Question

The energy of a particle of massbound by an unusual spring is${\beta }^{-}/2m+b{x}^4$.

(a) Classically. it can have zero energy. Quantum mechanically, however, though both$x$ and$p$are "on average" zero, its energy cannot be zero. Why?

(b) Roughly speaking.$\Delta x$is a typical value of the particle's position. Making a reasonable assumption about a typical value of its momentum, find the particle's minimum possible energy.

Short answer

(a). The reasonthe energy of the particle cannot be zero is by Heisenberg's law, the uncertainty in one of them might be low while in the corresponding value be higher that means$x$and$p$cannot be zero at the same time.

(b). The reasonable assumption about the typical value of momentum is that it$p\approx \frac{h}{x}$ such that energy of spring is obtained in terms of its displacement and the minimum possible energy of the particle is $\frac{3}{2}\sqrt[3]{\frac{b{h}^4}{2m^2}}$.

Step-by-step solution

The formula for the expression for the energy of the particle-bound by an unusual spring.

The expression for the energy of the particle-bound by an unusual spring is given by,

$E=\frac{p^2}{2m}+b{x}^4$

The expression for uncertainty in momentum and position of the particle is given by,

$\begin{array}{l}\Delta p_{\text{x}}\Delta x\approx \frac{h}{k}\\ k\Delta p_{\text{x}}\approx \frac{h}{\Delta x}\end{array}$

$\Delta p_{\text{x}}$is the uncertainty of momentum, $\Delta x$is the uncertainty of position ,$h$ is the Planck’s constant and$k$ is constant.

The expression for the energy of the particle bound by an unusual spring is given by,

$\begin{array}{l}E=\frac{p^2}{2m}+b{x}^4\\ E=\frac{{\left(\hslash lx\right)}^2}{2m}+b{x}^4\end{array}$

Use the formula E=p22m+bx4  for explanation.

(a)

Well, zero energy means setting both $x$and $p$to zero (simultaneously), which requires that their uncertainties to be zero as well (we will prove this in part b), consequently violating the uncertainty relation. Hence, we can't set the energy of the harmonic oscillator (HO) equal to zero according to quantum mechanics and the minimum energy isn't zero it's $\frac{\hslash \omega }{2}$(review problem 52).

Use the formula for the expression of uncertainty in momentum and position and the energy of the particle-bound.

b)

Starting from the classical definition of the classical $\text{HO}$,

$E=\frac{p^2}{2m}+b{x}^4$

Then, to account for the wave-nature of the particles, we are required to replace the typical values of the position$x$and momentum $p$, with the ones given by the uncertainty relation. This connection is derived through the basic definition of uncertainty as the square root of the difference between the average of$x^2$and that of the mean squared${\left(\overline{x}\right)}^2$

$\because {\left(\Delta x\right)}^2=\overline{x^2}-{\left(\overline{x}\right)}^2$

$\because \overline{x}\approx 0$This is a valid assumption in the case of a HO

$\therefore \Delta x\approx \sqrt{{\overline{x}}^2}=x$

The same goes for the momentum uncertainty $\Delta p\approx p$, where we have replaced the typical values of $x$and $p$ with their corresponding momentum and position uncertainties $\Delta x$and $\Delta p$, using the assumption that the average value of the position and momentum in $\text{HO}$is approximately equal to zero (which is reasonable if you consider the symmetry in its motion). Now, let's rewrite the energy relation in part a) and apply the uncertainty relation to make the energy a function in one unknown only.

$\begin{array}{l}\therefore E=\frac{{\left(\Delta p\right)}^2}{2m}+b{\left(\Delta x\right)}^4\\ \because \Delta p\Delta x=\frac{\hslash }{2}\Rightarrow \Delta p=\frac{\hslash }{2\Delta x}\\ \therefore E=\frac{{\hslash }^2}{\mathbf{8}\mathbf{m}{\left(\mathbf{\Delta x}\right)}^2}+\mathbf{b}{\left(\mathrm{\Delta x}\right)}^{\mathbf{4}}\end{array}$

Using the final expression for the energy, we could solve for the minimum value of$E$by taking the derivative with respect to$\Delta x$and equating this expression to zero.

Therefore,

role="math" localid="1659939191963" $\begin{array}{c}\frac{dE}{d\left(\Delta x\right)}=\frac{-2{\hslash }^2}{8m{\left(\Delta x\right)}^3}+4b{\left(\Delta x\right)}^3=0\\ \Delta x=\sqrt[6]{\frac{{\hslash }^2}{16mb}}\\ E_{\min }=\frac{{\hslash }^2}{8m{\left(\sqrt[6]{\frac{{\hslash }^2}{16mb}}\right)}^2}+b{\left(\sqrt[6]{\frac{{\hslash }^2}{16mb}}\right)}^4\\ E_{\min }=\frac{b^{\frac{1}{3}}\times {\hslash }^{\frac{4}{3}}}{2^{\frac{5}{3}}{m}^{\frac{2}{3}}}+b^{\frac{1}{3}}{\left(\frac{{\hslash }^2}{16m}\right)}^{\frac{2}{3}}\\ E_{\min }=(\frac{1}{{\mathbf{2}}^{\frac{5}{3}}}+\frac{1}{{\mathbf{2}}^{\frac{8}{3}}}){\mathbf{b}}^{\frac{1}{3}}{\left(\frac{{\hslash }^2}{m}\right)}^{\frac{2}{3}}=\frac{\mathbf{3}{\mathbf{b}}^{\frac{1}{3}}}{{\mathbf{2}}^{\frac{8}{3}}}{\left(\frac{{\hslash }^2}{m}\right)}^{\frac{2}{3}}\end{array}$

Hence, the final relation will have a different coefficient, however, the procedures are essentially the same.