Physics Asked by Nakshatra Gangopadhay on June 29, 2021
Let us assume, we are given the following potential,
$$V(x)=frac{1}{2}ax^2-2x+epsilon x^3$$
We need to find the energy levels of a particle bound in this potential
Let us think of the ground level for now. As we can see, the first term in the potential resembles a harmonic oscillator, so I can possibly use perturbation theory on the second and the third terms to come up with an approximate value of the ground level energy of this particle.
We’d get a ground state energy of the form,
$$ E_0 = frac{1}{2}hbar sqrt{frac{a}{m}} + E_x^1 + E_x^3$$
The two terms, in the end, come from the *perturbations due to the linear and cubic terms.
However, I’ve also recently read about small oscillations.
We are usually given an arbitrary potential, and we approximate it into an oscillator potential about the local minima, using
$$
begin{align}
left.frac{dV(x)}{dx}right|_{x_{0}} &= 0 [5pt]
left.frac{d^2V(x)}{dx^2}right|_{x_{0}} &= K [5pt]
omega = sqrt{frac{K}{m}}
end{align}
$$
This is how we obtain the value of angular frequency for a random potential, in small oscillations.
Now, the potential in our given problem can also be approximated like in small oscillations. We can obtain the minima at around $x=frac{sqrt{a^2+24 epsilon }-a}{6 epsilon }approx 2/a -12epsilon/a$, and by differentiating the potential twice, we get :
$$K=a+frac{12epsilon}{a}.$$
This equates to an angular frequency
$$omega = sqrt{frac{frac{a^2+12epsilon}{a}}{m}}$$
Using this frequency, we can also calculate the ground state energy level to be :
$$ E_0 = frac{1}{2}hbar sqrt{frac{frac{a^2+12epsilon}{a}}{m}}$$
Thus using two different methods, perturbation theory, and small oscillations, I’ve obtained two approximations of the ground state energy level.
Which of the above methods is wrong, and why can I not use that ?
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