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  1. When the Hamiltonian is Hermitian(i,e. beyond the effective mass approximation), generally under which conditions the eigenfunctions/wavefunctions are real?

  2. What happens in 1D case like the finite quantum well symmetric with respect to the origin might be an example. any general rule? further generalization into 2D case?

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    $\begingroup$ The two subquestions are essentially Problem 2.1b and Problem 2.1c, respectively, in Griffiths, Intro to QM. More on Problem 2.1: physics.stackexchange.com/q/44003/2451 and links therein. $\endgroup$ – Qmechanic Sep 19 '13 at 17:00
  • $\begingroup$ The answer to 1 actually depends upon how you define reality structure on the complex space of states. One can always choose a basis of eigenstates of the Hamiltonian and call them real. $\endgroup$ – user10001 Sep 20 '13 at 5:16
  • $\begingroup$ Sorry, now i understand that the question is not that trivial since you are asking about reality of wavefunctions (i.e. position representation of eigenstates) and not the eigenstates. $\endgroup$ – user10001 Sep 20 '13 at 5:27
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All bound states can typically be chosen to have real-valued wavefunctions. The reason for this is that their wavefunction obeys a real differential equation, $$ -\frac{\hbar^2}{2m}\nabla^2\psi(\mathbf r)+V(\mathbf r)\psi(\mathbf r)=E\psi(\mathbf r)$$ and therefore for any solution you can construct a second solution by taking the complex conjugate $\psi(\mathbf r)^\ast$. This second solution will either be

  • linearly dependent on $\psi$, in which case $\psi$ differs from a real-valued function by a phase, or
  • linearly independent, in which case you can "rotate" this basis into the two independent real-valued solutions $\operatorname{Re}(\psi)$ and $\operatorname{Im}(\psi)$.

For continuum states this also applies, but things are not quite as clear as the boundary conditions are not invariant under conjugation: incoming scattering waves with asymptotic momentum $\mathbf p$, for example, behave asymptotically as $e^{i\mathbf p\cdot \mathbf r/\hbar}$, and this changes into outgoing waves upon conjugation. Thus, while you can still form two real-valued solutions, they will be standing waves and the physics will be quite different.

In the second case, when you have a degeneracy, the physical characteristics of the real-valued functions are in general different to those of the complex-valued ones. For example, in molecular physics, $\Pi$ states typically have such a degeneracy: you can choose

  • $\Pi_x$ and $\Pi_y$ states, which are real-valued, have a node on the $x$ and $y$ plane, resp., have a corresponding factor of $x$ and $y$ in the wavefunction, and have zero expected angular momentum component along the $z$ axis, or
  • $\Pi_\pm$ states, which have a complex factor of $x\pm i y$ and no node, and have definite angular momentum of $\pm \hbar$ about the $z$ axis.

Thus: you can always choose a real-valued eigenstate, but it may not always be the one you want.

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In addition to Emilio's great answer, and in answer to your second question: Specifically in 1D potential problems (i.e. $\hat H = \frac{1}{2m} \hat p^2 + V(\hat x)$), all the bound states can simultaneously be made real. This is because of the theorem that bound states in 1D are non-degenerate; then, $\psi$ and $\psi^*$, which are both solutions in any dimensionality, must be linearly dependent.

The situation is different if you have a magnetic field; then the prescription ("minimal coupling") is to replace $\hat p \gets (\hat p - e A(\hat x))$, which results in a complex Hamiltonian ($A$ is the vector potential).

Also, the argument above holds for a "well-behaved" potential; see http://arxiv.org/abs/0706.1135 for a modern take on that.

Finally, regarding your comment about a band-structure calculation: I am not sure this is what you mean, but, at least in the context of certain codes, you will often hear it said that a calculation for a centrosymmetric structure without spin-orbit coupling is "real", but this does not mean that the true eigenfunctions $\psi_{nk}$ are real-valued (remember, they contain a plane-wave factor $e^{ikr}$). Rather, the coefficients in the basis used for the calculation can be chosen to be real.

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Since Hamiltonian is a linear operator, one could always choose real eigenfunctions. This is because if f is a complex eigenfunction of any linear operator O with eigenvalue e, then its both real and imaginary parts are also O's eigenfunction with the same eigenvalue e.

It does not require the operator to be Hermitian. Moreover, this is true for all dimensions.

Edit: It's a wrong statement. I made a mistake there.

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  • $\begingroup$ Let us say an example, giving a Dirac operator for Dirac fermions, can you always write down the Dirac fermion's eigenfunctions in terms of a REAL representation? (I thought only the Majorana operator you have that the always REAL representation.) $\endgroup$ – wonderich Jun 29 '14 at 4:04
  • $\begingroup$ Yes, you can always choose to work with real eigenfunctions for a linear operator, which is Dirac Hamiltonian in you case. I would repeat that eigenfunction can be complex, in general. If the Hamiltonian is invariant under time-reversal and has non-degenerate spectrum, then only real eigenfunctions are allowed. Maybe for Majorana operator, this happens. $\endgroup$ – Mohit Pandey Jun 29 '14 at 14:22
  • $\begingroup$ I think Pandey's statement is not always correct. Assume your hamiltonian is the matrix $\sigma_y=\left(\begin{matrix} 0& i\\ -i&0 \end{matrix}\right)$. The eigenfunction of this matrix is always imaginary. A better criterium would be H = transpose(H) which is equivalent to saying that H is real, since H is hermitian. In this case you can show that there is always a real valued eigenvector of this matrix by following Pisantys suggestion above. $\endgroup$ – physicsGuy Feb 11 '15 at 20:06

protected by Qmechanic Jun 29 '14 at 5:32

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