# Some Queries about Eigenvalues and vectors

1. Jan 5, 2014

### Shan K

Hi,
I am studying quantum mechanics right now and I can't able to understand some questions about Eigenvalues and Eigenvectors.
1. What does the eigenvalue tell us about the quantum mechanical operators i.e. if we operate a momentum operator on ψ what does the Eigen value of that equation tell us about the momentum of the particle for which is the wave function ?
2. I have read that if we operate any kind of operator on ψ, it will collapse to an Eigen state . What does it mean ?
3. Does the meaning the Eigen function and the Eigen state is same ?
Sorry if these questions are asked before . Then just provide the link .

2. Jan 5, 2014

### Staff: Mentor

For an eigenfunction, the eigenvalue is "the" value of the corresponding observable - here the momentum.

If you measure momentum (as an example), you get a single value for the momentum and the wavefunction afterwards is an eigenfunction of the momentum operator, even if the initial wavefunction is not.

Momentum is not a good example as mathematically, there are no proper eigenfunctions of it, but that is a technical detail that does not matter here.

An eigenfunction is a possible state for the particle.

3. Jan 5, 2014

### Shan K

What is meant by the "value of the corresponding observable" ? Is it means that the particle has a momentum equal to its Eigen value ?

If the initial wave function is not an Eigen function of the momentum operator then how can we measure the momentum?

4. Jan 5, 2014

### Staff: Mentor

In QM, a "state" is an abstract thing that can be represented in various ways, one of which is by using a "[wave] function": a solution of Schrödinger's equation in terms of position, ψ(x). "Eigenstate" and "eigenfunction" have the same relationship. They are special cases of the more general "state" and "[wave] function".

(By the way, a friendly tip : in English we don't put a space before the punctuation mark, e.g. a period or question mark, at the end of a sentence.)

5. Jan 5, 2014

### Staff: Mentor

Here it has the same meaning as in classical mechanics.
After a momentum measurement, it has a momentum equal to one of the momentum eigenvalues (there is not just a single one).

The shape of the initial wavefunction does not matter for momentum measurements. You can do them similar to classical mechanics - measure the curvature in a magnetic field, measure the velocity, or something similar.

6. Jan 6, 2014

### Shan K

Are you trying to say that before the measurement of the momentum of the particle, the particle may not have the momentum equal to the Eigenvalue?
And is there any kind of possibility of having multiple eigenvalues for the same operator and the same Eigen function?

I can not able to understand this point can you please explain it in some more detail?

7. Jan 6, 2014

### vanhees71

Perhaps one has to always add the interpretation one follows, before answering that question. I'm a proponent of the minimal statistical interpretation (ensemble interpretation). I think the idea of a state collapse is inconsistent with Einstein causality and thus should be abandoned in the foundation of quantum mechanics from the very beginning. That said, I think the following set of postulates describes quantum theory pretty nicely:

(1) The state of a quantum mechanical system is an equivalence class of preparation procedures, reprsented by a Statistical Operator $\hat{R}$ on a (rigged) Hilbert space. A Statistical Operator is a self-adjoint positive semidefinite operator with trace 1.

The projection operarors
$$\hat{R}_{\psi}=|\psi \rangle \langle \psi|,$$
with a normalized Hilbert-space vector $|\psi \rangle$ represents a pure state. Any other kind of statistical operator is called a mixed state or mixture.

(2) Observables are represented by self-adjoint operators on Hilbert space. Their spectrum determines the possible outcomes of measurements. If $|a,b \rangle$ is a complete set of orthonormalized generalized eigenvectors, then the probability to find the value $a$ in the spectrum of the operator $\hat{A}$ representing an observable $A$ is given by
$$P(a|R)=\sum_b \langle a,b|\hat{R}|a,b \rangle.$$
If the $b$ also contains a continuous part one has to use the appropriate integral.

In addition you need further axioms to define the dynamics of the system, involving the Hamiltonian and (unitary) time evolution, but that's not necessary to discuss the meaning of states and observables.

8. Jan 6, 2014

### Shan K

Many Many thanks Vanhees71 for your post but linear vector space is not known to me. So can you explain it qualitatively? I am studying QM book written by Singh and Singh and there they have stated the term eigenvalue to explain the energy of the particle in a 1d box and in other applications of the Schrodinger equation. I have studied eigenvalues and eigenvectors in matrix but didn't know their physical meaning, after studying it in QM now I want clear those concept from its very basic. So it will be very useful if you explain it qualitatively.

9. Jan 6, 2014

### vanhees71

How does your book introduce states and observables then, if they don't use the Hilbert-space language?

10. Jan 6, 2014

### atyy

Collapse is not necessarily inconsistent with Einstein causality, provided that the observables are restricted to those consistent with causality, eg. http://arxiv.org/abs/hep-th/0110205.

11. Jan 6, 2014

### Shan K

They only have said that all the observables in classical mechanics are represented as an operator in QM. They mainly focuses on solving the Schrodinger equation for various potentials and analyzes the results.

12. Jan 6, 2014

### bhobba

Sigh.

Its obviously one of those books that show you how to solve problems rather than explain whats going on. Many beginning books in QM are like that. We all must start somewhere and rather than tackle the difficult problem of explaining exactly what's going on they just dive right in. But a thinking student can be left scratching their head.

Even though its at graduate level, THE book IMHO to get is Ballentine - Quantum Mechanics - A Modern Development:
https://www.amazon.com/Quantum-Mechanics-A-Modern-Development/dp/9810241054

See if you can dig up a copy from a library and have a look at the first 3 chapters. If you are just starting out its probably a bit advanced but providing you don't get too caught up in the detail you will likely pick up what's REALLY going on. Believe it or not QM basically depends on just 2 axioms. Once you understand this things are a lot clearer.

Later, once you have progressed a bit you can study that whole book. It had a big effect on me - there were many areas it cleared up issues with eg the real basis of Schrodinger's equation. It's not pulled out of the air or based on some loose analogies with Poisson brackets - if you don't get that one don't worry - learn it correctly from Ballentine and you wont have to worry about 'unlearning' the handwavey stuff elsewhere.

Thanks
Bill

Last edited by a moderator: May 6, 2017
13. Jan 6, 2014

### Staff: Mentor

Some introductory books start with the Schrödinger equation in position space and focus on solutions to it, and discuss the abstract Hilbert-space / vector-space stuff only in later chapters. Some books do the vector-space stuff more rigorously than others.

14. Jan 6, 2014

### JorisL

I dunno. Maybe it prevents having to unlearn stuff. But it comes at the cost of taking a lot of time to properly grasp the stuff said.

I myself had my first introduction using Griffiths book. In my opinion it is a nice starting point for a purely introductory course. The later chapters not as good as the first few (in my opinion).

Also I don't know how well of a job Ballentine does introducing symmetries (and the inevitable groups generating those). I kind of like Weinbergs book for that stuff. Except the 'funky' (as my lecturer called it) notation.* That's not something for an undergraduate text I reckon but a second course could use that book.

* Until now I've not encountered any example where I'd prefer to use dyads approach. Probably I'm biased toward Dirac's bra-ket notation because it's familiar but the reason he gives in one of his footnotes seems hand-wavey to say the least.

Sorry for the kind of off-topic reaction wanted to chime in regarding the books.

Last edited by a moderator: May 6, 2017
15. Jan 6, 2014

### atyy

The momentum operator has many eigenvalues and eigenstates (also known as eigenvectors). When you make a measurement of momentum, the result you get for the momentum is one of the eigenvalues of the momentum operator.

An operator can act on any state (also known as a vector). In general, an operator will change the state in a complicated way. However, for each operator, there are some states for which the action of the operator is simply multiplication by a number. These are the eigenstates of the operator. The number by which the operator multiplies an eigenstate is the eigenvalue of that eigenstate. In general, an operator has many eigenstates, each with its own eigenvalue.

The collapse to an eigenstate means that when you measure the momentum of a state, the resulting momentum is one of the eigenvalues of the momentum operator, and the state after the measurement is an eigenstate of the momentum operator. (This is not strictly right, as mfb says above, but it's conceptually ok.)

Yes, eigenfunction and eigenstate and eigenvector are the same (again, oversimplified, but conceptually ok at this level).

A vector can be represented as a column of numbers, eg: (f1,f2,f3).
In this example, the indices of the vector are 1,2,3
We can rewrite the column of numbers as (f(1),f(2),f(3)).

Similarly, a function f(x) can be thought of as a representing a vector whose index is continuous.

Last edited: Jan 6, 2014
16. Jan 6, 2014

### WannabeNewton

In my opinion he does a brilliant job with it. He is very explicit with the calculations as well which is always a plus.

17. Jan 6, 2014

### WannabeNewton

In a sense yes but this is a conceptual rabbit hole. Do you remember from linear algebra how a finite-dimensional vector space (equipped with an inner product) always has an orthonormal basis and that any vector in the space can be written as a finite sum of these basis vectors (if you don't then recall that the basis part is a consequence of Zorn's lemma and the orthonormal part is a consequence of the Gram-Schmidt algorithm)? For example a vector $\vec{v}\in \mathbb{R}^{3}$ can be represented as $\vec{v} = v_x e_x + v_y e_y + v_z e_z$ where the triple $\{e_x,e_y,e_z \}$ is the standard Cartesian orthonormal basis.

Furthermore, a self-adjoint matrix (that is, a matrix which is equal to its complex conjugate transpose) always admits a set of eigenvectors that form an orthonormal basis for the vector space it acts on (this is a consequence of the finite-dimensional spectral theorem). Moreover the eigenvalues of a self-adjoint matrix are always real-valued.

Well these things carry over to the infinite-dimensional vector space formulation of QM, in loose terms (I say this because I don't want to go into the mathematical subtleties). This vector space (again in a loose sense) contains the information about states of a physical system sort of like in classical mechanics where we use $\mathbb{R}^{6}$ to talk about the state $(x,p)$ of a free particle, except in QM state refers to something more abstract (see vanhees' post #7).

Anyways, given a self-adjoint operator that corresponds to a dynamical variable (e.g. momentum operator) we can find its eigenvectors (eigenfunctions) and use them appropriately to form an orthonormal basis for the underlying vector space. Any general vector in the space can now be written as a unique linear combination of these eigenvectors except that unlike in the finite dimensional case, now the set of eigenvectors can be countably infinite or even continuous so the linear combinations can be infinite sums or integrals depending on the physical system and the associated boundary conditions.

If we do this with the momentum operator, for example, then any general state vector in the space can be written as a linear combination of momentum eigenvectors. The associated eigenvalues correspond to the possible momentum values that you can get upon measurement, each carrying an associated probability of showing up. But before measurement this state is still in a linear combination of momentum eigenvectors (in general) or, in other words, is in a superposition of various momentum eigenstates each with a probability weighted momentum eigenvalue that can potentially show up upon measurement.

Sure.

EDIT: I think delving too much into the abstract isn't going to help you much. Just take a look at the standard concrete examples. Chapters 1 and 2 of Griffiths should do the trick. Good luck.

Last edited: Jan 6, 2014
18. Jan 6, 2014

### Staff: Mentor

What do you mean with the eigenvalue?
The one that comes out of the measurement?

Before the measurement, you cannot say "the particle has a momentum of [some value]", right.

No, an eigenfunction has exactly one eigenvalue.

19. Jan 6, 2014

### bhobba

Griffiths is a good book - but its a bit pricey so would not be my first choice starting out.

That said Griffith states explicitly he will teach you to basically solve problems and leave the difficult issues of interpretation to one side. Of itself that is fine, and a reasonable way to proceed, but it does leave the thinking student with unanswered questions.

I am not suggesting studying Ballentine - unless your math background it really good its not the book to start with. And even then it's probably better to warm up with something like Griffiths first.

I am simply suggesting getting the GIST of the first 3 Chapters, and probably Chapter 9 as well, not a detailed study. It means reading it, but not going through the detailed proofs such as the symmetry ones that are rather detailed - simply understand the result - you can do the chug and glug down the road. That way you know that symmetry is the real basis of Schrodinger's equation, why the system is in an eigenstate of the observable after observation (to the OP its got to do with Born's Rule and physical continuity - it not something pulled out of a hat - it follows from other physical principles - but only for observations where the original system is not destroyed - beginning books sometimes don't make clear it only applies to the special case where it's not destroyed), that wavefunction collapse is a total non issue with the minimal statistical interpretation, and other stuff the beginning student has difficulty grasping. Even the idea that observation requires an observer - it doesn't - and Ballentine makes that abundantly clear - but beginning books leave that one up in the air and I have replied to guys that have studied QM from Griffiths who think an actual organic conscious observer is required.

Thanks
Bill

20. Jan 7, 2014

### Shan K

.
Yes the eigenvalue which can be got from the eigenvalue equation of the momentum operator.