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Finding the quantized energies of a particle

  • #1
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0

Homework Statement



Okay, so the question i'm trying to solve is to find the quantized energies for a particle in the potential:

$$V(x)=V_0 \left ( \frac{b}{x}-\frac{x}{b} \right )^2$$

for some constant b.

The Attempt at a Solution


I am following along with the derivation of the quantized energies for the SHO in griffiths quantum(2nd ed) from pages 51-56. I used the schrodinger equation with the substitution $$\psi=\frac{\tilde \psi}{\sqrt{x}}$$ as well as $\tilde x=bx$(to make it unitless) and substituted it into the schrodinger equation and obtained after much simplification
$$ \tilde \psi''- \tilde \psi' x^{-1}+ \tilde \psi (\frac{3/4+V_0}{x^2}+V_0x^2-(E+2V_0)=0 $$

Then, I followed along with the derivation of Griffiths, by finding an integrating factor and trying to expand $\psi$ as a power series which led me to the following
$$\psi(\tilde x)=f(x)e^-\frac{\tilde x^2}{2} \quad \text{Eq 2.77 in Griffiths 2nd ed}$$
$$f(x)=\sum_{n=0}^{\infty}C_nx^n$$
$$f'(x)=\sum_{n=0}^{\infty}nC_nx^{n-1}$$
$$f''(x)=\sum_{n=0}^{\infty}n(n-1)C_nx^{n-2}$$

This of course leads to(dropping the tilde's on the x for simplicity) $$\tilde \psi= f( x)e^{- x^2/2}$$
$$\tilde \psi '=e^{- x^2/2} (f'(x)-xf(x)) $$
$$\tilde \psi ''=e^{-x^2/2}(f''(x)-2xf'(x)+f(x)(x^2-1))$$

I believe these are the correct $\psi$'s however my problem now is when I sub them back into the schrodinger equation I found above, I dont get the same cancellation that happens for the SHO, so I am struggling to find the recurrence relations in order to find the quantized energies. If anyone can help me find the energies I would be much appreciative.
 

Answers and Replies

  • #2
264
70
In your differential equation you have a term proportional to ##1/x^2##. Your solution must be regular at the origin.
Therefore, you need to use an expansion of the form
$$f(x)=x^2\sum_{n=0}^{\infty}C_n x^n$$.

Homework Statement



Okay, so the question i'm trying to solve is to find the quantized energies for a particle in the potential:

$$V(x)=V_0 \left ( \frac{b}{x}-\frac{x}{b} \right )^2$$

for some constant b.

The Attempt at a Solution


I am following along with the derivation of the quantized energies for the SHO in griffiths quantum(2nd ed) from pages 51-56. I used the schrodinger equation with the substitution $$\psi=\frac{\tilde \psi}{\sqrt{x}}$$ as well as $\tilde x=bx$(to make it unitless) and substituted it into the schrodinger equation and obtained after much simplification
$$ \tilde \psi''- \tilde \psi' x^{-1}+ \tilde \psi (\frac{3/4+V_0}{x^2}+V_0x^2-(E+2V_0)=0 $$

Then, I followed along with the derivation of Griffiths, by finding an integrating factor and trying to expand $\psi$ as a power series which led me to the following
$$\psi(\tilde x)=f(x)e^-\frac{\tilde x^2}{2} \quad \text{Eq 2.77 in Griffiths 2nd ed}$$
$$f(x)=\sum_{n=0}^{\infty}C_nx^n$$
$$f'(x)=\sum_{n=0}^{\infty}nC_nx^{n-1}$$
$$f''(x)=\sum_{n=0}^{\infty}n(n-1)C_nx^{n-2}$$

This of course leads to(dropping the tilde's on the x for simplicity) $$\tilde \psi= f( x)e^{- x^2/2}$$
$$\tilde \psi '=e^{- x^2/2} (f'(x)-xf(x)) $$
$$\tilde \psi ''=e^{-x^2/2}(f''(x)-2xf'(x)+f(x)(x^2-1))$$

I believe these are the correct $\psi$'s however my problem now is when I sub them back into the schrodinger equation I found above, I dont get the same cancellation that happens for the SHO, so I am struggling to find the recurrence relations in order to find the quantized energies. If anyone can help me find the energies I would be much appreciative.
 

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