Hi Steve,

This is very simple to prove, and I am rather taken aback that nobody has yet given you a proof. Quantum states are generally represented as vectors. The reason we use "eigenfunction" instead of say "eigenvector" if because it is often convenient to use functions which form a vector space, rather than, say, column vectors. However, you can reason about them in the same way as column vectors, and indeed for finite dimensional systems such as spins we do often use column vectors. So, I'll prove this using column vectors, but exactly the same argument can be applied with functions.

Take two solutions to an eigenvalue equation, u and v which correspond to the same eigenvalue e. So we have Mu = eu and Mv = ev. In this case, if we take any linear combination w=au+bv, we can see that it must also be an eigenvector of M with eigenvalue e: M(au+bv)= Mau + Mbv = aMu + bMv = aeu + bev = e(au+bv). Thus Mw = ew. So any linear combination of eigenvectors with the same eigenvalue simply results in another eigenvector with the same eigenvalue. This is not true if the eigenvalues are different, and this fact accounts for all dynamics.

This is, in fact, nothing to do with quantum physics, but rather a simple consequence of linear algebra, so it shows up in lots of areas of physics (such as electromagnetism). As Alexander mentions, however, in quantum mechanics, in order to preserve probability, the only physical solutions are linear combinations which have been normalised to be unit vectors.

Hello

First, is correct that eigenfunctions are those functions that satisfy eigenvalue equations. but dont is correct that eigenvalue equation is linear. I attach the demonstration [1]. In quantum mechanics the schrodinger equation defined the evolution of the quantum systems. We say that the schrodinger equation is linear, then all linear combinations of its solutions are also solutions. I attach the demonstration [2].

Steve, good question; it always confuses student when they study QM. OK, in mathematical sense, yes, the Schroedinger eqn is linear and any linear combination of its eigenfunctions mathematically speaking is supposed to be also a valid solution  --  superposition principle. Yet almost in the same breath you've got to tell them "BUT!!!" -- here is a restriction! Linear combination is the the sum of some (or all) eigenfunctions with ARBITRARY coefficients. And here is where a red light is flashing; NOT ANY coefficients are good for QM! A properly normalized eigenfunction has a property whereby a full integral of the square of it absolute value over the entire space is equal "1", meaning that it is a pure state, and no other states are excited, which is equivalent to the statement that the particle has a probability "1" to be in that state. In general, when we have a mixed state, i. e. the particle may be found in other eigenstates too, the probability of it to be found in the "i"-th  state is simply the square of absolute of the coefficient "A_i"("amplitude") assigned to the respective NORMALISED eigenfunction, i. e. "probability_i = IA_i|^2"   (it is often called "populations of the state").

Here comes a crunch imposed by QM -- since the total probability for the particle to be found SOMEWHERE is 1,  the limitation on those coefficients is that their total sum must be "1"; "sum (|A_i|^2 over all "i" = 1 ". So, the probability nature of QM imposes a harsh limit on the "freedom of  linear combinations" or "freedom of superposition principle". Unfortunately, this little terminological contradiction is sort of tackled under the rug in most (if not all)  of QM textbooks.

Note that this situation occurs in QM only; in classical mechanics or electrodynamics, a  superposition principle is a straightforward and 'honest" principle -- ANY combinations of eigenmodes in a linear system is physically allowed (unless of course, the linearity is broken).

Dear Steve

I think superposition principle is valid in classical physics which can be describe in linear equations. However we should distrust superposition principle even if the equations are linear in QM.

I encourage you to read following papers.

Article Unobservable Potentials to Explain a Quantum Eraser and a De...

Hi Steve,

This is very simple to prove, and I am rather taken aback that nobody has yet given you a proof. Quantum states are generally represented as vectors. The reason we use "eigenfunction" instead of say "eigenvector" if because it is often convenient to use functions which form a vector space, rather than, say, column vectors. However, you can reason about them in the same way as column vectors, and indeed for finite dimensional systems such as spins we do often use column vectors. So, I'll prove this using column vectors, but exactly the same argument can be applied with functions.

Take two solutions to an eigenvalue equation, u and v which correspond to the same eigenvalue e. So we have Mu = eu and Mv = ev. In this case, if we take any linear combination w=au+bv, we can see that it must also be an eigenvector of M with eigenvalue e: M(au+bv)= Mau + Mbv = aMu + bMv = aeu + bev = e(au+bv). Thus Mw = ew. So any linear combination of eigenvectors with the same eigenvalue simply results in another eigenvector with the same eigenvalue. This is not true if the eigenvalues are different, and this fact accounts for all dynamics.

This is, in fact, nothing to do with quantum physics, but rather a simple consequence of linear algebra, so it shows up in lots of areas of physics (such as electromagnetism). As Alexander mentions, however, in quantum mechanics, in order to preserve probability, the only physical solutions are linear combinations which have been normalised to be unit vectors.

Thank you to you all.

I still think there is contradiction in this.  An eigenvector is any vector that is mapped by  the operator, to itself (upto some scaling which we call the eigenvalue).  Am I not correct that, generally, linear combinations of the eigenvectors do not map like this.  That is to say: they do not map like eigenvectors.  If they did, they would be eigenvectors.  They do not satisfy the eigenvalue equation.  Is there something I am missing here?

You are confused a bit. Superposition of the eigenfunctions multiplied by time exponents is a solution of time dependent equations of motion. This is general for linear  equation. The way you formulated it, it is obviously incorrect

Dear All Colleagues

Really, I am very much glad to read all nice comments. I am able to understand a lot of doubts related to eigen-values and eigen functions.

Because I am teaching Phys. 354 (Quantum Mechanics-1) since last 5 years.

I am not saying linear combinations of eigenvectors are not meaningful. The Superposition Principle is clearly important.  I am saying it is misleading to teach students, as we have always done, that the linear combinations are solutions. To call the general linear combination the general solution,  I don't know what is meant by this. And I wonder where this terminology came from.

Thank you.  I see now I misunderstood the concept.  Pity, though, that I attracted a remark intend on ending conversation.  I don't see there is place for that here.

Hello, for some reasons my browser won't let me see the other answers that have already been given, so I apologize if what I'm gonna say is redundant.

In Quantum Physics, the state of a system belongs to a Hilbert space. Mathematically speaking, this state can be seen as a N-dimensional vector, with the possibility of N being infinite. But, for simplicity, it is sufficient to consider that N is finite. Now let us consider the example of a position vector, R, in the 3-dimensional Euclidean space. As you know, we can choose a basis of 3 linearly independent vectors of this space (usually a set of 3 orthogonal vectors in the x, y, z directions, but not necessarily), and express R as a linear combination of these basis vectors. Then the components of R are just the coefficients of this linear combination. The point is that the choice of the basis of states is purely arbitrary. You can always decompose R in any basis you choose.

It is the same for a quantum state. You can always express it as a linear combination of basis states of the Hilbert space. The only difference with the case of the position vector R is that the number of basis states is N, and the coefficients are complex numbers. So it is mathematically exact to decompose any quantum state in any arbitrary basis.

Now, the Hamiltonian of a system is a Hermitian operator, which implies that the set of eigenstates of the Hamiltonian constitutes a basis of the Hilbert space. In conclusion, it is correct and exact to decompose any quantum state in the basis of eigenstates, as well as it is in any other basis.

The answer is pure and simple:  Yes.  This is true for any linear equation, according to the principle of linear superposition.  Schrodinger's equation is no exception.  There is no distinction here between classical and quantum.  However, re the general solution for psi in Schrodinger's wave equation, one must then normalize it to satisy the "probability = one" condition.

I second Joseph Fitzsimmons'  reply:  any linear combination of two eigenfunctions belonging to the same eigenvalue is also an eigenfunction of a linear operator belonging to that eigenvalue.  A nontrivial linear combination of eigenfunctions belonging to different eigenvalues is not an eigenfunction.  There may be some confusion here with solutions of the time-dependent Schroedinger equation: being linear, a superposition of any two solutions is also a solution.  Solving the time-dependent Schroedinger equation is not normally construed as an eigenvalue problem;  however,  by grouping all the nonzero terms on (say) the left side of the equation with zero on the right, any solution is, in a sense, an eigenvalue with eigenvalue zero.  Hence all solutions are degenerate, i.e., belong to the same (zero) eigenvalue.

Erratum:  read "...an eigenfunction with eigenvalue zero." in the second-to-last sentence.

Thank you Gerhard.  You say exactly the point I had missed.  

Use a trial wave function constructed from the linear combination of (basis)wave functions with variational parameter(s) and then do a minimization of the eigenvalue obtained to get the actual variational parameter that minimizes the expression. Plug this into the original trial wave function to get the best one.

The usuall answer, unless I misunderstand the question:

Before measurement the state is in a linear superposition of different

eigenstates. (not generally determinable)

After measurement, the system jumps into some definite eigenstate, and

one talks about wave function collapse. It gives as result one of the possible

eigenvalues., with a probability which depends on the initial superposition.

The initial superposition is suppose to obey the schrodinger equation, and in this sense is a solution-

Horrible answer, but this is not my fault.