Chapter 3: 3.20P (page 118)

Show that the energy-time uncertainty principle reduces to the "your name" uncertainty principle (Problem 3.14), when the observable in question is x.

Short Answer

Expert verified

The energy-time uncertainty principle reduces to $σ_{H}^{2} σ_{x}^{2} = \frac{ℏ^{2}}{4 m^{2}} ⟨ p ⟩^{2}$

Step by step solution

The generalized uncertainty principle for two operators.

The generalized uncertainty principle for two operators is:

$σ_{A}^{2} σ_{B}^{2} ⩾ {(\frac{1}{2 i} ⟨ [\hat{A}, \hat{B}] ⟩)}^{2}$

Show that the energy-time uncertainty principle reduces to the "your name" uncertainty principle.

From the energy-time uncertainty principle, an operator Q satisfies the equation:

$\frac{d}{d t} ⟨ Q ⟩ = \frac{i}{ℏ} ⟨ [H, Q] ⟩ + <\frac{\partial Q}{\partial t}>$

for the position operator, $Q = x$ , the time derivative of the position is zero, since $x$ and $t$ are independent variables,

$\frac{d}{d t} ⟨ x ⟩ = \frac{i}{ℏ} ⟨ [H, x] ⟩$

Let $\hat{A} = \hat{H}$ and $\hat{B} = \hat{x}$ and use (2), to get:

$σ_{H}^{2} σ_{x}^{2} ⩾ {(- \frac{ℏ}{2} \frac{d ⟨ x ⟩}{d t})}^{2} = \frac{ℏ^{2}}{4 m^{2}} {(m \frac{d ⟨ x ⟩}{d t})}^{2} = \frac{ℏ^{2}}{4 m^{2}} ⟨ p ⟩^{2} σ_{H}^{2} σ_{x}^{2} = \frac{ℏ^{2}}{4 m^{2}} ⟨ p ⟩^{2}$

Thus, the energy-time uncertainty principle reduces to $σ_{H}^{2} σ_{x}^{2} = \frac{ℏ^{2}}{4 m^{2}} ⟨ p ⟩^{2}$ .

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Show that the energy-time uncertainty principle reduces to the "your name" uncertainty principle (Problem 3.14), when the observable in question is x.

Short Answer

Step by step solution

The generalized uncertainty principle for two operators.

Show that the energy-time uncertainty principle reduces to the "your name" uncertainty principle.

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