[supaman5]

http://lnk.nu/google.co.uk/28fr

Are results that come from a spectrometer such as the image above continuous? I ask because I need to find the area under a spectrum. If the data points are continuous then I would use integration, if it's discrete then I would use a summation.

I've confused myself as even though a continuum of wavelengths in an interval are applied to a molecule, surely it is impossible to record the absorption at every value in a that interval.

[Borek]

Technically spectrum is continuous, but it is recorded as series of points. No way to integrate other than numerically, by summation.

[curiouscat]

Technically spectrum is continuous, but it is recorded as series of points. No way to integrate other than numerically, by summation.

Isn't the spectrum fundamentally discrete? I remember doing Gaussian calculations and ending up with a discrete number of fundamental modes of vibration. I think it was ( 3×n - 2 ) modes where n is the number of atoms.

What am I getting wrong?

[Borek]

Solution of the Schroedinger equation is discrete, but recorded spectrum is not. Detector noise, Doppler effect and so on.

[curiouscat]

Solution of the Schroedinger equation is discrete, but recorded spectrum is not. Detector noise, Doppler effect and so on.

Gotcha! Thanks!

It's an interesting pipeline:

Discrete (Schrodinger)  Continuous (Noise etc.)  Discrete (Measurement)  Continuous (As displayed by instrument)

[discodermolide]

Solution of the Schroedinger equation is discrete, but recorded spectrum is not. Detector noise, Doppler effect and so on.

Gotcha! Thanks!

It's an interesting pipeline:

Discrete (Schrodinger)  Continuous (Noise etc.)  Discrete (Measurement)  Continuous (As displayed by instrument)

Pipeline is better than most Pharma companies!
But don't forget old Heisenberg, I'm uncertain where he fits in here, but it must be somewhere.

[Jorriss]

Solution of the Schroedinger equation is discrete, but recorded spectrum is not. Detector noise, Doppler effect and so on.

Gotcha! Thanks!

It's an interesting pipeline:

Discrete (Schrodinger)  Continuous (Noise etc.)  Discrete (Measurement)  Continuous (As displayed by instrument)

If yo're interested, you should read about broadening in spectra, particularly natural lifetime and doppler broadening.

[curiouscat]

Solution of the Schroedinger equation is discrete, but recorded spectrum is not. Detector noise, Doppler effect and so on.

Gotcha! Thanks!

It's an interesting pipeline:

Discrete (Schrodinger)  Continuous (Noise etc.)  Discrete (Measurement)  Continuous (As displayed by instrument)

If yo're interested, you should read about broadening in spectra, particularly natural lifetime and doppler broadening.

I am interested. Thanks for that lead!