The wave functions for the ground and first excited states of a simple harmonic oscillator are ${\mathbf{Ae}}^{\mathbf{-}{\mathbf{bx}}^{\mathbf{2}}\mathbf{/}\mathbf{2}}$and${\mathbf{Bxe}}^{\mathbf{-}{\mathbf{bx}}^{\mathbf{2}}\mathbf{/}\mathbf{2}}$. Suppose you have two particles occupying these two states.

(a) If distinguishable, an acceptable wave function would berole="math" localid="1659955524302" ${\mathbf{Ae}}^{\mathbf{-}{\mathbf{bx}}_{\mathbf{1}}^{\mathbf{2}}\mathbf{/}\mathbf{2}}{\mathbf{Bx}}_{\mathbf{2}}{\mathbf{e}}^{\mathbf{-}{\mathbf{bx}}_{\mathbf{2}}^{\mathbf{2}}\mathbf{/}\mathbf{2}}$. Calculate the probability that both particles would be on the positive side of the origin and divide by the total probability for both being found over all values of${\mathbf{x}}_{\mathbf{1}}$,${\mathbf{x}}_{\mathbf{2}}$. (This kind of normalizing-as-we-go will streamline things.)

(b) Suppose now that the particles are indistinguishable. Using the$\mathbf{\pm }$symbol to reduce your work. calculate the same probability ratio, but assuming that their multiparticle wave function is either symmetric or antisymmetric. Comment on your results.

Step by step solution

01

Given information:

The ground state of simple harmonic oscillator is ${\mathrm{Ae}}^{-\frac{{\mathrm{bx}}^2}{2}}$.

The first excited state of simple harmonic oscillator is ${\mathrm{Bxe}}^{-\frac{{\mathrm{bx}}^2}{2}}$.

02

Concept of probability ratio

(a) The expression for probability ratio is given by,

$\mathbf{R}\mathbf{=}\frac{\left(A^2{\displaystyle {\int }_0^{\infty }{e}^{-{\mathrm{bx}}_1^2}}{\mathrm{dx}}_1\right)\left(B^2{x}_2^2{\displaystyle {\int }_0^{\infty }{e}^{-{\mathrm{bx}}_2^2{\mathrm{dx}}_2}}\right)}{\left(\left(A^2{\displaystyle {\int }_{-=}^{\infty }{e}^{-{\mathrm{tx}}_1^2}}{\mathrm{dx}}_1\right)\right)\mathbf{}\left(B^2{x}_2^2{\displaystyle {\int }_{-\infty }^{\infty }{e}^{-{\mathrm{bx}}_2^2{\mathrm{dx}}_2}}\right)}$

03

Evaluate probability ratio

(a) The probability ratio is calculated as,

$\mathrm{R}=\frac{\left({\mathrm{A}}^2{\displaystyle {\int }_0^{\mathrm{z}}{\mathrm{e}}^{-{\mathrm{bx}}_1^2{\mathrm{dx}}_1}}\right)\left({\mathrm{B}}^2{\mathrm{x}}_2^2{\displaystyle {\int }_0^{\mathrm{z}}{\mathrm{e}}^{-{\mathrm{bx}}_2^2{\mathrm{dx}}_2}}\right)}{\left({\mathrm{A}}^2{\displaystyle {\int }_{-\mathrm{\infty }}^{\mathrm{z}}{\mathrm{e}}^{-{\mathrm{bx}}_1^2{\mathrm{dx}}_1}}\right)\left({\mathrm{B}}^2{\mathrm{x}}_2^2{\displaystyle {\int }_{-\mathrm{\infty }}^{\mathrm{w}}{\mathrm{e}}^{-{\mathrm{bx}}_2^2{\mathrm{dx}}_2}}\right)}$

$\mathrm{R}=\frac{\left({\mathrm{A}}^2{\displaystyle {\int }_0^{\mathrm{w}}{\mathrm{e}}^{-{\mathrm{bx}}_1^2{\mathrm{dx}}_1}}\right)\left({\mathrm{B}}^2{\mathrm{x}}_2^2{\displaystyle {\int }_0^{\mathrm{w}}{\mathrm{e}}^{-{\mathrm{bx}}_2^2{\mathrm{dx}}_2}}\right)}{\left(2{\mathrm{A}}^2{\displaystyle {\int }_0^{\mathrm{m}}{\mathrm{e}}^{-{\mathrm{bx}}_1^2{\mathrm{dx}}_1}}\right)\left(2{\mathrm{Bx}}_2^2{\displaystyle {\int }_0^{\mathrm{w}}{\mathrm{e}}^{-{\mathrm{bx}}_2^2{\mathrm{dx}}_2}}\right)}$

$\mathrm{R}=\frac{1}{4}=0.25$

04

Evaluate probability ratio

(b)

The probability ratio is calculated as,

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