# Density of states for non-interacting systems

1. Apr 9, 2010

### Niles

Hi guys

In some articles I've read, they all mention that the (local) density of states is related to the retarded Greens function for a non-interacting system by

-(1/π)Im[G(r,ω)] = LDOS(r,ω),

i.e. the imaginary part of the Greens function. The above relation holds because in k-space the retarded Greens function for translational invariant systems is given by

Gretarded = 1/(ω-Ek+iΓ),

where Γ is the self-energy (and Ek is the self-energy). In the limit where Γ goes to zero, we can take the imaginary part, and then we obtain the above relation - so far so good. But this is just for the case of a translational invariant systems - i.e. we have not assumed any impurities.

Now lets say I place an impurity in my system. Now my retarded Greens function is not given as above. How do we know that the LDOS is still just proportional to the imaginary part of the retarded Greens function for the system?

Last edited: Apr 9, 2010
2. Apr 9, 2010

### ZapperZ

Staff Emeritus
It depends on how "complex" you want to handle this. In many cases, the scattering to impurities only adds to the imaginary part of the self energy, i.e. it causes a constant broadening term to Im(G). This translates to a decrease in the mean free path.

Zz.

3. Apr 9, 2010

How does one even define the local density of states generally? As far as I know, one usually refers to the spectral function $$\rho_{\alpha}(\omega) \sim \textrm{Im} G^R_{\alpha\alpha}(\omega)$$ loosely as the ''local density of states". This is natural, since the spectral function satisfies

$$\langle c_{\alpha}^{\dagger} c_{\alpha} \rangle = \int d\omega n(\omega) \rho_{\alpha} (\omega),$$

i.e. the spectral function measures the availability of the state $$|\alpha \rangle$$ at energy $$\omega$$. Here $$n(\omega)$$ is the Fermi function. Note that this equation holds even in the presence of interactions (or impurities).

The relation $$\rho_{\alpha}(\omega) \sim \textrm{Im} G^R_{\alpha\alpha}(\omega)$$ holds in thermal equilibrium, in non-equilibrium the definition is a little different.

4. Apr 9, 2010

### ZapperZ

Staff Emeritus
Typically, if you have the spectral function, i.e. ~Im(G), the DOS is simply the momentum average, i.e. integrate Im(G) over all k values.

Zz.

5. Apr 9, 2010

That seems clear for non-interacting systems, yes. Taking the Hamiltonian to be $$H=\sum \epsilon_{\alpha} c_{\alpha}^{\dagger} c_{\alpha},$$ the density of states is $$d(\omega) = \sum_{\alpha} \delta(\omega-\epsilon_{\alpha}),$$ and since for a non-interacting system $$\rho_{\alpha}(\omega) \sim \delta(\omega-\epsilon_{\alpha}),$$ we get $$d(\omega) \sim \sum_{\alpha} \rho_{\alpha}(\omega)$$. But I don't see, how this would hold for an interacting system. But I guess we were talking about impurities, not interactions.

6. Apr 9, 2010

### kanato

The general definition of density of states is
$$g(\omega) = \sum_{k} \delta(\omega - E_k)$$
k can represent any set of quantum numbers you like, it does not have to be restricted to crystal momentum. Of course in a crystal, you need a band index as well as the pseudomomentum, but this definition is general for any system regardless of whether there is any physical interpretation of a Fourier transform of the quantum numbers.

For any non-interacting system the retarded Green's function is $$G(\omega) = \sum_k [\omega - E_k + i\delta]^{-1}$$ where again, k is just a placeholder for an arbitrary set of quantum numbers. The spectral function is defined as
$$A(\omega) = -\frac1\pi \mathrm{Im\,} G(\omega) = \frac1\pi \sum_k \frac{\delta}{(\omega-E_k)^2 + \delta^2}$$
In the limit that $$\delta$$ goes to zero, that is a representation of the Dirac delta function, and the spectral function reduces to the definition of the density of states.

7. Apr 9, 2010

### kanato

Yes, for an interacting system, there is a problem since the density of states is defined in terms of single particle energy levels which don't formally exist in an interacting system. This is why the imaginary part of the Green's function is called the spectral function instead of the density of states.

8. Apr 9, 2010

### Niles

First of all - just to be clear - placing a single impurity in a lattice still makes the system non-interacting, right?

Second, e.g. take a look at http://books.google.dk/books?id=v5v...resnum=5&ved=0CCwQ6AEwBA#v=onepage&q=&f=false, equation 8.55. This is the equation you wrote above, and here they have derived it from the translational invariant retarded Green's function (equation 8.41c). So I can't see why

$$A(\omega) = -\frac1\pi \mathrm{Im\,} G(\omega)$$

would still gives us the (local) DOS in the limit δ→0 when we have an impurity in our lattice (because in this case it is not translational invariant).

BTW your post is very clearly formulated, unlike my original post. Thanks.

9. Apr 9, 2010

### kanato

Well yes, whether or not you have interactions is independent of whether you have impurities.

I don't see anything there that is really specific to a translationally invariant system. They don't explicitly use the functional form of E_k, all that's done in 8.55 is to Fourier transform the time variable to get a frequency dependent G instead. Ok, there might be an implicit assumption of time translational invariance, but you will get into much harder problems if you object to that.

Notice below (around 8.57) they say that it's true for any quadratic Hamiltonian, and they even use a different label for the eigenstates. If you calculate the Green's function from the definition in the first line of 8.45 for the Hamiltonian in 8.57 and then do the time Fourier transform in 8.55 you should find the exact expression I gave for the Green's function above.

10. Apr 9, 2010

### Niles

Say I start of by using the very definition of the retarded Greens function in real space (I do not include spin)

$$G^R(r,r';t,t') = -i\theta(t-t')\left\langle {\left[ {c\left( {r,t} \right),c^\dag \left( {r',t'} \right)} \right]{}_{fermion}} \right\rangle$$

Here θ(t-t') is the Heaviside step function, c and c-dagger are my second-quantization field operators. First I have written it in a diagonal basis in real-space (i.e. made a transformation) and then I have Fourier-transformed it - and I do not end up with the same expression as in equation 8.55. In my expression, the norm squared of the corresponding eigenstates are multiplied on to 1/(ω-Ek+iδ).

How can this be explained? (I have triple-checked my calculations - they are correct).

11. Apr 9, 2010

### kanato

Can you show a little more of what you did and what you got? Finding the norm squared seems appropriate, since you should assume your states are normalized. If you do this stuff in a non-orthogonal basis you will get projections between states when doing this, so getting the norm of your states seems ok.

12. Apr 9, 2010

### Niles

I get (for a site n in real space)

$$G(n,\omega ) = \sum\limits_k {\frac{{\left| {S_{nk} } \right|^2 }}{{\omega - E_k + i\delta }}}$$

Here S is the matrix containing the (normalized) eigenstates, i.e. it satisfies

$$E = S^\dagger H S,$$

where H is the Hamiltonian. BTW, I need some sleep, so I will answer in 7-8 hours, when I get up. Thanks for participating.

13. Apr 9, 2010

### kanato

Ok, so your state is normalized, so $$|S_{nk}|^2 = 1$$, right?

14. Apr 10, 2010

### Niles

What I have done is first construct my tight-binding Hamiltonian in MATLAB. Then I have diagonalized it, and the normalized eigenvectors I get from MATLAB I have put in the matrix S. So in my matrix, the element Snk is just a number. So each column in S is a normalized eigenvector - but I guess that means that Snk is not 1.

Last edited: Apr 10, 2010
15. Apr 10, 2010

I'm sorry to jump in between, but you're correct. In the diagonal basis, the Green's function is

$$G= \sum_k \frac{|k \rangle \langle k |}{\omega -E_k + i\delta}$$,

so projecting on site n gives

$$G_{nn}(\omega) = \sum_k \frac{|S_{nk}|^2}{\omega -E_k + i\delta},$$

where $$S_{nk}$$ is the amplitude of the wavefunction k at the site n. Thus the local density of states is

$$\rho_n(\omega) = \sum_k |S_{nk}|^2 \delta (\omega-E_k).$$

This is as far as you get. Summing over n gives of course the density of states, since $$\sum_n |S_{nk}|^2=1$$. So unless the Hamiltonian is diagonal in the position basis, in which case would be $$|S_{nk}|=\delta_{nk}$$, you will not get an equation such as

$$\rho_n(\omega) = \delta (\omega-E_n).$$

16. Apr 10, 2010

### Niles

You are more than welcome to join in as well for my sake. With the way I have defined the matrix S above, is it correct that "Snk is the amplitude of the wavefunction k at the site n"?

And thank you for your post. I really appreciate it when you guys spend time helping me; it is thoroughly appreciated.

17. Apr 10, 2010

Yes, Snk = <n|k> is perfectly analogous to the "usual" wavefunctions ψ(x)=<x|ψ>. And this discussion is just fun, so no need to thank, I'm just glad to if I can be of any help!

18. Apr 11, 2010

### Niles

I have a follow-up question on this matter. For the past hour I've been sitting with paper and pen trying to figure this out, but no luck. In http://books.google.dk/books?id=v5v...&resnum=5&ved=0CCwQ6AEwBA#v=onepage&q&f=false they show say that in the eigenbasis of the Hamiltonian, the Green's function is given by

$$G(\omega) = \sum_k [\omega - E_k + i\delta]^{-1}$$

Now, what I got was

$$G_{nn}(\omega) = \sum_k \frac{|S_{nk}|^2}{\omega -E_k + i\delta},$$

What is the difference in my case, since I am not getting the proper form?

19. Apr 13, 2010

This equation is not really correct. On the right-hand side, you have Tr[G(\omega)], since you are summing over all single-particle eigenvalues. The retarded Green's function for a non-interacting system is either
$$G(\omega) = \sum_k \frac{|k\rangle \langle k |}{\omega - E_k + i\delta},$$
where G is written as a matrix, or one can write
$$G_{kk}(\omega) = \frac{1}{\omega - E_k + i\delta},$$
by taking <k|G|k> from the equation above (changing the summation index first).
Your result is absolutely correct. Since the Hamiltonian is not diagonal in the n-basis (position basis, I presume), you will NOT get a result such as
$$G_{nn}(\omega) = \frac{1}{\omega - E_n + i\delta}.$$
As I wrote above, if {|k>} is the eigenbasis, you have
$$G(\omega) = \sum_k \frac{|k\rangle \langle k |}{\omega - E_k + i\delta},$$
from which it follows that
$$G_{nn}(\omega) = \langle n|G|n \rangle = \sum_k \frac{\langle n |k\rangle \langle k|n\rangle }{\omega - E_k + i\delta}= \sum_k \frac{S_{nk} S_{nk}^*}{\omega - E_k + i\delta}=\sum_k \frac{|S_{nk}|^2}{\omega - E_k + i\delta}.$$
Hope this helps.

20. Apr 13, 2010

### Niles

Thanks for being patient, I got it now.

21. Apr 15, 2010

### Niles

Ok, starting from the general expression

$$G^R(r,r ';t,t ') = -i\theta(t-t ')\left\langle {\left[ {c\left( {r,t} \right),c^\dag \left( {r ',t '} \right)} \right]{}_{fermion}} \right\rangle ,$$

I have made a transformation to a diagonal basis using

$${\cal H} = \mathbf a^\dagger\mathsf H \mathbf a = \mathbf a^\dagger \mathsf S\mathsf S^\dagger\mathsf H \mathsf S\mathsf S^\dagger\mathbf a = \mathbf b^\dagger \mathsf D \mathbf b.$$

This gives me the usual

$$G_{nn}(\omega) = \sum_k \frac{|S_{nk}|^2}{\omega -E_k + i\delta},$$

which is not the same as

$$G_{kk}(\omega) = \frac{1}{\omega - E_k + i\delta}.$$

But it should be since I wrote my general expression for the retarded Greens function in a diagonal basis. So does this mean that

$$\frac{1}{\omega - E_k + i\delta} = \sum_k \frac{|S_{nk}|^2}{\omega -E_k + i\delta}?$$

I know you guys might be thinking that it is getting silly by now with all my questions, but this is the last problem on this matter. After this I would say I have a fair understanding of the topic. So any help will again be greatly appreciated.

Last edited: Apr 15, 2010
22. Apr 19, 2010

No, you must notice that this equality would not even make any sense, since the indices do not match. Suppose your Hamiltonian in position basis is e.g. $$H=\hat \sigma_x$$, i.e. with zeros on diagonal and ones off-diagonal. Since the Hamiltonian is not diagonal, the Green's function in this basis will not be diagonal. But of course, it is a simple matter to diagonalize the Hamiltonian. But diagonalization is simply a change of basis, the Hamiltonian operator remains the same! So even after the change of basis, the Green's function cannot become diagonal in the position basis. Does this make sense?

Last edited: Apr 19, 2010
23. Apr 19, 2010

### Niles

But if the Hamiltonian is diagonal, then I can write my operator as

$$\hat H = \sum_{i}E_ic^\dagger_ic_i,$$

which is also diagonal?

24. Apr 19, 2010

$$G_{ij}(\omega)=\frac{1}{\omega - E_i + i\delta} \delta_{ij}.$$
In position basis, the Green's function would read (with $$\psi_i(x)=\langle x |i \rangle$$)
$$G_{xy}(\omega)=\sum_i \frac{\psi_i^*(x) \psi_i(y)}{\omega - E_i + i\delta}.$$