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1. Jan 27, 2017

Gianmarco

Hey everyone, I've been doing some quantum mechanics but I think I have yet to fully grasp the meaning of eigenstate. What I mean is, I understand that an eigenstate $x$ is such that, if we have an operator $\hat{A}$, it satisfies $\hat{A} x=\lambda x$ and so $\hat{A}$ represents a "dilation" of the eigenstate in the mathematical sense. Is there a way to have a physical interpretation of this though? I'm thinking about the operators $\hat{x}, \hat{p}$ and $\hat{H}$.
I've also read that when we measure an observable on a state, the process "transforms" the state into an eigenstate and if we measure it again we will obtain the same answer. What does it mean in practice?
Thanks to anyone who will bother reading this :)

2. Jan 27, 2017

PeroK

You may get a better answer, but this information is available in any introductory QM text and freely available online in many reliable sources.

Someone on here may take the time to explain it all, but it will only be information that is already available to you.

Better to ask specific questions here that may help clarify the material you are learning from.

3. Jan 27, 2017

Gianmarco

Hi PeroK, I understand what you are saying but the books that I'm using for this class are very formal about the whole thing. I understand the mathematics but what I'm looking for is a physical interpretation of it.

4. Jan 27, 2017

PeroK

Can you make any sense of this?

http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/Scheq.html#c5

5. Jan 27, 2017

hilbert2

If the eigenvalue $\lambda$ is 1, the operation doesn't change the norm. If the operator doesn't represent a physical measurable quantity, the eigenvalue can even be a complex number. Quantum mechanics contains a lot of complications that are related to the infinite-dimensionality of the state spaces. Think about a three-state system that has a Hamiltonian operator $H$ and eigenstates $|1>$, $|2>$ and $|3>$, which correspond to eigenvalues E=1,E=2 and E=3. Now if you have an arbitrary state $|\psi >=a|1> + b|2> + c|3>$ and you operate on it $N$ times with the Hamiltonian ($N$ is some very large number), the resultant state vector $a|1> + 2^{N}b|2> + 3^{N}c|3>$ is approximately a multiple of $|3>$ (if the number $N$ was large enough). In the case where there are infinitely many eigenstates of $H$, you can't just do that when you want to find the eigenstate with largest eigenvalue.

6. Jan 27, 2017

Truecrimson

I don't think there is any physical meaning to the dilation. The eigenvalues are just labels of the measurement outcomes. Note that I can shift the zero eigenvalue to any eigenstate I want, but the zero vector doesn't correspond to a state vector.

You should think of this condition as a characterization of repeatable measurements rather than of a general measurement. A standard counterexample is a photon counting which absorbs and destroys the photon(s).