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View Full Version : Quantum Eigenfunctions
Mr Tulip 03-10-04, 05:20 PM I mucked up Psi and psi earlier, Psi1=psi1... Theres only 2 possible psi's in the problem psi1, and psi2 - this changes a few things I think [unless you presumed there were only 2 all along - thanks for the help sofar]
Hi All.
Got another tricky one.
psi 1 and psi2 are normalized eigenfunctions of the observable A, which are degenerate and hence not necessarily orthogonal. If {psi1|psi2}=c and c is real, find linear combinations of psi1 and psi2 which are normalised and orthogonal to:
a) psi1
b) psi1+psi2
- I used curly brackets for the dirac notation, as didnt want to drop into HTML. Is theres a limit to the number of solutions? or is there a more general solution?
Orthogonality requires {psi1|psi2}=0, and normalisation {psi1|psi1}=1.
Any hints ppl?
Thanks.
Helloween 03-10-04, 05:47 PM bracket psi1 and psi2 can be viewed as a dot product of the two basis vectors. the fact that you get a constant out tells you that they are not orthogonal, but that they are at some angle to each other.
you know this:
{psi1|psi2}=c
now find this:
{psi1|(a|psi1} +/- b|psi2})=0
a{psi1|psi1} +/- b{psi1|psi2}=0
= a +/- b*c = 0
the second part is similar. hope that helps!
Hi All.
Got another tricky one.
Psi 1 and Psi2 are normalized eigenfunctions of the observable A, which are degenerate and hence not necessarily orthogonal. If {psi1|psi2}=c and c is real, find linear combinations of psi1 and psi2 which are normalised and orthogonal to:
a) Psi1
You should write |v<sub>1</sub>> = a|Ψ<sub>1</sub>> + b|Ψ<sub>2</sub>> and write that <Ψ<sub>1</sub>|v<sub>1</sub>> = 0. You find that b = -a/c (we suppose that c is not zero)
And using the fact that |v<sub>1</sub>> is normalized (that is |a|<sup>2</sup> + |b|<sup>2</sup> = 1) you find that:
a = c/sqrt(1 + c<sub>2</sub>) and b = -1/sqrt(1 + c<sub>2</sub>)
b) Psi1+Psi2
You do the same as for the first case (using the fact that c is not -1, which must be treated separately).
You write:
|v<sub>2</sub>> = a|Ψ<sub>1</sub>> + b|Ψ<sub>2</sub>> and when you write that (<Ψ<sub>1</sub>| + <Ψ<sub>2</sub>|)|v<sub>2</sub>> = 0; you get:
(a+b)(1+c) = 0.
If c is not -1, then b = -a, and using the normalisation of |v<sub>2</sub>> you find |v<sub>2</sub>> = (|Ψ<sub>1</sub>> - |Ψ<sub>2</sub>>)/sqrt(2).
If c = -1; this means that |Ψ<sub>2</sub>> = - |Ψ<sub>1</sub>>
and then |Ψ<sub>1</sub>> + |Ψ<sub>2</sub>> = 0, and the solution is that every vector is orthogonal to |Ψ<sub>1</sub>> + |Ψ<sub>2</sub>> .
errandir 03-10-04, 06:07 PM ... find linear combinations of psi1 and psi2 which are normalised and orthogonal to:
a) Psi1
b) Psi1+Psi2This is the GrammSchmidt procedure. There's nothing tricky about it, especially since the ambiguity has been removed by specifying the starting point.
Is theres a limit to the number of solutions?Yes and no. There are infinitely many solutions, but they are all scaled multiples of each other, so they are limited in that respect.
... is there a more general solution?More general that what?
Orthogonality requires {Psi1|Psi2}=0, ...Well, this is a little confusing, since you know that ψ<sub>1</sub> and ψ<sub>2</sub> are <i>not</i> orthogonal.
Well, this is a little confusing, since you know that ψ<sub>1</sub> and ψ<sub>2</sub> are <i>not</i> orthogonal.
He meant just to explain what is orthogonal
Mr Tulip 03-11-04, 04:49 PM Ive only just started using Dirac notation...
But for the first one, we basically say
{apsi1+bpsi2|psi1}=0 and we can expand this to a + bc=0
btw: how do I get this edit to let me use pointy brackets? the code html command wont do it. Greek lettering would be nice too.
Mr Tulip 03-11-04, 04:54 PM As an aside, I find it very interesting/annoying that the QM course has tons of ppl in it who havent done linear algebra. We are struggling a bit. One of the guys in the subject went through GrammSchmidt yesterday with a few of us.
As an aside, I find it very interesting/annoying that the QM course has tons of ppl in it who havent done linear algebra. We are struggling a bit. One of the guys in the subject went through GrammSchmidt yesterday with a few of us.
Ouch! Linear Algebra gets Crisp's "Must have seen" label before starting QM ;).
But seriously, how can you understand quantum mechanics if you do not know what a Hilbert space is (or a vectorspace for a start), or what the role of operators, eigenvalues, eigenvectors is... Teaching it along with QM seems not a good idea either, because then the QM gets obscured even more by mathematics ;)
Mr Tulip 03-12-04, 03:56 AM Ive only ever used operators in a QM course.
And hilbert spaces were a shock to the system! function spaces indeed.
errandir 03-12-04, 12:32 PM Teaching it along with QM seems not a good idea either, because then the QM gets obscured even more by mathematics ;)I have always learned the required math during the physics course. I.e. I learned calculus from freshmen physics, and I learned what a hilbert space, eigenwhatsit, and vector are from Graduate QM. Maybe this is the main source of my problem ;) .
I have always learned the required math during the physics course. I.e. I learned calculus from freshmen physics, and I learned what a hilbert space, eigenwhatsit, and vector are from Graduate QM. Maybe this is the main source of my problem ;) .
i always knew the requisite math <i>before</i> taking the physics class, and i feel that this is a much better approach.
James R 03-12-04, 11:50 PM It's very common in Australia to learn the relevant maths at the same time as you're doing the physics. It does make things difficult though.
PhysMachine 03-13-04, 11:44 AM First off, it's possible to learn quantum mechanics entirely as finding wave functions in the wave mechanics model. This is the approach that the Griffiths Introduction to Quantum Mechanics takes, and is a perfectly valid one for an introductory course. However, to really work with quantum mechanics you have to think in terms of state vectors, which is the approach that the Shankar Principles of Quantum Mechanics takes, and in my opinion is the better approach.
As for the problem, it is just gramm-schmidt orthonormalization of an inner product space, so yeah, not too hard.
Lemme pose to you a slightly more interesting problem:
PROVE that the eigenvectors of an Hermitian operator are automatically orthogonal (keeping in mind that in the degenerate case eigenvectors can always be chosen to be orthogonal, which is a distinct difference).
oxymoron 03-13-04, 08:03 PM I learnt linear algebra (eigenfunctions, vector spaces, hermitian, etc...) in my first semester of first year. I think it is better because now (2nd year) when we begin to really get into QM and GR the maths is already there and I can concentrate on just the physics.
First off, it's possible to learn quantum mechanics entirely as finding wave functions in the wave mechanics model. This is the approach that the Griffiths Introduction to Quantum Mechanics takes, and is a perfectly valid one for an introductory course. However, to really work with quantum mechanics you have to think in terms of state vectors, which is the approach that the Shankar Principles of Quantum Mechanics takes, and in my opinion is the better approach.
Griffiths also uses the very same state vectors as Shankar (I presume), and he also has a chapter two which introduces (very elimentary) some of the math required. However, it is very limited and simply not enough to give you a broad perspective of what is going on.
PROVE that the eigenvectors of an Hermitian operator are automatically orthogonal (keeping in mind that in the degenerate case eigenvectors can always be chosen to be orthogonal, which is a distinct difference).
A true classic :). (*) is used both as Hermitian and complex conjugate:
Assuming H|φ> = μ|φ> and H|ψ> = λ|ψ> are two _distinct_ eigenvectors/eigenvalues:
1) <ψ | H | φ> = <φ | H* | ψ>*
because of the properties of the scalar product.
2) <ψ | H | φ> = <φ | H | ψ>*
because H = H* Hermitian
3) μ <ψ | φ> = λ* <φ | ψ>*
because they are eigenvectors
4) (μ - λ*) <ψ | φ> = 0
4b) (μ - λ ) <ψ | φ> = 0
since eigenvalues of Hermitian operators are real
5) <ψ | φ> = 0
because μ is different from λ and the expression must equate zero.
I just love this proof, together with the proof that the eigenvalues of Hermitian operators are real, this is simply beautifully simple algebra ;)
Bye!
Crisp
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