91影视

In this method, the accelerations and velocities at a specific time point are taken to be constant during a time interval and are utilised to find the location of the following time point.

(a)

The function ${蠄}_0$is even, and ${蠄}_1$is odd. In either case $鈭�{蠄}^2$is even, so the value of $鉄�x鉄�=鈭�x|蠄{|}^{2}dx=0$. Therefore$鉄�p鉄�=\frac{md鉄�x鉄�}{dt}=0$

Thus, the value of is $鈭�x|蠄{|}^{2}dx-0$and the value of $鉄�p鉄�$is$\frac{md鉄�x鉄�}{dt}=0$

Now, evaluate the value of and .

The below values are known that

${蠄}_0=伪e^{-{尉}^2/2}$

${蠄}_1=\sqrt{2}伪尉e^{-{尉}^2/2}$

For $n=0$, the value of will be

....................(1)

For given equation-

$$\begin{gathered} 尉=\sqrt{\frac{m蝇}{\xcancel{鈻�}}}x \\ x=尉\sqrt{\frac{\xcancel{鈻�}}{m蝇}} \\ {x}^2={尉}^2\left(\frac{\xcancel{鈻�}}{m蝇}\right) \end{gathered}$$

And

$$\begin{gathered} x=尉\sqrt{\frac{\xcancel{鈻�}}{m蝇}} \\ dx=d尉脳\sqrt{\frac{\xcancel{鈻�}}{m蝇}} \end{gathered}$$

Substitute the value of $x^2$and $dx$in equation (1)

And the value of is

For $n=1$, the value of will be

Substitute the value of and in the above equation. So,

And the value of is

Therefore, the value of is $\frac{3\xcancel{鈻�}}{2m蝇}$and is $\frac{3m\xcancel{鈻�}蝇}{2}$

(b)

The value of will be for :

$$\begin{gathered} {蟽}_x{蟽}_P=\sqrt{\frac{\xcancel{鈻�}}{2m蝇}}\sqrt{\frac{m蝇\xcancel{鈻�}}{2}} \\ =\frac{\xcancel{鈻�}}{2} \end{gathered}$$

(Right t the uncertainty moment)

For $n=1$, the value will be :

$$\begin{gathered} {蟽}_x=\sqrt{\frac{3\xcancel{鈻�}}{2m蝇}} \\ {蟽}_P=\sqrt{\frac{3m蝇\xcancel{鈻�}}{2}} \\ {蟽}_x{蟽}_P=3\frac{\xcancel{鈻�}}{2}>\frac{\xcancel{鈻�}}{2} \end{gathered}$$

The uncertainty principal for these states is greater than$\frac{\xcancel{鈻�}}{2}$

(c)

The average kinetic energy is:

$=\left\{\begin{array}{l}\frac{1}{4}蝇\xcancel{鈻�}(n=0)\\ \frac{3}{4}蝇\xcancel{鈻�}(n=1)\end{array}\right\}$

The average potential energy is

$=\left\{\begin{array}{l}\frac{1}{4}蝇\xcancel{鈻�}(n=0)\\ \frac{3}{4}蝇\xcancel{鈻�}(n=1)\end{array}\right\}$

, as expected.

So, the value of the average kinetic energy is $\left\{\begin{array}{c}\frac{1}{4}蝇\xcancel{鈻�}(n=0)\\ \frac{3}{4}蝇\xcancel{鈻�}(n=1)\end{array}\right\}$and the average potential energy is $\left\{\begin{array}{c}\frac{1}{4}蝇\xcancel{鈻�}(n=0)\\ \frac{3}{4}蝇\xcancel{鈻�}(n=1)\end{array}\right\}$. The sum of these two is same as expected value.