Derive RMS around average of a hermitian operator

In summary: I would not be comfortable teaching myself quantum mechanics. :)In summary, the lecturer showed us how to calculate the RMS around the expectation of an operator by considering the E of a system in equal superposition of two energy eigenstates. He then says "This gives some measure of how far off we would be likely to be from the expectation value for an average measurement. The general case for this can be derived. For an operator Q, then the wavefunction can be expressed as an expansion in eigenstates of Q" and writes the wavefunction as a sum over the eigenstates with coefficients whose modulus squared represent the probabilities of each eigenstate.
  • #1
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I'm in my second year of a physics degree and my QM lecturer showed us how to calculate the RMS around the expectation of an operator by considering the E of a system in equal superposition of two energy eigenstates u_1 and u_2. He then says
"This gives some measure of how far off we would be likely to be from the expectation value for an average measurement. The general case for this can be derived. For an operator Q, then the wavefunction can be expressed as an expansion in eigenstates of Q " and writes the wavefunction as a sum over the eigenstates with coefficients whose modulus squared represent the probabilities of each eigenstate.

Could someone show me the derivation please?

Thanks!
 
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  • #2
Can you post the way that your lecturer derived it for the two-state case?
 
  • #3
Hi there. It's not really a derivation so much as an example but here goes:

ψ = 1/sqrt(2)*(u_1 + u_2)

so <E> is 0.5(E_1 + E_2)

E_1 - <E> = 0.5(E_1 - E_2)
E_2 - <E> = 0.5(E_2 - E_1)

so (∆E)^2 = ∆E^2 = 0.5{0.5[E_1 - E_2]^2 + 0.5[E_2 - E_1]^2}

which gives an rms of:

∆E = 0.5|E_2 - E_1|
 
  • #4
Ok. There were a few tricks done in there that were possible due to the fact that the state was an equal superposition of the two states--a lot of things canceled out that wouldn't necessarily do so otherwise. So probably the best step to do next is to try to work it out in a case where the system isn't in an equal superposition. So try [tex]\Psi = a_1 U_1 + a_2 U_2[/tex], and see if you can work out what happens then. Once that's done, it's easy to generalize the sum to summation or integral, and then you've proven the general case.

(BTW, this smells like a homework problem. If so, you might want to look at the rules regarding homework questions https://www.physicsforums.com/showthread.php?t=414380".)
 
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  • #5
I assure you this isn't a homework questions, don't worry, I like solving problems too much to ever cheat on a homework assignment :).

Thanks for your help but I'm now a little confused. In my notes, my lecturer seems to be finding <E>. E is the notation he has used previously for the eigenvalue associated with an eigenstate of the Hamiltonian operator. I have understood however, that in QM we are interested in the expectation of an operator so if we're talking about the spread of measurements of energy we would want to find <H>. Do you think he has used inconsistent notation or have I misunderstood something?
 
  • #6
Yes, [tex]\langle H\rangle[/tex] would probably be more appropriate. Expectation values are calculated for operators (or, more accurately still, operators applied to a given state.) Quantum mechanics is rife with notational inconsistencies like that--sometimes through laziness, other times because it's hard to know what a good notation even is, and still other times because deliberately bending the notation can make it easier to write the equations down.

For instance, one will often see an eigenvalue variable used inside of a bra or ket, to refer to a state which has that eigenvalue. This leads to WTF-inducing equations like [tex]\hat{H}|E\rangle = E|E\rangle[/tex]. However, despite the fact that it takes the brain a minute to decode the notation, it ends up being useful, because then you can refer to a whole set of states easily. For instance, if there are n energy levels [tex]E_n[/tex], you could say something like [tex]\Psi = \sum{a_n|E_n\rangle}[/tex], and easily denote that you're talking about the expansion of the state into the energy basis. In fact, you'll probably end up doing exactly that when you get to the final stages of generalizing your original question.
 
  • #7
I see. Thanks for your help! I managed to convince myself that |a_n|^2 represent probabilities and got through the lecture notes. I'm not sure if you'd want to give away this information but are you a professor? I'm curious as I'd like to find out just how good you need to be to get to that level and asking my professors in person would be more than embarrassing. I'm still learning mathematics that I find conceptually difficult but the PhD students that help out at some of our sessions perform these manipulations with such ease it's almost embarrassing! How hard did you find Hilbert spaces and contour integration when you were an undergrad?
 
  • #8
Cool, glad you figured it out.

No, I'm not a professor--I'm actually not even a physics major, I'm a computer engineer. I've been a huge math and physics nerd for years, though, and have gradually been trying to see how much of this stuff I can teach myself. A sick hobby, yes, but to each their own I guess. :)

So I'm still learning some of this too, but I can tell you with some confidence that all the basic math does make sense eventually. Hilbert spaces were really hard for me to wrap my head around at first (I still don't think I completely get them), but in essence they're just vector spaces, so it pays to know your linear algebra pretty well. The key insight to understanding any vector space is that you never actually know what the elements themselves are, the only thing that matters is how they relate to each other. Once you really, truly grok that, the rest is downhill. As far as contour integration goes, I haven't really done all that much of it so far, and am actually trying to learn a little bit about it right now in order to understand some stuff in quantum field theory. A course in complex analysis back in school would have made things a lot easier. :/
 

Related to Derive RMS around average of a hermitian operator

What is a hermitian operator?

A hermitian operator is a mathematical term used in quantum mechanics to describe a type of operator that is self-adjoint. This means that the operator is equal to its own conjugate transpose.

Why is it important to calculate the RMS around the average of a hermitian operator?

Calculating the RMS around the average of a hermitian operator is important because it allows us to determine the spread or uncertainty of the operator's values. This is useful in quantum mechanics when studying the behavior of particles and their properties.

How is the RMS around the average of a hermitian operator calculated?

The RMS around the average of a hermitian operator is calculated by taking the square root of the average of the squared differences between each value and the average. Mathematically, it can be represented as: RMS = √(1/N * Σ|xi - x_avg|^2).

What does the RMS value tell us about a hermitian operator?

The RMS value of a hermitian operator tells us the width or spread of the operator's values around its average. A higher RMS value indicates a larger spread, while a lower RMS value indicates a smaller spread.

Can the RMS around the average of a hermitian operator ever be negative?

No, the RMS around the average of a hermitian operator cannot be negative. This is because the squared differences used in the calculation are always positive, resulting in a non-negative value for the RMS.

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