91影视

Schwarz inequality is $\mathbf{|}<a鈭�b>{\mathbf{|}}^{\mathbf{2}}\mathbf{鈮�}<a鈭�a><b鈭�b>\mathbf{.}$.

In statistics, the variance is expressed as 鈥渢he average of the square of the difference from the mean鈥�. So, for an observable $\hat{\mathbf{A}}$we have:${\mathbf{蟽}}_{\mathbf{A}}^{\mathbf{2}}\mathbf{鈮�}\mathbf{鉄�}\mathbf{(}\hat{\mathbf{A}}\mathbf{-}<A>{\mathbf{)}}^{\mathbf{2}}\mathbf{鉄�}$

(a)

For an observable$\hat{\mathrm{A}}$ the variance can be written as:

Further solving above equation,

And for an observable we can write:

From above two equations of variance,

${\mathrm{蟽}}_{\mathrm{A}}^2{\mathrm{蟽}}_{\mathrm{B}}^2=鉄�\mathrm{f}鈭�\mathrm{f}鉄�鉄�\mathrm{g}鈭�\mathrm{g}鉄�鈮�|鉄�\mathrm{f}鈭�\mathrm{g}鉄�{|}^2$ 鈥︹�� (1)

Where used Schwarz inequality . For any complex number $\mathrm{z}=\mathrm{x}+\mathrm{iy},$we have:

$$\begin{gathered} |\mathrm{z}{|}^2=[\mathrm{Re}\left(\mathrm{z}\right){]}^2+\left[\mathrm{Im}\right(\mathrm{z}){]}^2 \\ ={\left[\frac{1}{2}(\mathrm{z}+{\mathrm{z}}^{*})\right]}^2+{\left[\frac{1}{2\mathrm{i}}(\mathrm{z}-{\mathrm{z}}^{*})\right]}^2鈥�鈥�(2) \end{gathered}$$

Let$\mathrm{z}=鉄�\mathrm{f}鈭�\mathrm{g}鉄�$; thus$|鉄�\mathrm{f}鈭�\mathrm{g}鉄�{|}^2=|\mathrm{z}{|}^2$. Substitute from (2) into (1) to get (note that$\mathrm{z}=鉄�\mathrm{f}鈭�\mathrm{g}鉄�$).

Find and , as:

For observable .

Adding equations (4) and (5) we can write:

$$\begin{gathered} 鉄�\mathrm{f}鈭�\mathrm{g}鉄�+鉄�\mathrm{g}鈭�\mathrm{f}鉄�=鉄�\hat{\mathrm{A}}\hat{\mathrm{B}}鉄�-鉄�\mathrm{A}鉄�鉄�\mathrm{B}鉄�+鉄�\hat{\mathrm{B}}\hat{\mathrm{A}}鉄�-鉄�\mathrm{A}鉄�鉄�\mathrm{B}鉄�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥� \\ 鈥�鈥�鈥�鈥�鈥�鈥�鈥�=鉄�\hat{\mathrm{A}}\hat{\mathrm{B}}+\hat{\mathrm{B}}\hat{\mathrm{A}}-2鉄�\mathrm{A}鉄�鉄�\mathrm{B}鉄�鉄�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥� \\ 鈥�鈥�=鉄�\mathrm{D}鉄� \end{gathered}$$

Subtracting equations and we can write:

$鉄�\mathrm{f}鈭�\mathrm{g}鉄�-鉄�\mathrm{g}鈭�\mathrm{f}鉄�=鉄�\hat{\mathrm{A}}\hat{\mathrm{B}}鉄�-鉄�\hat{\mathrm{B}}\hat{\mathrm{A}}鉄�=鉄�[\hat{\mathrm{A}},\hat{\mathrm{B}}]鉄�$

Substitute into (3) we get:

Thus, the generalized uncertainty principle can be strengthened to read role="math" localid="1656317038528" .

(b)

For the case of $\hat{\mathrm{A}}=\hat{\mathrm{B}}$the values of$\hat{\mathrm{C}}\&\hat{\mathrm{D}}$ are:

Further solving above values of role="math" localid="1656320627569" $\hat{\mathrm{C}}\&\hat{\mathrm{D}}$as:

$鉄�\mathrm{C}鉄�=0鈥�鉄�\mathrm{D}鉄�=2(鉄�{\hat{\mathrm{A}}}^2鉄�-鉄�\mathrm{A}{鉄�}^2)=2{\mathrm{蟽}}_{\mathrm{A}}^2$

Substitute into the result of part (a) we get:

$$\begin{gathered} {\mathrm{蟽}}_{\mathrm{A}}^2{\mathrm{蟽}}_{\mathrm{A}}^2鈮�(1/4)4{\mathrm{蟽}}_{\mathrm{A}}^4={\mathrm{蟽}}_{\mathrm{A}}^4 \\ {\mathrm{蟽}}_{\mathrm{A}}^2{\mathrm{蟽}}_{\mathrm{A}}^2鈮�{\mathrm{蟽}}_{\mathrm{A}}^4 \end{gathered}$$

Thus, the is true, but not tell you anything since both sides are equal. Therefore, it is not very informative${\mathrm{蟽}}_{\mathrm{A}}^2{\mathrm{蟽}}_{\mathrm{A}}^2鈮�{\mathrm{蟽}}_{\mathrm{A}}^4$