Question

Extended uncertainty principle.The generalized uncertainty principle (Equation 3.62) states that

${\mathbf{\sigma }}_{\mathbf{A}}^{\mathbf{2}}{\mathbf{\sigma }}_{\mathbf{B}}^{\mathbf{2}}\mathbf{\ge }\frac{\mathbf{1}}{\mathbf{4}}{<C>}^{\mathbf{2}}$

where$\mathbf{}\hat{\mathbf{C}}\mathbf{\equiv }\mathbf{-}\mathbf{i}\mathbf{[}\hat{\mathbf{A}}\mathbf{}\mathbf{,}\hat{\mathbf{B}}\mathbf{}\mathbf{̂}\mathbf{]}\mathbf{.}$.

(a) Show that it can be strengthened to read

${\mathbf{\sigma }}_{\mathbf{A}}^{\mathbf{2}}{\mathbf{\sigma }}_{\mathbf{B}}^{\mathbf{2}}\mathbf{\ge }\frac{\mathbf{1}}{\mathbf{4}}\mathbf{(}{<C>}^{\mathbf{2}}\mathbf{+}{<D>}^{\mathbf{2}}\mathbf{)}$ [3.99]

where$\hat{\mathbf{D}}\mathbf{\equiv }\hat{\mathbf{A}}\mathbf{}\hat{\mathbf{B}}\mathbf{+}\hat{\mathbf{B}}\mathbf{}\hat{\mathbf{A}}\mathbf{-}\mathbf{2}\mathbf{⟨}\mathbf{A}\mathbf{⟩}\mathbf{⟨}\mathbf{B}\mathbf{⟩}\mathbf{.}$. Hint: Keep the term in Equation 3.60

(b) Check equation 3.99 for the case$\mathbf{B}\mathbf{=}\mathbf{A}$(the standard uncertainty principle is trivial, in this case, since$\hat{\mathbf{C}}\mathbf{=}\mathbf{0}$; unfortunately, the extended uncertainty principle doesn't help much either).

Short answer

Step-by-step solution

Schwarz inequality

Schwarz inequality is $\mathbf{|}<a\mid b>{\mathbf{|}}^{\mathbf{2}}\mathbf{\le }<a\mid a><b\mid b>\mathbf{.}$.

In statistics, the variance is expressed as “the average of the square of the difference from the mean”. So, for an observable $\hat{\mathbf{A}}$we have:${\mathbf{\sigma }}_{\mathbf{A}}^{\mathbf{2}}\mathbf{\equiv }\mathbf{⟨}\mathbf{(}\hat{\mathbf{A}}\mathbf{-}<A>{\mathbf{)}}^{\mathbf{2}}\mathbf{⟩}$

Solve for z and apply Schwarz inequality

(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,

${\mathrm{\sigma }}_{\mathrm{A}}^2{\mathrm{\sigma }}_{\mathrm{B}}^2=⟨\mathrm{f}\mid \mathrm{f}⟩⟨\mathrm{g}\mid \mathrm{g}⟩\ge |⟨\mathrm{f}\mid \mathrm{g}⟩{|}^2$ …… (1)

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

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

Solve for σA2σB2

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

Find and , as:

For observable .

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

$$\begin{gathered} ⟨\mathrm{f}\mid \mathrm{g}⟩+⟨\mathrm{g}\mid \mathrm{f}⟩=⟨\hat{\mathrm{A}}\hat{\mathrm{B}}⟩-⟨\mathrm{A}⟩⟨\mathrm{B}⟩+⟨\hat{\mathrm{B}}\hat{\mathrm{A}}⟩-⟨\mathrm{A}⟩⟨\mathrm{B}⟩\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em} \\ \hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}=⟨\hat{\mathrm{A}}\hat{\mathrm{B}}+\hat{\mathrm{B}}\hat{\mathrm{A}}-2⟨\mathrm{A}⟩⟨\mathrm{B}⟩⟩\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em} \\ \hspace{0.33em}\hspace{0.33em}=⟨\mathrm{D}⟩ \end{gathered}$$

Subtracting equations and we can write:

$⟨\mathrm{f}\mid \mathrm{g}⟩-⟨\mathrm{g}\mid \mathrm{f}⟩=⟨\hat{\mathrm{A}}\hat{\mathrm{B}}⟩-⟨\hat{\mathrm{B}}\hat{\mathrm{A}}⟩=⟨[\hat{\mathrm{A}},\hat{\mathrm{B}}]⟩$

Substitute into (3) we get:

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

Solve for σA2σA2

(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:

$⟨\mathrm{C}⟩=0\hspace{1em}⟨\mathrm{D}⟩=2(⟨{\hat{\mathrm{A}}}^2⟩-⟨\mathrm{A}{⟩}^2)=2{\mathrm{\sigma }}_{\mathrm{A}}^2$

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

$$\begin{gathered} {\mathrm{\sigma }}_{\mathrm{A}}^2{\mathrm{\sigma }}_{\mathrm{A}}^2\ge (1/4)4{\mathrm{\sigma }}_{\mathrm{A}}^4={\mathrm{\sigma }}_{\mathrm{A}}^4 \\ {\mathrm{\sigma }}_{\mathrm{A}}^2{\mathrm{\sigma }}_{\mathrm{A}}^2\ge {\mathrm{\sigma }}_{\mathrm{A}}^4 \end{gathered}$$

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