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

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In summary, the conversation discusses the question of whether the wavefunction, \Psi, in the Schrodinger Equation can be assumed to be real if the Hamiltonian, H, is real and Hermitian. The conclusion is that while the energy, E, is guaranteed to be real, the wavefunction is not necessarily real. It is possible for the wavefunction to be entirely real, but this is rare and not likely to stay that way under time evolution. It is also mentioned that in the case of a non-degenerate eigenvalue, the eigenfunctions are both real and can be expressed as linear combinations of each other. However, for infinite dimensional cases, more careful considerations are needed.
  • #1
AxiomOfChoice
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If we're attempting to solve

[tex]
H\psi = E\psi
[/tex]

where [itex]H[/itex] is real and Hermitian, are we allowed to assume [itex]\psi[/itex] is real? Why or why not? My gut tells me the answer is "yes," since we know [itex]E[/itex] is real, but I can't make my idea rigorous.
 
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  • #2
No, in fact it's very likely that [tex]\Psi[/tex] is not real, at least not everywhere. Consider a simple solution of the free-space Schrodinger Equation, the plane wave:

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

Because [tex]H = i\hbar\frac{\partial}{\partial{t}}[/tex], we have:

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

So [tex]\psi[/tex] is an eigenvector of [tex]H[/tex], and [tex]E[/tex] is the eigenvalue. [tex]E[/tex] is guaranteed to be real, because [tex]H[/tex] is Hermitian, but [tex]\psi(x, t)[/tex] 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.
 
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  • #3
Chopin said:
No, in fact it's very likely that [tex]\Psi[/tex] is not real, at least not everywhere. Consider a simple solution of the free-space Schrodinger Equation, the plane wave:

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

Because [tex]H = i\hbar\frac{\partial}{\partial{t}}[/tex], we have:

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

So [tex]\psi[/tex] is an eigenvector of [tex]H[/tex], and [tex]E[/tex] is the eigenvalue. [tex]E[/tex] is guaranteed to be real, because [tex]H[/tex] is Hermitian, but [tex]\Psi(x, t)[/tex] 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 [itex]H[/itex] is not real. Thanks anyway :)
 
  • #4
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?
 
  • #5
Here's an argument I've just devised for why [itex]\Psi[/itex] can be real: Suppose [tex]\Psi = \psi + i \phi[/itex], where [itex]\psi,\phi[/itex] are real, and [itex](H - E)\Psi = 0[/itex] for [itex]H[/itex] real, Hermitian. Then

[tex]
(H - E)(\psi + i \phi) = 0,
[/tex]

so [itex](H - E)\psi = i (H-E) \phi = 0[/itex]. Hence [itex]\psi[/itex] 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?
 
  • #6
Chopin said:
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?
 
  • #7
Makes sense, as long as [tex]H[/tex] 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.
 
  • #8
Chopin said:
Makes sense, as long as [tex]H[/tex] 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 :)
 
  • #9
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?
 
  • #10
MrHigh said:
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.
 
  • #11
thanks, and i agree: yep, it is quite odd.
 
  • #12
Yes, if H is real and hermitian, then the eigenfunctions are real. To prove this, start with [itex]H\psi=E\psi[/itex], and take the complex (not hermitian!) conjugate: [itex]H^*\psi^*=E\psi^*[/itex]. (Here I have used that the eigenvalue [itex]E[/itex] of the hermitian operator [itex]H[/itex] is real.) Now assume [itex]H[/itex] is real (that is, contains no explicit factors of [itex]i[/itex]). Then we have [itex]H\psi^*=E\psi^*[/itex]. So, [itex]\psi[/itex] and [itex]\psi^*[/itex] are both eigenfunctions with the same eigenvalue. Hence, so are any linear combinations of them, including [itex]\psi+\psi^*[/itex] and [itex]i(\psi-\psi^*)[/itex], 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).
 
  • #13
Avodyne said:
Yes, if H is real and hermitian, then the eigenfunctions are real. To prove this, start with [itex]H\psi=E\psi[/itex], and take the complex (not hermitian!) conjugate: [itex]H^*\psi^*=E\psi^*[/itex]. (Here I have used that the eigenvalue [itex]E[/itex] of the hermitian operator [itex]H[/itex] is real.) Now assume [itex]H[/itex] is real (that is, contains no explicit factors of [itex]i[/itex]). Then we have [itex]H\psi^*=E\psi^*[/itex]. So, [itex]\psi[/itex] and [itex]\psi^*[/itex] are both eigenfunctions with the same eigenvalue. Hence, so are any linear combinations of them, including [itex]\psi+\psi^*[/itex] and [itex]i(\psi-\psi^*)[/itex], 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 [itex]H[/itex] 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.
 
  • #14
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 [tex]-\partial_x^2[/tex], and have it act on a [tex]e^{ipx}[/tex]. 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".
 
  • #15
DrFaustus said:
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 [tex]-\partial_x^2[/tex], and have it act on a [tex]e^{ipx}[/tex]. 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: [tex]e^{ipx} + e^{-ipx}[/tex], which is also an eigenvector of the operator [tex]-\partial_x^2[/tex].

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

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

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

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.
 

1. What is a real Hermitian eigenvalue problem?

A real Hermitian eigenvalue problem is a type of mathematical problem in linear algebra where a square matrix is multiplied by a vector and the resulting vector is equal to a scalar multiple of the original vector. The scalar multiple is known as an eigenvalue and the corresponding vector is known as an eigenvector.

2. How do you determine if a real Hermitian eigenvalue problem has a real solution?

A real Hermitian eigenvalue problem will always have a real solution if the matrix is a real symmetric matrix. This means that the matrix is equal to its transpose and all of its elements are real numbers. If these conditions are met, then the eigenvalues and eigenvectors will also be real numbers.

3. What is the significance of a real solution in a real Hermitian eigenvalue problem?

A real solution in a real Hermitian eigenvalue problem indicates that the corresponding eigenvectors are orthogonal (perpendicular) to each other. This property is useful in many applications, such as in physics and engineering, where orthogonal vectors are needed for calculations and analysis.

4. Can a real Hermitian eigenvalue problem have multiple real solutions?

Yes, a real Hermitian eigenvalue problem can have multiple real solutions. This means that there may be multiple eigenvectors that satisfy the eigenvalue equation for a given eigenvalue. However, these solutions will always be orthogonal to each other.

5. How are real Hermitian eigenvalue problems used in real-world applications?

Real Hermitian eigenvalue problems are used in a wide range of applications, such as in quantum mechanics, signal processing, and data analysis. They are also used in machine learning algorithms, such as principal component analysis, to reduce the dimensionality of data and extract important features.

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