Recent content by loginorsinup

That's exactly it. The "DC" term comes from the square-to-double-argument trigonometric identity. Got it. Thank you very much! The rest of it is straightforward. :)

You're right. I will fix that.

Hello everyone. I'm trying to better understand structured illumination microscopy and in the literature, I keep coming across bits of text like this. Source: http://www.optics.rochester.edu/workgroups/fienup/PUBLICATIONS/SAS_JOSAA09_PhShiftEstSupRes.pdf From Fourier analysis, if I take the...

Hi. Thanks. Interesting that you also had the same problem! I was able to go through the calculus/algebra to show (and derive) how to work the cosh and sinh expansions, but, I ended up with $$a(\lambda) = a\cosh(2\lambda) -a^{\dagger}\sinh(2\lambda)$$ Just a quick look at the last post where...

I am pretty sure that is enough to solve what I have been missing. Thank you very much. I wish I could have seen that as I was solving it. I felt like I was in the dark when I first did it alone, prior to asking here. How did you know how to solve these? Does it come from experience?

\begin{align*} e^{\lambda W} a e^{-\lambda W} &= a + \lambda[W,a] + \frac{\lambda^2}{2!}[W,[W,a]] + \frac{\lambda^3}{3!}[W,[W,[W,a]]] + \frac{\lambda^4}{4!}[W,[W,[W,[W,a]]]] + \ldots\\ [W,a] &= -2a^{\dagger}\\ [W,a^{\dagger}] &= -2a\\ [W,[W,a]] &= [W,-2a^{\dagger}]\\ &= W(-2a^{\dagger}) -...

\begin{align*} [a^{\dagger}a^{\dagger}-aa,a^{\dagger}] &= [a^{\dagger}a^{\dagger},a^{\dagger}]-[aa,a^{\dagger}]\\ &= (a^{\dagger}a^{\dagger}a^{\dagger}-a^{\dagger}a^{\dagger}a^{\dagger})-(aaa^{\dagger}-a^{\dagger}aa)\\ &= -aaa^{\dagger}+a^{\dagger}aa\\ &= a^{\dagger}aa-aaa^{\dagger}\\ &=...

From \begin{align*} [a^{\dagger}a^{\dagger},a] &= a^{\dagger}(a^{\dagger}a) - aa^{\dagger}a^{\dagger}\\ &= a^{\dagger}a^{\dagger}a + 0 - aa^{\dagger}a^{\dagger}\\ &= a^{\dagger}a^{\dagger}a - a^{\dagger}aa^{\dagger} + a^{\dagger}aa^{\dagger} - aa^{\dagger}a^{\dagger}\\ &=...

Hi, I really tried to see a pattern that led to the inclusion of hyperbolic trigonometric functions, but I am stuck. I calculated ##[\lambda W,a]## and ##[\lambda W,a^{\dagger}]## for ##W = a^{\dagger}a^{\dagger} - aa## and got the following. For ##[\lambda W,a]##...

I can't edit, but I meant exponential version of the momentum operator.

Hi, thanks for the response. I'm not sure I follow all of it. I should mention that I come from an engineering background and never took formal physics courses aside from the introductory series before taking this quantum course. I am not really sure what a symmetry transformation is, but from...

Hey, thanks for that. You helped me understand the mechanics of (A), but I am very much stuck on (B). I don't even know where to begin other than to compute ##e^{\lambda W}Qe^{-\lambda W}## for ##W = a^{\dagger}a^{\dagger} - aa##, but for what ##Q##? What does any of this stuff even physically...

So, ##e^{\lambda a}ae^{-\lambda a}## would require me to compute a few commutation relations. ##\left[\lambda a,a\right]## \begin{align*}\left[\lambda a,a\right] &= \lambda aa - a\lambda a\\ &= \lambda aa - \lambda aa\\&= 0\end{align*} Any commutation relation that depends on this commutation...

Hey, thanks for that. I knew I was on the right track with that. So I just have to compute a ton of commutators, basically for part A. At what point do I get to truncate this power series relation you gave me (which is very similar to the one I had presented earlier)?

Homework Statement Virtually all quantum mechanical calculations involving the harmonic oscillator can be done in terms of the creation and destruction operators and by satisfying the commutation relation \left[a,a^{\dagger}\right] = 1 (A) Compute the similarity transformation...