Question

Show that the energy of a simple harmonic oscillator in the n = 2 state is 5ℏω/2 by substituting the wave functionψ₂ = A(2αx²- 1)e^-αx2/2 directly into the Schroedinger equation, as broken down in the following steps.
First, calculate dψ₂/dx, using A, x, and α.
dψ₂/dx = ..........................

Second, calculate d²ψ₂/dx², using A, x, and α.
d²ψ₂/dx² = .........................

Third, calculate α²x²ψ₂ - d²ψ₂/dx², using A, x, and α.
α²x²ψ₂ - d²ψ₂/dx² = .......................

Fourth, calculate (α²x²ψ₂ - d²ψ₂/dx²)/ψ₂, using A, x, and α.
(α²x²ψ₂ - d²ψ₂/dx²)/ψ₂ = ...........................

Finally, calculate E ≡ [(α²x²ψ₂ - d²ψ₂/dx²)/ψ₂]ℏ²/(2m), using A, x, α, m, ω, and ℏ.
E ≡ .............................

please answer the question as requested  

Answer 1

The given wave function is
   $\psi_2(x)=A(2a x^2 -1)e^{-\frac{ax^2}{2}}$
where,
    $a=\frac{m\omega}{\hbar}$
And so, we have

Step 1 :

   $\frac{\mathrm{d} \psi_2}{\mathrm{d} x}=\frac{\mathrm{d} }{\mathrm{d} x}\left [A(2a x^2 -1)e^{-\frac{ax^2}{2}} \right ]$
   $\Rightarrow \frac{\mathrm{d} \psi_2}{\mathrm{d} x}=A\left [4ax-ax(2a x^2 -1) \right ]e^{-\frac{ax^2}{2}}$
   $\Rightarrow \frac{\mathrm{d} \psi_2}{\mathrm{d} x}=A\left [5ax-2a^2 x^3 \right ]e^{-\frac{ax^2}{2}}$
Step 2 :
    $\Rightarrow \frac{\mathrm{d}^2 \psi_2}{\mathrm{d} x^2}=\frac{\mathrm{d} }{\mathrm{d} x}\left [A\left (5ax-2a^2 x^3 \right )e^{-\frac{ax^2}{2}} \right ]$
   $\Rightarrow \frac{\mathrm{d}^2 \psi_2}{\mathrm{d} x^2}=A\left [(5a-6a^2x^2)-ax\left (5ax-2a^2 x^3 \right ) \right ]e^{-\frac{ax^2}{2}}$
    $\Rightarrow \frac{\mathrm{d}^2 \psi_2}{\mathrm{d} x^2}=A\left [5a-6a^2x^2-5a^2x^2+2a^3 x^4 \right ]e^{-\frac{ax^2}{2}}$
    $\Rightarrow \frac{\mathrm{d}^2 \psi_2}{\mathrm{d} x^2}=A\left [5a-11a^2x^2+2a^3 x^4 \right ]e^{-\frac{ax^2}{2}}$

Step 3 :
    $a^2x^2 \psi_2 - \frac{\mathrm{d}^2 \psi_2}{\mathrm{d} x^2}=a^2x^2A(2ax^2-1)e^{-\frac{ax^2}{2}}-A\left [5a-11a^2x^2+2a^3 x^4 \right ]e^{-\frac{ax^2}{2}}$
   $\Rightarrow a^2x^2 \psi_2 - \frac{\mathrm{d}^2 \psi_2}{\mathrm{d} x^2}=A\left [a^2x^2(2ax^2-1)-5a+11a^2x^2-2a^3 x^4 \right ]e^{-\frac{ax^2}{2}}$
    $\Rightarrow a^2x^2 \psi_2 - \frac{\mathrm{d}^2 \psi_2}{\mathrm{d} x^2}=A\left [10a^2x^2-5a \right ]e^{-\frac{ax^2}{2}}$
   $\Rightarrow a^2x^2 \psi_2 - \frac{\mathrm{d}^2 \psi_2}{\mathrm{d} x^2}=5a \left (A\left [2ax^2-a \right ]e^{-\frac{ax^2}{2}} \right )$
Step 4 :
    $\frac{a^2x^2 \psi_2 - \frac{\mathrm{d}^2 \psi_2}{\mathrm{d} x^2}}{\psi_2}=5a \frac{\left (A\left [2ax^2-a \right ]e^{-\frac{ax^2}{2}} \right )}{A\left [2ax^2-a \right ]e^{-\frac{ax^2}{2}} }$
   $\Rightarrow \frac{a^2x^2 \psi_2 - \frac{\mathrm{d}^2 \psi_2}{\mathrm{d} x^2}}{\psi_2}=5a$
Step 5 :
    $E=\left ( \frac{a^2x^2 \psi_2 - \frac{\mathrm{d}^2 \psi_2}{\mathrm{d} x^2}}{\psi_2} \right )\frac{\hbar^2}{2m}=5a \frac{\hbar^2}{2m}$
   $\Rightarrow E= 5\frac{\hbar^2}{2m}a$
Now we use the definition of the parameter a
   $a=\frac{m\omega}{\hbar}$
And so,
    $\Rightarrow E=5 \frac{\hbar^2}{2m} \times \frac{m\omega}{\hbar}$
    $\Rightarrow E=\frac{5}{2} \hbar \omega$