Eigenvalues and density of states

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SUMMARY

The discussion centers on the relationship between the density of states (DOS) function f(x) and the histogram of Hamiltonian eigenvalues. The user expected the histogram to be an exact copy of f(x) but noted discrepancies in scale, specifically that f(x) peaks at approximately 100 while the histogram peaks at around 60. The conversation highlights the importance of bin size in histograms, emphasizing that the units of a histogram are unitless and require division by the bin width to achieve proper scaling. Participants confirmed that discrepancies could arise from bin size choices and potential data convolutions.

PREREQUISITES
  • Understanding of density of states (DOS) in quantum mechanics
  • Familiarity with Hamiltonian eigenvalues
  • Knowledge of histogram construction and bin size effects
  • Basic grasp of unit conversions in physical measurements
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  • Research the mathematical formulation of density of states (DOS) in quantum systems
  • Learn about histogram normalization techniques in data analysis
  • Explore the implications of bin size selection on histogram accuracy
  • Investigate unit conversion methods between meV and cm-1 in physical contexts
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Physicists, data analysts, and researchers working with quantum mechanics, particularly those involved in analyzing eigenvalues and density of states in Hamiltonian systems.

Niles
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Hi guys

I have an analytical expression f(x) for my density of states, and I have plottet this. Now, I also have a complete list of my Hamiltonians eigenvalues.

When I make a histogram of these eigenvalues, I thought that I should get an exact (non-continuous) copy of my plot of f(x). They have the same form, but they are not to scale, i.e. f(x) has a maximum at ~100, while the histogram has a maximum at ~60.

Is there an error somewhere, or am I wrong to expect that they should be "exact" copies?
 
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I am not sure whether i am correct!
some scale factor for eigenvaues or for DOS?
Also check with units of y-axis (i often get these errors when i convert meV to cm-1)
or your data is convoluted already?
 
Last edited:
The units are the same (I have checked).

So you believe they should be the same?
 
The units of a numerical histogram depend on the bin size chosen. Are you remembering to divide by the width of the bin?
 
Manchot said:
The units of a numerical histogram depend on the bin size chosen. Are you remembering to divide by the width of the bin?

Isn't a histogram unit-less?
 
Last edited:
Yes, a histogram is unitless. That's why you need to divide by the bin size to get it in units of 1/energy (or 1/energy/volume, if that's what you're dealing with).
 
I see. Thanks - I'll play around with it to see if it works. Thanks.
 

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