Chapter 3: Q14P (page 113)

Prove the famous "(your name) uncertainty principle," relating the uncertainty in position $A = x$ to the uncertainty in energy $B = p^{2} / 2 m + v$ :
$σ_{x} σ_{H} \geq \frac{h}{2 m} | |$
For stationary states this doesn't tell you much-why not?

Short Answer

Expert verified

The uncertainty principle is $σ_{x} σ_{H} \geq \frac{h}{2 m} ||$

Step by step solution

Concept used

The generalized uncertainty principle for two observables A and B is given by:

$σ_{A}^{2} σ_{B}^{2} \geq {(\frac{1}{2 i} <[\hat{A}, \hat{B}]>)}^{2}$

Calculate the uncertainty principle

The generalized uncertainty principle for two observables A and B is given by:

$σ_{A}^{2} σ_{B}^{2} \geq {(\frac{1}{2 i} <[\hat{A}, \hat{B}]>)}^{2}$

The position-energy uncertainty relation is:

$σ_{x}^{2} σ_{H}^{2} \geq {(\frac{1}{2 i} <[\hat{x}, \hat{H}]>)}^{2}$ ....(1)

So, we need to find the commutator $[\hat{x}, \hat{H}]$ as:

$[\hat{X}, \hat{H}] g = - \frac{h^{2}}{2 m} \times \frac{\partial^{2} g}{\partial X^{2}} + xVg + - \frac{h^{2}}{2 m} \frac{\partial^{2}}{\partial x^{2}} (xg) - x (Vg) = \frac{h^{2}}{2 m} (- x \frac{\partial^{2} g}{\partial X^{2}} + 2 \frac{\partial g}{\partial x} + x \frac{\partial^{2} g}{\partial X^{2}}) = \frac{h^{2}}{m} \frac{\partial g}{\partial x} = \frac{i h}{m} (pg)$

Substitute in equation 1:

$σ_{x}^{2} σ_{H}^{2} \geq {(\frac{1}{2 i} <\hat{x}, \hat{H}]>)}^{2} = \frac{h^{2}}{4 m^{2}} {}^{2}$

So, the uncertainty principle here becomes

$σ_{x} σ_{H} \geq \frac{h}{2 m} ||$

For stationary states, this doesn't tell you much because the average position of the particle doesn't change, $σ_{H} = 0 a n d = 0 .$

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Prove the famous "(your name) uncertainty principle," relating the uncertainty in position $A = x$ to the uncertainty in energy $B = p^{2} / 2 m + v$ :
$σ_{x} σ_{H} \geq \frac{h}{2 m} | <p> |$
For stationary states this doesn't tell you much-why not?

Short Answer

Step by step solution

Concept used

Calculate the uncertainty principle

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Prove the famous "(your name) uncertainty principle," relating the uncertainty in positionA=x to the uncertainty in energyB=p2/2m+v:σxσH≥h2m|p|For stationary states this doesn't tell you much-why not?

Short Answer

Step by step solution

Concept used

Calculate the uncertainty principle

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Physics Textbooks

Kinematics Physics

Wave Optics

Astrophysics

Force

Medical Physics

Torque and Rotational Motion

Study anywhere. Anytime. Across all devices.

Prove the famous "(your name) uncertainty principle," relating the uncertainty in position $A = x$ to the uncertainty in energy $B = p^{2} / 2 m + v$ :
$σ_{x} σ_{H} \geq \frac{h}{2 m} | <p> |$
For stationary states this doesn't tell you much-why not?