How to Derive an Exact Time-Energy Uncertainty Answer?

[Leonhard]

Homework Statement

In laser physics a common approximation for the time-energy uncertainty is given by.

$$\Delta \omega \cdot \tau \geq \approx 2 \pi$$

The problem is to use the Energy-Time Uncertainty relationship to derive a more exact answer than this.

Homework Equations

$$\sqrt{\left\langle E^2 \right\rangle \left\langle t^2 \right\rangle} \geq \frac{\hbar}{2}$$

It should be noted that we are supposed to use FWHM for the uncertainties, so the following relationship is supplied.

$$\Delta_{FWHM} x = 2\sqrt{2 ln(2)}\sqrt{\left\langle x^2 \right\rangle}$$

The Attempt at a Solution

$$\sqrt{\left\langle E^2 \right\rangle \left\langle t^2 \right\rangle} = \frac{\hbar}{2}$$

I substitute

$$E = \hbar \omega$$

Giving

$$\sqrt{\left\langle \hbar^2 \omega^2 \right\rangle \left\langle t^2 \right\rangle} = \frac{\hbar}{2}$$

I divide with $$\hbar$$ on both sides

$$\sqrt{\left\langle \omega^2 \right\rangle \left\langle t^2 \right\rangle} = \frac{1}{2}$$

We then substitute

$$\tau = 2\sqrt{2 ln(2)} \sqrt{\left\langle t^2 \right\rangle}$$

And

$$\Delta \omega = 2\sqrt{2 ln(2)} \sqrt{\left\langle \omega^2 \right\rangle}$$

Giving

$$\Delta \omega \cdot \tau = 4 ln(2)$$

I don't know where I've done anything wrong, but I would appreciate some help a lot :3

 

[Jhero]

Hello,

Your attempt at a solution is on the right track, but there are a few errors that need to be corrected.

Firstly, in the given equation for the energy-time uncertainty, the \Delta symbols represent the standard deviation, not the full width at half maximum (FWHM). So the equation should be:

\Delta E \cdot \Delta t \geq \frac{\hbar}{2}

Next, when substituting E = \hbar \omega, you should use the standard deviation for E, which is \Delta E = \hbar \Delta \omega. So the equation becomes:

\sqrt{\left\langle \hbar^2 \Delta \omega^2 \right\rangle \left\langle t^2 \right\rangle} = \frac{\hbar}{2}

Dividing by \hbar on both sides gives:

\sqrt{\left\langle \Delta \omega^2 \right\rangle \left\langle t^2 \right\rangle} = \frac{1}{2}

Next, you need to substitute for the FWHM uncertainties, so the equation becomes:

\sqrt{\left\langle \frac{\Delta \omega_{FWHM}^2}{4 ln(2)} \right\rangle \left\langle t_{FWHM}^2 \right\rangle} = \frac{1}{2}

Simplifying gives:

\sqrt{\frac{\left\langle \Delta \omega_{FWHM}^2 \right\rangle}{4 ln(2)} \cdot \left\langle t_{FWHM}^2 \right\rangle} = \frac{1}{2}

Finally, substituting for the FWHM uncertainties using the given equation, we get:

\sqrt{\frac{\left\langle \Delta \omega^2 \right\rangle}{16 ln(2)} \cdot \left\langle \Delta t^2 \right\rangle} = \frac{1}{2}

And solving for \Delta \omega \cdot \Delta t gives:

\Delta \omega \cdot \Delta t \geq 4 ln(2)

Which is the same result you got. So your method was correct, but there were some errors in your substitutions. I hope this helps!