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Complete sets and eigenvalues question

by VortexLattice
Tags: eigenvalues, sets
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VortexLattice
#1
Nov9-12, 11:40 AM
P: 146
Let's say I'm looking at the infinite square well. Typically, given some arbitrary initial (normalized) wavefunction, we can decompose it into a linear combination of components of the complete set (on the interval [-a,a] or whatever) of sin's and cos's. Then, if you measure something like the energy, you get one of the eigenstates (one of the sin's or cos's), and you measure the energy associated with that eigenstate.

But there are many complete sets, sin and cos are just one of them. So, let's say we chose some other one. Obviously, because it's complete, you could decompose the initial wavefunction into a linear combo of this set with the same average energy. But this set might have a different spectrum of energy eigenvalues. But this seems like a contradiction, because nature doesn't care what math you're using.

Could this happen? If not with the infinite square well, with an unbound particle?
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Bill_K
#2
Nov9-12, 01:19 PM
Sci Advisor
Thanks
Bill_K's Avatar
P: 4,160
There's only one complete basis of states which are the eigenstates of energy. You can decompose the given state using another complete basis, but since they aren't the eigenstates, each of them will have some probability amplitude of having any one of the energy eigenvalues.

Let |ψ> be the initial state, and let |n> be the energy eigenstate with energy En. Then |ψ> = Σ |n><n|ψ>, showing that <n|ψ> is the probability amplitude to measure energy En.

Now let |α> be the other complete set of states. Then we can expand |ψ> in this basis also, |ψ> = Σ |α><α|ψ>, showing that <α|ψ> is the probability amplitude that the original state will be in state |α>.

But |α> are not eigenstates of energy. To find out how they are related, we must expand |α> again, |α> = Σ |n><n|α>, where <n|α> is now the probability amplitude that state |α> will have energy En. Putting the two things together, |ψ> = Σ |α><α|ψ> = Σ Σ|n><n|α><α|ψ>. So the probability of measuring energy En is Σ <n|α><α|ψ> = <n|ψ>.
VortexLattice
#3
Nov9-12, 02:36 PM
P: 146
Quote Quote by Bill_K View Post
There's only one complete basis of states which are the eigenstates of energy. You can decompose the given state using another complete basis, but since they aren't the eigenstates, each of them will have some probability amplitude of having any one of the energy eigenvalues.

Let |ψ> be the initial state, and let |n> be the energy eigenstate with energy En. Then |ψ> = Σ |n><n|ψ>, showing that <n|ψ> is the probability amplitude to measure energy En.

Now let |α> be the other complete set of states. Then we can expand |ψ> in this basis also, |ψ> = Σ |α><α|ψ>, showing that <α|ψ> is the probability amplitude that the original state will be in state |α>.

But |α> are not eigenstates of energy. To find out how they are related, we must expand |α> again, |α> = Σ |n><n|α>, where <n|α> is now the probability amplitude that state |α> will have energy En. Putting the two things together, |ψ> = Σ |α><α|ψ> = Σ Σ|n><n|α><α|ψ>. So the probability of measuring energy En is Σ <n|α><α|ψ> = <n|ψ>.
Ahhhh, right. Thanks!

Thoros
#4
Nov10-12, 06:17 AM
P: 21
Complete sets and eigenvalues question

Quote Quote by Bill_K View Post
So the probability of measuring energy En is Σ <n|α><α|ψ> = <n|ψ>.
Actually the probability of measuring energy E is the real number associated with the probability amplitude |<n|ψ>|2 = <n|ψ><n|ψ>*.

Sorry if it's a little too picky, it's not really relevant to the question at hand.


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