Question

Consider the observables$\mathbf{A}\mathbf{=}{\mathbf{x}}^{\mathbf{2}}$and$\mathbf{B}\mathbf{=}{\mathbf{L}}_{\mathbf{z}}$ .

(a) Construct the uncertainty principle for${\mathbf{\sigma }}_{\mathbf{A}}{\mathbf{\sigma }}_{\mathbf{B}}\mathbf{}$

(b) Evaluate${\mathit{\sigma }}_{\mathbf{B}}$ in the hydrogen state${\mathbf{\psi }}_{\mathbf{n}\mathbf{/}\mathbf{m}}\mathbf{}$ .

(c) What can you conclude about$\mathbf{<}\mathbf{xy}\mathbf{>}$in this state?

Short answer

(a) The uncertainty principle for ${\sigma }_A{\sigma }_B\ge ħ|⟨xy⟩|.$.

(b) in the hydrogen state ${\mathrm{\psi }}_{\mathrm{nim}}$is${\mathrm{\sigma }}_{\mathrm{B}}=0$

(c) The conclusion about in the given state is 0.

Step-by-step solution

Definition of the uncertainty

The uncertainty principle is a set of mathematical inequalities that establish a basic limit on the precision with which the values for certain pairs of physical properties of a particle, such as position and momentum, can be anticipated from initial conditions.

$\mathit{\Delta }\mathit{x}\mathit{\Delta }\mathit{p}\mathbf{\ge }\frac{\mathbf{h}}{\mathbf{4}\mathbf{\pi }}$

(a) Determination of the uncertainty principle for σAσB

Calculate the commutator of the two operators, A and B in the following way,

$$\begin{gathered} \left[\mathrm{A},\mathrm{B}\right]=\left[{\mathrm{X}}^2,{\mathrm{L}}_{\mathrm{z}}\right] \\ =\left[{\mathrm{x}}^2,{\mathrm{yp}}_{\mathrm{x}}-{\mathrm{xp}}_{\mathrm{y}}\right] \\ =\left[{\mathrm{x}}^2,{\mathrm{yp}}_{\mathrm{x}}\right]-\left[{\mathrm{x}}^2,{\mathrm{xp}}_{\mathrm{y}}\right] \\ =\left[{\mathrm{x}}^2,\mathrm{y}\right]{\mathrm{p}}_{\mathrm{x}}+\mathrm{y}\left[{\mathrm{x}}^2,{\mathrm{p}}_{\mathrm{x}}\right]-\left[{\mathrm{x}}^2,\mathrm{x}\right]{\mathrm{p}}_{\mathrm{y}}-\mathrm{x}\left[{\mathrm{x}}^2,{\mathrm{p}}_{\mathrm{y}}\right] \end{gathered}$$

Evaluate the above expression further.

$$\begin{gathered} \left[\mathrm{A},\mathrm{B}\right]=0+\mathrm{y}\left\{\mathrm{x}\left[\mathrm{x},{\mathrm{p}}_{\mathrm{x}}\right]+\left[\mathrm{x},{\mathrm{p}}_{\mathrm{x}}\right]\mathrm{x}\right\}-0-\mathrm{x}\left\{\mathrm{x}\left[\mathrm{x},{\mathrm{p}}_{\mathrm{y}}\right]+\left[\mathrm{x},{\mathrm{p}}_{\mathrm{y}}\right]\mathrm{x}\right\} \\ =\mathrm{y}\left\{\mathrm{x}\left(-\mathrm{i}\sout{\mathrm{h}}\right)+\left(-\mathrm{i}\sout{\mathrm{h}}\right)\mathrm{x}\right\}-\mathrm{x}\left\{0+0\right\} \\ =-2\mathrm{i}\sout{\mathrm{h}}\mathrm{yx} \end{gathered}$$

It is known that .

Apply it in expression.

Thus, the uncertainty principle for ${\mathrm{\sigma }}_{\mathrm{A}}{\mathrm{\sigma }}_{\mathrm{B}}\ge \mathrm{ħ}|⟨\mathrm{xy}⟩|$.

(b) Determination of σB  the hydrogen state

Now get the expectation values of both $L_z$ and $L_z^2$, in some arbitrary hydrogenic normalized state$|{\psi }_{n}lm⟩$ .

Evaluate the variation in $B,{\sigma }_B$,

Similarly, evaluate the variation in ${\mathrm{B}}^2,{\mathrm{\sigma }}_{\mathrm{B}}$

role="math" localid="1658137986506"

Find the value of${\sigma }_B$.

Thus,${\sigma }_B$ in the hydrogen state ${\psi }_{nim}$ is ${\mathrm{\sigma }}_{\mathrm{B}}=0$.

(c) Conclusion about <xy>  in the given state

Since the variation in$\mathrm{B},{\mathrm{\sigma }}_{\mathrm{B}},$ is zero and the right hand side of the uncertainty principle, $\hslash |⟨xy⟩$, is always positive , then the right hand side must also be zero. Hence,it is concluded that $⟨\mathrm{xy}⟩$ is also 0.

Thus, the conclusion about $⟨\mathrm{xy}⟩$ in the given state is 0.