QM - Eigenvalues of different States

[Anonymous No.6621136]

(7:40) https://www.youtube.com/watch?v=zNMzUf5GZsQ&feature=youtu.be&t=7m40s
"We've established that eigenvalues of different states are different."
I can't find that anywhere.
Why would different states/vectors imply different eigenvalues?

[Anonymous No.6621178]

Bamp

[Anonymous No.6621188]

I don't know anything about QM, but surely you see why different vectors have different eigenvalues?

[Anonymous No.6621190]

>>6621188
0/10

[Anonymous No.6621191]

>states
>kets
>bras
what the hell are you physicists doing?

[Anonymous No.6621202]

I'm not very knowledgeable about QM, but I think there is nothing preventing a generic operator, with discrete spectrum, from having eigenvalues with multiplicities in a quantum model. So it's more a case-by-case basis.

[Anonymous No.6621204]

>>6621191
No idea...
Here H is hermitian for physical reasons. But that's not enough
What other property of H would imply this 1 to 1 vector-value correspondence?
Any random guesses would help...

[Anonymous No.6621211]

>>6621204
I've never studied physics, but wiki says it's a consequence of some physical principle:
http://en.wikipedia.org/wiki/Orthogonality#Orthogonal_states_in_quantum_mechanics

[Anonymous No.6621222]

>>6621202
There are a lot of operators with degenerate eigenspaces. The two-dimensional harmonic oscillator hamiltonian is a simple example.

[Anonymous No.6621223]

>>6621211
Can you please stop posting if you don't even understand the question?

[Anonymous No.6621225]

>>6621211
This follows from (...) OR that observables are given by hermitian operators.
Welp hermitian matrices are hermitian operators. And hermitian matrices can have two have eigenvalues with 2 eigenvectors right?
I suppose wiki's wrong then?

Still very relevant. Thank you anon.

[Anonymous No.6621230]

>>6621222
>degenerate
http://en.wikipedia.org/wiki/Degenerate_energy_levels
1Dimension
"Since the number of independent eigenfunctions for a given energy E is maximally equal to two, the degree of degeneracy never exceeds two. It can be proved that in one dimension, there are no degenerate bound states for normalizable wave functions. This is true provided the potential under consideration is bounded from below and piecewise continuous."

[Anonymous No.6621266]

>>6621136
>Why would different states/vectors imply different eigenvalues?

It is indeed not necessairily the case. Take for instance the identity operator, 1. Every single nonzero vector of the Hilbert space is an eigenvector with eigenvalue 1.

I have not watched your video to its full extend so I don't know why the speaker says that <span class="math">\lambda_a \neq \lambda_b[/spoiler] but when teaching QM, a simplification often made is to consider only operators with non-degenerate eigenvalues, i.e. such that for any eigenvalue a, the corresponding subspace is of dimension 1. It does simplify proofs and discussion, yet it does not significantly changes the results (as in, about everything demonstrated within this hypothesis can be proved with a little more work in the more general setting).

Now, if you consider a general operator A, and two vectors x and y associated to different eigenvalues a and b, then they are linearly independent (in geometric terms they are not colinear).

If, in addition, you assume A hermitian, x and y are in addition orthogonal.

>>6621191

These are simply physical names for mathematical quantities.
A quantum system is described by a vector (of unit norm) of an Hilbert space so these vectors are called "states" (when you know the state of a physical system you know everything about it).

Bras and Kets are useful notations for vectors of the Hilbert space and its dual space (it allows to quicky write operators and such and to conveniently perform operations).

I don't see where the problem is.

>>6621230

This is valid only for the Hamiltonian (and in a very specific setting), you can still be interested in other operators which may or may not have degenerate eigenvalues.

[Anonymous No.6621272]

>>6621266
>This is valid only for the Hamiltonian
Forgot to mention that.
I believe it's due to it being a second degree operator: which gives a normalizable and a non normalizable solution.

[Anonymous No.6621291]

>>6621272
Thanks anyway for poining that result out, I did not know it.

You can also have 2 non normalizable solutions, for instance for free particles (or electrons in a 1d cristal)