Mean Frequency and Frequency Spread of a Laser Pulse

[CDL]

Homework Statement

Laser probes are being used to examine the states of atoms and molecules at high temporal resolution. A laser operating at a wavelength of 400 nm produces a 1 femtosecond pulse. Compute the mean frequency and frequency spread, ∆ν, of this laser pulse.

Homework Equations

$$c = f \lambda, \ \Delta \nu \sim \frac{1}{\tau_c}, \ \text{where} \ \tau_c \ \text{is the coherence time.}$$
3. The Attempt at a Solution [/B]
At first glance, I decided to find the mean frequency using $$\bar{\nu} = \frac{c}{\bar{\lambda}}.$$ After further consideration, I don't think this necessarily correct since if we think of the wavelengths present in the pulse as following a statistical distribution with $\mathbb{E}(\lambda) = 400$ nm, then by Jensen's inequality, we would have $$\mathbb{E}\left(\frac{c}{\lambda}\right) = \mathbb{E}(\nu) \geq \frac{c}{\mathbb{E}(\lambda)}.$$ I.e I expect the mean frequency $\bar{\nu}$ to be greater than $\frac{c}{\bar{\lambda}}.$

I am not sure how to proceed, as I can't compute the exact mean frequency without knowing the distribution of the wavelengths, and I have only found a lower bound for the mean frequency. I could assume that the wavelength is gaussian distributed, but I'd also need to specify the variance of the wavelength in that case.

I took the coherence time $\tau_c$ to be $10^{-15}$ s, so I expect the frequency spread to approximately be $\Delta \nu \sim 10^{15}$ Hz. Using $\bar{\nu} = \frac{c}{\bar{\lambda}}$, I get $\bar{\nu} = 7.5 \times 10^{14}$, which is less than the spread, meaning that according to this calculation, the minimum frequency is $0$.

 

[DrClaude]

I think you are overthinking the problem. From the lack of details given in the problem statement, the appears to be a simple introductory exercise, so you can take the pulse to be symmetric around its peak frequency and to be transform-limited and thus use the time-energy uncertainty relation to get a lower bound on the frequency spread.

 

[RPinPA]

While you are correct that E[1/X] is not 1/E[X], it's an OK approximation to use when the associated probability distribution is narrow, as you're supposed to assume here.

 

[RPinPA]

While you are correct that E[1/X] is not 1/E[X], it's an OK approximation to use when the associated probability distribution is narrow, as you're supposed to assume here.

It's a good thing to know when you're using an approximation and how good the approximation is. So I explored that a little. Let's assume for simplicity in the calculation that $\lambda \sim U[a, b]$, that $\lambda$ is uniformly distributed between some values $a$ and $b$, and we'll define a fractional width $\epsilon$ by $b = a(1 + \epsilon)$.

So $b - a = a\epsilon$ and $b + a = a(2 + \epsilon)$.

Clearly the mean value of $\lambda$ is halfway between a and b, $E[\lambda] = \frac{b + a}{2} = a(1 + \frac{\epsilon}{2})$.

Now what is $E[1/\lambda]$?

$$E \left [\frac{1}{\lambda} \right ] = \frac{1}{b - a} \int_a^b \frac{d\lambda}{\lambda} = \frac{\ln(b) - \ln(a)}{b-a} = \frac{\ln(b/a)}{b - a} \\
= \frac{\ln(1 + \epsilon)}{a\epsilon} = \frac{1}{a \epsilon} \left (\epsilon - \frac{\epsilon^2}{2} + \frac{\epsilon^3}{3} - \frac{\epsilon^4}{4} + ...\right )\\
= \frac{1}{a} \left (1 - \frac{\epsilon}{2} + \frac{\epsilon^2}{3} - \frac{\epsilon^3}{4} + ... \right )$$
For comparison,
$$\frac {1}{E[\lambda]} = \frac{1}{a} \frac{1}{1 + \frac{\epsilon}{2}} \\
= \frac{1}{a} \left (1 - \frac{\epsilon}{2} + \frac{\epsilon^2}{4} - \frac{\epsilon^3}{8} + ... \right )$$
So as you can see, they are the same to first order in $\epsilon$. If $\epsilon^2$ is a lot smaller than $\epsilon$ (as it would be for, say, $\epsilon < 0.01$) then these expressions are approximately the same.