Complex Conjugate of Fourier Transform

Epoch12000BC
Messages
3
Reaction score
0
Hello All,

As I understand it, the wavefunction Psi(x) can be written as a sum of all the particle's momentum basis states (which is the Fourier transform of Psi(x)). I was woundering if the wavefunction's complex conjugate Psi*(x) can be written out in terms of momentum basis states, similar to the transform of Psi(x), and if so, what form would it take in order to make make the probability density of Psi(x)Psi*(x) the correct real-valued number.

In other words, if both Psi(x) and Psi*(x) were written out in their Fourier transform forms, and multiplied together in these forms, what would the transform of Psi*(x) look like?

Any insight into this would be greatly appriciated.

-Andrew
 
Physics news on Phys.org
No magic here. If
\psi(x)= \frac{1}{\sqrt{2\pi}} \int\!dk\, \psi(k) e^{\frac{i k x}{\hbar},
then
\psi^*(x)= \frac{1}{\sqrt{2\pi}} \int\!dk\, \psi^*(k) e^{\frac{-i k x}{\hbar},
 
Last edited:
Thanks for the help, lbrits!

I figured the complex conjugate would involve making negative all the imaginary parts of the transform, but it seemed like doing the ol' FOIL method when multiplying Psi(x) and Psi*(x) would result in a number that would generally be complex, not the real number that the probability density should be. I guess I'm a bit rusty on my math :-)

Thanks again for the help,
-droo
 
Yeah I was a bit rusty on the whole distributivity property of products over sums. Anyway, things are quite forgiving:
\psi^*(x) \psi(x) = \frac{1}{2\pi \hbar} \int\!dk\,dq\, \psi^*(k) \psi(q) e^{i (k-q)x/hbar}
I believe you can show this to be the Fourier transform of f(k) = \int\!dq\, \psi^*(k) \psi(k-q), and to show that the thing is real, is to remember that in the Fourier transform things always come in pairs, one multiplied by e^{ikx} and one by e^{-ikx}. The previous expression exhibits a symmetry under k -> -k.
 
I read Hanbury Brown and Twiss's experiment is using one beam but split into two to test their correlation. It said the traditional correlation test were using two beams........ This confused me, sorry. All the correlation tests I learnt such as Stern-Gerlash are using one beam? (Sorry if I am wrong) I was also told traditional interferometers are concerning about amplitude but Hanbury Brown and Twiss were concerning about intensity? Isn't the square of amplitude is the intensity? Please...
I am not sure if this belongs in the biology section, but it appears more of a quantum physics question. Mike Wiest, Associate Professor of Neuroscience at Wellesley College in the US. In 2024 he published the results of an experiment on anaesthesia which purported to point to a role of quantum processes in consciousness; here is a popular exposition: https://neurosciencenews.com/quantum-process-consciousness-27624/ As my expertise in neuroscience doesn't reach up to an ant's ear...
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. Towards the end of the first lecture for the Qiskit Global Summer School 2025, Foundations of Quantum Mechanics, Olivia Lanes (Global Lead, Content and Education IBM) stated... Source: https://www.physicsforums.com/insights/quantum-entanglement-is-a-kinematic-fact-not-a-dynamical-effect/ by @RUTA
Back
Top