91影视

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,

$\begin{array}{l}螖p_{\text{x}}螖x鈮�\frac{h}{k}\\ k螖p_{\text{x}}鈮�\frac{h}{螖x}\end{array}$

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

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

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

(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 $\frac{\mathrm{鈩�}蝇}{2}$(review problem 52).

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${\left(\stackrel{炉}{x}\right)}^2$

$鈭�{\left(螖x\right)}^2=\stackrel{炉}{x^2}鈭�{\left(\stackrel{炉}{x}\right)}^2$

$鈭�\stackrel{炉}{x}鈮�0$This is a valid assumption in the case of a HO

$鈭�螖x鈮�\sqrt{{\stackrel{炉}{x}}^2}=x$

The same goes for the momentum uncertainty $螖p鈮�p$, where we have replaced the typical values of $x$and $p$ with their corresponding momentum and position uncertainties $螖x$and $螖p$, 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.

$\begin{array}{l}鈭�E=\frac{{\left(螖p\right)}^2}{2m}+b{\left(螖x\right)}^4\\ 鈭�螖p螖x=\frac{\mathrm{鈩�}}{2}鈬�螖p=\frac{\mathrm{鈩�}}{2螖x}\\ 鈭�E=\frac{{\mathrm{鈩�}}^2}{\mathbf{8}\mathbf{m}{\left(\mathbf{螖虫}\right)}^2}+\mathbf{b}{\left(\mathrm{螖虫}\right)}^{\mathbf{4}}\end{array}$

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

Therefore,

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

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