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Zero modes ~ zero eigenvalue modes ~ zero energy modes?

+ 7 like - 0 dislike
13323 views

There have been several Phys.SE questions on the topic of zero modes. Such as, e.g.,

Here I would like to understand further whether "Zero Modes" may have physically different interpretations and what their consequences are, or how these issues really are the same, related or different. There at least 3 relevant issues I can come up with:

(1) Zero eigenvalue modes

By definition, Zero Modes means zero eigenvalue modes, which are modes $\Psi_j$ with zero eigenvalue for some operator $O$. Say,

$$O \Psi_j = \lambda_j \Psi_j,$$ with some $\lambda_a=0$ for some $a$.

This can be Dirac operator of some fermion fields, such as $$(i\gamma^\mu D^\mu(A,\phi)-m)\Psi_j = \lambda_j \Psi_j$$ here there may be nontrivial gauge profile $A$ and soliton profile $\phi$ in spacetime. If zero mode exists then with $\lambda_a=0$ for some $a$. In this case, however, as far as I understand, the energy of the zero modes may not be zero. This zero mode contributes nontrivially to the path integral as $$\int [D\Psi][D\bar{\Psi}] e^{iS[\Psi]}=\int [D\Psi][D\bar{\Psi}] e^{i\bar{\Psi}(i\gamma^\mu D^\mu(A,\phi)-m)\Psi } =\det(i\gamma^\mu D^\mu(A,\phi)-m)=\prod_j \lambda_j$$ In this case, if there exists $\lambda_a=0$, then we need to be very careful about the possible long range correlation of $\Psi_a$, seen from the path integral partition function (any comments at this point?).

(2) Zero energy modes

If said the operator $O$ is precisely the hamiltonian $H$, i.e. the $\lambda_j$ become energy eigenvalues, then the zero modes becomes zero energy modes: $$ H \Psi_j= \lambda_j \Psi_j $$ if there exists some $\lambda_a=0$.

(3) Zero modes $\phi_0$ and conjugate momentum winding modes $P_{\phi}$

In the chiral boson theory or heterotic string theory, the bosonic field $\Phi(x)$ $$ \Phi(x) ={\phi_{0}}+ P_{\phi} \frac{2\pi}{L}x+i \sum_{n\neq 0} \frac{1}{n} \alpha_{n} e^{-in x \frac{2\pi}{L}} $$ contains zero mode $\phi_0$.


Thus: Are the issues (1),(2) and (3) the same, related or different physical issues? If they are the same, why there are the same? If they're different, how they are different?

I also like to know when people consider various context, which issues they are really dealing with: such as the Jackiw-Rebbi model, the Jackiw-Rossi model and Goldstone-Wilczek current computing induced quantum number under soliton profile, Majorana zero energy modes, such as the Fu-Kane model (arXiv:0707.1692), Ivanov half-quantum vortices in p-wave superconductors (arXiv:cond-mat/0005069), or the issue with fermion zero modes under QCD instanton as discussed in Sidney Coleman's book ``Aspects of symmetry''.

ps. since this question may be a bit too broad, it is totally welcomed that anyone attempts to firstly answer the question partly and add more thoughts later.


This post imported from StackExchange Physics at 2014-03-12 15:56 (UCT), posted by SE-user Idear

asked Aug 13, 2013 in Theoretical Physics by Idear (1,195 points) [ revision history ]
edited May 7, 2014 by dimension10
You seem to have answered your own question . 2 is a type of 1 where the $A=H$ ... But (3) doesn't seem to be the same as 1 or 2 though . . . P.S. By no means is this question too broad . . .

This post imported from StackExchange Physics at 2014-03-12 15:56 (UCT), posted by SE-user Dimensio1n0
Welcome on Phys.SE Idear. I do not fully understand your question. It seems you have already answered it, ad @Dimension10 mentioned. Do you want an introduction to the index theorem, which relates the zero-energy modes to soliton solutions in quantum field ? What does your next-to-last paragraph mean ? (Do you want an introduction to Jackiw-Rebbi/Rossi and Goldstone-Wilczek models ?)

This post imported from StackExchange Physics at 2014-03-12 15:56 (UCT), posted by SE-user FraSchelle
Otherwise, to directly answer your question, a zero-energy mode indeed verify the secular equation $Af=0$, with $f$ a vector, and $A$ an operator. The Hamiltonian $H$ is indeed a (differential) operator, so points (1) and (2) are equivalent. The field you noted $\Phi$ seems to be the vector I noted $f$ of a given Hamiltonian. I have no idea of the Hamiltonian/Lagrangian of the heterotic string theory, sorry for that.

This post imported from StackExchange Physics at 2014-03-12 15:56 (UCT), posted by SE-user FraSchelle
@Oaoa: The heteorotic string lagrangian densitiyie's pretty simple, actually. just add 32 majorana - weyl fermions.

This post imported from StackExchange Physics at 2014-03-12 15:56 (UCT), posted by SE-user Dimensio1n0

1 Answer

+ 0 like - 0 dislike

From $H|\psi\rangle=E|\psi \rangle$, then, if $E=0$ (a zero-energy mode), the eigen - value of $H$ is 0. Therefore, this is a special type of a zero - eigenvalue mode. Of course, being a zero - eigenvalue mode doesn't force it to be a zero - energy mode, but the reverse is true.

As for a zero - mode, no... It doesn't seem related to me. For all those who didn't understand the rather opaque notation used in the question, it's just the mode expansion of the bosonic field $X^\mu $...

$$X^\mu=X_0^\mu+P_{\phi} \frac{2\pi}{\ell_s}\sigma +i \sum_{n\neq 0} \frac{1}{n} \alpha_{n} e^{-in \sigma \frac{2\pi}{\ell_s }}$$

answered Aug 15, 2013 by dimension10 (1,950 points) [ revision history ]
edited May 21, 2014 by dimension10
Hi Dimension 10. You are partly right. From what I gave in the question, I already define three issues clearly. What I really want to probe the Phys.SE. here is the connection between the three issues. I do think someone here can explain deeper understanding than both of us written here so far. There still can be some room to connect the notions, though they are indeed different. Let us wait a bit. But thank you for reply!

This post imported from StackExchange Physics at 2014-03-12 15:57 (UCT), posted by SE-user Idear

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