Question

the energy of a particle of mass m bound by a special kind of spring is E=p^2/2m +kx^4 where k is a positive constant .use the heisenbergs uncertainty principle to calculate the minimum possible energy of the particle.

Answer 1

Answer on Question #73944, Physics / Other

the energy of a particle of mass $m$ bound by a special kind of spring is $E = p^2 / 2m + kx^4$ where $k$ is a positive constant. Use the heisenbergs uncertainty principle to calculate the minimum possible energy of the particle.

Solution:

From uncertainty principle let

$$p \times \frac{\hbar}{2}$$

So,

$$p = \frac{\hbar}{2x}$$

Then

$$E = \frac{\hbar^2}{8mx^2} + \frac{kx^2}{2}$$

Then we differentiate to find location of minimum $E$

$$\frac{dE}{dx} = -\frac{2\hbar^2}{8mx^3} + kx = 0$$ $$x = \left[ \frac{\hbar^2}{4mk} \right]^{\frac{1}{4}}$$

Substituting this into the energy equation to find minimum energy $E_0$

$$E_0 = \frac{\hbar^2}{8m\sqrt{\frac{\hbar^2}{4mk}}} + \frac{k\sqrt{\frac{\hbar^2}{4mk}}}{2} = \left(\frac{\hbar}{4}\right)\sqrt{\frac{k}{m}} + \left(\frac{\hbar}{4}\right)\sqrt{\frac{k}{m}} = \frac{\hbar}{2}\sqrt{\frac{k}{m}}$$

Answer: $\frac{\hbar}{2}\sqrt{\frac{k}{m}}$

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