Is there always a real solution for every real Hermitian eigevalue problem?

[AxiomOfChoice]

If we're attempting to solve

<br /> H\psi = E\psi<br />

where H is real and Hermitian, are we allowed to assume \psi is real? Why or why not? My gut tells me the answer is "yes," since we know E is real, but I can't make my idea rigorous.

 

[Chopin]

No, in fact it's very likely that \Psi is not real, at least not everywhere. Consider a simple solution of the free-space Schrodinger Equation, the plane wave:

\psi(x, t) = e^{\frac{i}{\hbar}(kx - E{t})}

Because H = i\hbar\frac{\partial}{\partial{t}}, we have:

H\psi(x, t) = i\hbar\frac{\partial}{\partial{t}}\psi(x, t) = Ee^{\frac{i}{\hbar}(kx - E{t})} = E\psi(x, t)

So \psi is an eigenvector of H, and E is the eigenvalue. E is guaranteed to be real, because H is Hermitian, but \psi(x, t) is not. In fact, because the Hamiltonian is defined to be the time derivative of the wavefunction, a particle's energy is basically defined by how fast its wavefunction rotates through the complex plane.

 

[AxiomOfChoice]

No, in fact it's very likely that \Psi is not real, at least not everywhere. Consider a simple solution of the free-space Schrodinger Equation, the plane wave:

\psi(x, t) = e^{\frac{i}{\hbar}(kx - E{t})}

Because H = i\hbar\frac{\partial}{\partial{t}}, we have:

H\psi(x, t) = i\hbar\frac{\partial}{\partial{t}}\Psi(x, t) = Ee^{\frac{i}{\hbar}(kx - E{t})} = E\psi(x, t)

So \psi is an eigenvector of H, and E is the eigenvalue. E is guaranteed to be real, because H is Hermitian, but \Psi(x, t) is not. In fact, because the Hamiltonian is defined to be the time derivative of the wavefunction, a particle's energy is basically defined by how fast its wavefunction rotates through the complex plane.

Yes, of course, but my question was about *real* Hermitian operators; your H is not real. Thanks anyway :)

 

[Chopin]

Yeah, sorry, I just re-read your question and realized you were probably asking something a little more complicated than I thought. How exactly are you defining a 'real' operator? Just one that maps any incoming real number to an outgoing real?

 

[AxiomOfChoice]

Here's an argument I've just devised for why \Psi can be real: Suppose \Psi = \psi + i \phi[/itex], where \psi,\phi are real, and (H - E)\Psi = 0 for H real, Hermitian. Then<br /> <br /> &lt;br /&gt; (H - E)(\psi + i \phi) = 0,&lt;br /&gt;<br /> <br /> so (H - E)\psi = i (H-E) \phi = 0. Hence \psi itself is a (real) solution, so for every (nontrivial) complex solution there is a real solution. So we might as well have assumed the solution was real to begin with. What do you think?

 

[AxiomOfChoice]

Yeah, sorry, I just re-read your question and realized you were probably asking something a little more complicated than I thought. How exactly are you defining a 'real' operator? Just one that maps any incoming real number to an outgoing real?

Well, I was thinking (from a QM perspective) of an operator that doesn't contain any i's. Is there something deficient about that definition?

 

[Chopin]

Makes sense, as long as H is linear, but I think all quantum mechanical operators are. Out of curiosity, though, why are you wondering about this? It's pretty rare to have a wavefunction be entirely real, isn't it? And even if it is, it's not going to stay that way under time evolution.

 

[AxiomOfChoice]

Makes sense, as long as H is linear, but I think all quantum mechanical operators are. Out of curiosity, though, why are you wondering about this? It's pretty rare to have a wavefunction be entirely real, isn't it? And even if it is, it's not going to stay that way under time evolution.

I agree with your point about time evolution. This is related to a time-independent Born-Oppenheimer approximation problem I'm solving :)

 

[MrHigh]

sorry to interrupt, i was following you guys...
for me the probabilities don't change so being real or complex is just a matter of a phase that... as Mr AxiomOfChoice said will indeed change... am i lost?

 

[Chopin]

sorry to interrupt, i was following you guys...
for me the probabilities don't change so being real or complex is just a matter of a phase that... as Mr AxiomOfChoice said will indeed change... am i lost?

Yeah, but usually the wavefunction has a different phase in different places...see my plane wave example above. The key point here is that the wavefunction is real everywhere, meaning everybody has the same phase. Still don't quite understand what usefulness of a wavefunction like that is, though.

 

[MrHigh]

thanks, and i agree: yep, it is quite odd.

 

[Avodyne]

Yes, if H is real and hermitian, then the eigenfunctions are real. To prove this, start with H\psi=E\psi, and take the complex (not hermitian!) conjugate: H^*\psi^*=E\psi^*. (Here I have used that the eigenvalue E of the hermitian operator H is real.) Now assume H is real (that is, contains no explicit factors of i). Then we have H\psi^*=E\psi^*. So, \psi and \psi^* are both eigenfunctions with the same eigenvalue. Hence, so are any linear combinations of them, including \psi+\psi^* and i(\psi-\psi^*), which are both real. In practice, the eigenvalue is often nondegenerate, and so these two real eigenfunctions are the same, up to an overall constant factor (or one of them is zero).

 

[fermi]

Yes, if H is real and hermitian, then the eigenfunctions are real. To prove this, start with H\psi=E\psi, and take the complex (not hermitian!) conjugate: H^*\psi^*=E\psi^*. (Here I have used that the eigenvalue E of the hermitian operator H is real.) Now assume H is real (that is, contains no explicit factors of i). Then we have H\psi^*=E\psi^*. So, \psi and \psi^* are both eigenfunctions with the same eigenvalue. Hence, so are any linear combinations of them, including \psi+\psi^* and i(\psi-\psi^*), which are both real. In practice, the eigenvalue is often nondegenerate, and so these two real eigenfunctions are the same, up to an overall constant factor (or one of them is zero).

I agree with you up to a fairly minor technicality. What you say above is true for any finite dimensional linear space with a bi-linear product. For the infinite dimensional case (which is nearly always the case in QM) you need some additional input. For example it is sufficient (but not necessary) that the Hamiltonian H be compact. In QFT (and even in QM) one sometimes has to deal with non-compact Hermitian operators. The eigenvectors of these operators cannot always be rendered real.

 

[DrFaustus]

AxiomOfChoice -> Your argument above does not hold. If A-B=0 it does not follow that A=B=0, but only A=B.

Also, consider the 1D free Hamiltonian, which is simply -\partial_x^2, and have it act on a e^{ipx}. You have a "real" Hermitian operator acting on a complex function giving you a real eigenvalue. This situation is really just that of the hydrogen atom. The time-independent solution comes in terms of spherical harmonics, which are not real, yet the time-independent Hamiltonian is "real".

 

[xepma]

AxiomOfChoice -> Your argument above does not hold. If A-B=0 it does not follow that A=B=0, but only A=B.

Also, consider the 1D free Hamiltonian, which is simply -\partial_x^2, and have it act on a e^{ipx}. You have a "real" Hermitian operator acting on a complex function giving you a real eigenvalue. This situation is really just that of the hydrogen atom. The time-independent solution comes in terms of spherical harmonics, which are not real, yet the time-independent Hamiltonian is "real".

Yea, but you can construct a real eigenbasis by taking the combination: e^{ipx} + e^{-ipx}, which is also an eigenvector of the operator -\partial_x^2.

The same holds for the spherical harmonics: in the usual convention the functions Y^m^_l(\theta,\phi) are complex, and they are simultaneous eigenfunctions of the (real) Hamiltonian and the angular momentum operators L^2 and L^z.

However, you can construct a real-valued basis of functions through:

Y_{lm} \sim Y^{|m|}_l + \textrm{sign}(m)(-1)^m Y_l^{-{|m|}

for m not equal to zero (and up to a normalization which may include a factor of i)

These are still eigenfunctions of the Hamiltonian. They are less convienent, because they are not eigenfunctions of the angular momentum operator.