}_{\mathbf{n}}\mathbf{\cdot }\mathbf{\left(}{\mathbf{x}}_{\mathbf{1}}\mathbf{\right)}{\mathbf{\psi }}_{\mathbf{n}}\mathbf{\left(}{\mathbf{x}}_{\mathbf{2}}\mathbf{\right)}$ but it is not normalized.

(a) In applying quantum mechanics, we usually deal with quantum states that are "orthonormal." That is, if we integrate over all space the square of any individual-particle function, such as${\mathbf{\psi }}_{\mathbf{n}}\mathbf{\left(}{\mathbf{x}}_{\mathbf{2}}\mathbf{\right)}{\mathbf{\psi }}_{\mathbf{n}}\mathbf{\left(}{\mathbf{x}}_{\mathbf{2}}\mathbf{\right)}\mathbf{\pm }{\mathbf{\psi }}_{\mathbf{n}}\mathbf{\cdot }\mathbf{\left(}{\mathbf{x}}_{\mathbf{1}}\mathbf{\right)}{\mathbf{\psi }}_{\mathbf{n}}\mathbf{\left(}{\mathbf{x}}_{\mathbf{2}}\mathbf{\right)}$, we get 1, but for the product of different individual-particle functions, such as${\mathbf{\psi }}_{\mathbf{n}}^{\mathbf{\phi }}\mathbf{\left(}\mathbf{x}\mathbf{\right)}{\mathbf{\psi }}_{\mathbf{n}}\mathbf{,}\mathbf{\left(}\mathbf{x}\mathbf{\right)}$, we get 0. This happens to be true for all the systems in which we have obtained or tabulated sets of wave functions (e.g., the particle in a box, the harmonic oscillator, and the hydrogen atom). Assuming that this holds, what multiplicative constant would normalize the symmetric and antisymmetric functions?

(b) What value$\mathit{A}$gives the vector$\mathbf{V}\mathbf{=}\mathbf{A}\mathbf{(}\widehat{\mathbf{x}}\mathbf{\pm }\widehat{\mathbf{y}}\mathbf{)}$unit length?

(c) Discuss the relationship between your answers in (a) and (b)?

Short answer

(a) The multiplicative constant that would normalize symmetric and anti-symmetric function is$\frac{1}{\sqrt{2}}$

(b) The value of A that will give the vector unit length is$\frac{1}{\sqrt{2}}$

(c) The solutions in parts () and would be almost identical.

Step-by-step solution

Given information:

(a) General form of symmetric and anti-symmetric wave function${\mathrm{\psi }}_{\mathrm{n}}\left({\mathrm{x}}_1\right){\mathrm{\psi }}_{{\mathrm{n}}^{\text{'}}}\left({\mathrm{x}}_2\right)\pm {\mathrm{\psi }}_{{\mathrm{n}}^{\text{'}}}\left({\mathrm{x}}_1\right){\mathrm{\psi }}_{\mathrm{n}}\left({\mathrm{x}}_2\right)$

(b)$\mathrm{V}=\mathrm{A}(\widehat{\mathrm{x}}\pm \widehat{\mathrm{y}})$

Concept of multiplicative constant

$\mathbf{\psi }\mathbf{(}{\mathbf{x}}_{\mathbf{1}}\mathbf{,}{\mathbf{x}}_{\mathbf{2}}\mathbf{)}\mathbf{=}\mathbf{A}\mathbf{[}{\mathbf{\psi }}_{\mathbf{n}}\mathbf{\left(}{\mathbf{x}}_{\mathbf{1}}\mathbf{\right)}{\mathbf{\psi }}_{{\mathbf{n}}^{\mathbf{\text{'}}}}\mathbf{\left(}{\mathbf{x}}_{\mathbf{2}}\mathbf{\right)}\mathbf{\pm }{\mathbf{\psi }}_{{\mathbf{n}}^{\mathbf{\text{'}}}}\mathbf{\left(}{\mathbf{x}}_{\mathbf{1}}\mathbf{\right)}{\mathbf{\psi }}_{\mathbf{n}}\mathbf{\left(}{\mathbf{x}}_{\mathbf{2}}\mathbf{\right)}\mathbf{]}$

Determine the probability particle

To do so, we need the probability of the particle to be found to be 1 , therefore

$\iint {\left|\mathrm{\psi }({\mathrm{x}}_1,{\mathrm{x}}_2)\right|}^2{\mathrm{dx}}_1{\mathrm{dx}}_2=1$

Hence, probability of particle to be found will now be

${\mathrm{A}}^2\iint \Vert [{\mathrm{\psi }}_{\mathrm{n}}\left({\mathrm{x}}_1\right){\mathrm{\psi }}_{{\mathrm{n}}^{\text{'}}}\left({\mathrm{x}}_2\right)\pm {\mathrm{\psi }}_{{\mathrm{n}}^{\text{'}}}\left({\mathrm{x}}_1\right){\mathrm{\psi }}_{\mathrm{n}}\left({\mathrm{x}}_2\right){\Vert }^2{\mathrm{dx}}_1{\mathrm{dx}}_2=1$

Expand the equation

${\mathrm{A}}^2\left(\int {\left|{\mathrm{\psi }}_{\mathrm{n}}\left({\mathrm{x}}_1\right)\right|}^2{\mathrm{dx}}_1)(\int {\left|{\mathrm{\psi }}_{{\mathrm{n}}^{\text{'}}}\left({\mathrm{x}}_2\right)\right|}^2{\mathrm{dx}}_2\right)$

$\pm {\mathrm{A}}^2\left[\begin{array}{l}(\int {\mathrm{\psi }}_{\mathrm{n}}{\left({\mathrm{x}}_1\right)}^{*}{\mathrm{\psi }}_{{\mathrm{n}}^{\text{'}}}\left({\mathrm{x}}_1\right){\mathrm{dx}}_1)(\int {\mathrm{\psi }}_{\mathrm{n}}{\left({\mathrm{x}}_2\right)}^{*}{\mathrm{\psi }}_{{\mathrm{n}}^{\text{'}}}\left({\mathrm{x}}_2\right){\mathrm{dx}}_1)\\ +(\int {\mathrm{\psi }}_{\mathrm{n}}\left({\mathrm{x}}_1\right){\mathrm{\psi }}_{{\mathrm{n}}^{\text{'}}}^{*}\left({\mathrm{x}}_1\right){\mathrm{dx}}_1)(\int {\mathrm{\psi }}_{\mathrm{n}}\left({\mathrm{x}}_2\right){\mathrm{\psi }}_{{\mathrm{n}}^{\text{'}}}\left({\mathrm{x}}_2\right){\mathrm{dx}}_1)\end{array}\right]$$+{\mathrm{A}}^2\left(\int {\left|{\mathrm{\psi }}_{{\mathrm{n}}^{\text{'}}}\left({\mathrm{x}}_1\right)\right|}^2{\mathrm{dx}}_1)(\int {\left|{\mathrm{\psi }}_{\mathrm{n}}\left({\mathrm{x}}_2\right)\right|}^2{\mathrm{dx}}_2\right)=1$

Use the following conditions in above equation

$\int {\mathrm{\psi }}_{\mathrm{n}}{\left(\mathrm{x}\right)}^{*}{\mathrm{\psi }}_{{\mathrm{n}}^{\text{'}}}\left(\mathrm{x}\right)\mathrm{dx}=\left\{\begin{array}{l}0\text{if}\mathrm{n}\ne {\mathrm{n}}^{\text{'}}\\ 1\text{if}\mathrm{n}={\mathrm{n}}^{\text{'}}\end{array}\right.$

The equation simplifies to

${\mathrm{A}}^2\left(1\right)\pm {\mathrm{A}}^2\left(0\right)+{\mathrm{A}}^2\left(1\right)=12{\mathrm{A}}^2=1\mathrm{A}=\frac{1}{\sqrt{2}}$

Determine the unit length of vector

(b) For the vector $\mathrm{V}=\mathrm{A}(\widehat{\mathrm{x}}\pm \widehat{\mathrm{y}})$we have the value of A as

$\mathrm{V}=\mathrm{A}(\widehat{\mathrm{x}}\pm \widehat{\mathrm{y}})\mathrm{V}\cdot \mathrm{V}=\mathrm{A}(\widehat{\mathrm{x}}\pm \widehat{\mathrm{y}})\cdot \mathrm{A}(\widehat{\mathrm{x}}\pm \widehat{\mathrm{y}})\mathrm{V}\cdot \mathrm{V}={\mathrm{A}}^2(\widehat{\mathrm{x}}\cdot \widehat{\mathrm{x}}\pm 2\widehat{\mathrm{x}}\cdot \widehat{\mathrm{y}}+\widehat{\mathrm{y}}\cdot \widehat{\mathrm{y}})={\mathrm{A}}^2(1\pm 0+1)=2{\mathrm{A}}^2$.

For unit length of the vector we have

$\begin{array}{c}\mathrm{V}\cdot \mathrm{V}=2{\mathrm{A}}^2\\ 1=2{\mathrm{A}}^2\\ \mathrm{A}=\frac{1}{\sqrt{2}}\end{array}$

Determine the relation between two answers

(c) The calculation for the two one-particle wave functions in part (a) proceeds similarly to the corresponding calculation in (b) for perpendicular vectors${\overline{\mathrm{v}}}_{\mathrm{n}}$ and${\overline{\mathrm{v}}}_{{\mathrm{n}}^{\text{'}}}$ with the dot product, or inner product, of the vectors corresponding to an integration of the product of a wave function times its complex conjugate${\overline{\mathrm{v}}}_{{\mathrm{n}}^{\text{'}}}\leftrightarrow {\mathrm{\psi }}_{{\mathrm{n}}^{\text{'}}}{\overline{\mathrm{v}}}_{\mathrm{n}}\cdot {\overline{\mathrm{v}}}_{{\mathrm{n}}^{\text{'}}}\leftrightarrow \int {\mathrm{\psi }}_{\mathrm{n}}\left(\mathrm{x}\right){\mathrm{\psi }}_{{\mathrm{n}}^{\text{'}}}\left(\mathrm{x}\right)\mathrm{dx}$

Then by taking ${\overline{\mathrm{v}}}_{\mathrm{n}}=\widehat{\mathrm{x}}$ and ${\overline{\mathrm{v}}}_{{\mathrm{n}}^{\text{'}}}=\widehat{\mathrm{y}}$ the solutions in parts $\left(a\right)$and$\left(b\right)$ would be almost identical.