91影视

Beyond conventional turning points, the stationary states (states of definite energy) have nonzero values. A classical oscillator is least likely to be discovered in the ground state at the location of the minimum of the potential well, where a quantum oscillator is most likely to be found.

The value of $鉄�x鉄�$and $鉄�蚁鉄�$will be

So,

$鉄�x鉄�=\sqrt{\frac{\xcancel{鈻�}}{2m蝇}}鈭�{蠄}_n^{*}\left(a_{-}+a_{-}\right){蠄}_{n}dx$

But, $a{蠄}_n=\sqrt{n+1}{蠄}_{n+1},a{蠄}_n-\sqrt{n}{蠄}_{n-1}$

$鉄�x鉄�=\sqrt{\frac{\xcancel{鈻�}}{2m蝇}}\left[\sqrt{n+1}鈭�{蠄}_n^{*}{蠄}_{n1}dx+\sqrt{n}鈭�{蠄}_n^{*}{蠄}_{n-1}dx\right]$

$=0$

And the value of$鉄�p鉄�=m\frac{d鉄�x鉄�}{dt}=0$

According to the value of $x$, the value of

$x^2=\frac{\xcancel{鈻�}}{2m蝇}{\left(a_{+}+a^2\right)}^2$

$=\frac{\xcancel{\xcancel{鈻�}}}{2m蝇}\left({a_{+}}^2+a_{+}{a}_{-}+a_{-}{a}_{+}+{a_{-}}^2\right)$

$=\frac{\xcancel{鈻�}}{2m蝇}(2n+1)$

$=\left(n+\frac{1}{2}\right)m\xcancel{鈻�}蝇$

Hence, the value is

Again, the value of,

$p^2=-\frac{m\xcancel{鈻�}蝇}{2}{\left(a_{-}-a_{-}\right)}^2$

$=-\frac{m\xcancel{鈻�}蝇}{2}\left(a_{-}^2-a_{-}{a}_{-}-a_{-}{a}_{+}+a_{-}^2\right)$

So,

$=\frac{m\xcancel{鈻�}蝇}{2}(2n+1)$

$=\left(n+\frac{1}{2}\right)m\xcancel{鈻�}蝇$

Hence,

The value of $鉄�T鉄�$will be evaluated.

So,

$鉄�T鉄�=\left(\frac{p^2}{2m}\right)$

$=\frac{1}{2}\left(n+\frac{1}{2}\right)\xcancel{鈻�}蝇$

Thus, the value of$鉄�T鉄�=\frac{1}{2}\left(n+\frac{1}{2}\right)\xcancel{鈻�}蝇$

And to check the satisfaction of uncertainty principal,

$=\sqrt{n+\frac{1}{2}\sqrt{\frac{\xcancel{鈻�}}{m蝇}}}$

$=\sqrt{n+\frac{1}{2}}\sqrt{m\xcancel{鈻�}蝇}$

${蟽}_x{蟽}_p=\left(n+\frac{1}{2}\right)\xcancel{鈻�}鈮�\frac{\xcancel{鈻�}}{2}$

Thus, the uncertainty principal is satisfied.