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Consider the wave packet defined by

$\mathbf{唯}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}{\mathbf{鈭�}}_{\mathbf{0}}^{\mathbf{鈭�}}\mathbf{鈥�}\mathbf{B}\mathbf{\left(}\mathbf{k}\mathbf{\right)}\mathbf{cos}\mathbf{鈦�}\mathbf{kxdk}$

Let $\mathit{B}\mathbf{\left(}\mathit{k}\mathbf{\right)}\mathbf{=}{\mathit{e}}^{\mathbf{-}{\mathbf{伪}}^{\mathbf{2}}{\mathbf{k}}^{\mathbf{2}}}$ . (a) The function B(k) has its maximum value at k= 0. Let k_h be the value of kat which B(k) has fallen to half its maximum value, and define the width of B(k) as w_k= k_h . In terms of a, what is w_k? (b) Use integral tables to evaluate the integral that gives $\mathit{蠄}\mathbf{\left(}\mathit{x}\mathbf{\right)}$. For what value of xis $\mathit{蠄}\mathbf{\left(}\mathit{x}\mathbf{\right)}$maximum? (c) Define the width of $\mathit{蠄}\mathbf{\left(}\mathit{x}\mathbf{\right)}$ as w_x= x_h, where x_h is the positive value of xat which $\mathit{蠄}\mathbf{\left(}\mathit{x}\mathbf{\right)}$ has fallen to half its maximum value. Calculate w_xin terms of a. (d) The momentum pis equal to hk/2蟺, so the width of Bin momentum is w_p= hw_k/2蟺. Calculate the product w_p w_xand compare to the Heisenberg uncertainty principle.

Step by step solution

01

(a) Determination of the value of wk.

Given,

$B\left(k\right)=e^{-a^2{k}^2}$

Now, putting k=0,

$\begin{array}{r}B\left(0\right)=B_{max}\\ =1\end{array}$

Putting, k = k_h

$\begin{array}{r}\mathrm{B}\left({\mathrm{k}}_{\mathrm{h}}\right)=\frac{1}{2}\\ ={\mathrm{e}}^{鈭�{伪}^2{\mathrm{k}}_h^2}\end{array}$

Taking log 鈥榣n鈥� both sides, we get

$$\begin{gathered} \ln \left(\frac{1}{2}\right)=鈭�{伪}^2{\mathrm{k}}_{\mathrm{h}}^2 \\ {\mathrm{k}}_{\mathrm{h}}=\frac{1}{\mathrm{伪}}\sqrt{\ln \left(2\right)} \\ ={\mathrm{w}}_{\mathrm{k}} \end{gathered}$$

Thus, it is concluded that at w_k= k_h, B(k) has fallen to half its maximum value.

02

(b) Determination of the value of x for which ψ(x) is minimum.

According to a standard integral,

${鈭�}_0^{\mathrm{鈭�}}鈥�e^{鈭�a^2{k}^2}\cos \left(kx\right)dk=\frac{\sqrt{蟺}}{2伪}{e}^{鈭�x^2/4{伪}^2}$

Now, in given question, the wave function is defined as,

$$\begin{gathered} \mathrm{唯}\left(x\right)={鈭�}_0^{\mathrm{鈭�}}鈥�e^{鈭�a^2{k}^2}\cos kxdk \\ =\frac{\sqrt{蟺}}{2伪}{e}^{鈭�x^2/4a^2} \end{gathered}$$

At x = 0,

$e^0=1$, so, $\mathit{蠄}\mathbf{\left(}\mathit{x}\mathbf{\right)}$is maximum.

03

(c) Determination of the value of wx.

At $x=x_h,$

$\mathrm{唯}\left(x\right)=\frac{\sqrt{蟺}}{4伪}$

This occurs when,

$$\begin{gathered} {\mathrm{e}}^{鈭�x^2/4a^2}=\frac{1}{2} \\ 鈭�\frac{x_h^2}{4a^2}=\ln \left(\frac{1}{2}\right) \\ {x}_h=2a\sqrt{\ln 2} \\ =w_x \end{gathered}$$

The value of x = x_h is $2伪\sqrt{\ln 2}$.

04

(d) Determination of the value of product wp wx.

The product,

$$\begin{gathered} {\mathrm{w}}_{\mathrm{p}}{\mathrm{w}}_{\mathrm{x}}=\left(\frac{{\mathrm{hw}}_{\mathrm{k}}}{2蟺}\right){\mathrm{w}}_{\mathrm{x}} \\ =\frac{\mathrm{h}}{2\mathrm{蟺}}\left(\frac{1}{\mathrm{伪}}\sqrt{\ln 2}\right)\left(2\mathrm{伪}\sqrt{\ln 2}\right) \\ =\frac{\mathrm{h}}{2\mathrm{蟺}}(2\ln 2) \\ =\frac{\mathrm{hln}2}{\mathrm{蟺}} \\ =(2\ln 2)\mathrm{魔} \end{gathered}$$

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