Second quantization and Hamiltonian diagonalization

Question

So I want to diagonalize my Hamiltonian (it is bosonic hamiltonian) which is:

$H=(E+\Delta)a^{\dagger}a + 1/2\Delta(a^{\dagger}a^{\dagger} + aa)$

My class didn't cover this material so I don't really know how to procede. I would be grateful for any literature which covers this topics and a problem book with solutions would be great too.

What I tried to do was writing my Hamiltonian in matrix form which would be: $\begin{pmatrix} 1/2 \Delta & 1/2(E+\Delta) \\ 1/2(E+\Delta) & 1/2 \Delta \\ \end{pmatrix}$

And then diagonalize it, find eigenstates etc. Is this the correct way?

They are of course annihilation and creation operators (they are bosonic - forgot to mention that). I don't know the dimension but hamiltonian acts on any state $|\phi>$. — Caims, Dec 15 '15 at 22:21
You want to find a new annihilation operator $b$, which is a linear combination of $a$ and $a^\dagger$, such that $[b, H]=E b$ where $E$ will be the eigenenergy. — Meng Cheng, Dec 15 '15 at 22:35
@MengCheng That is a very unnatural way to define the eigestates of an operator, don't you think? — gented, Dec 15 '15 at 23:21
@GennaroTedesco This is called Bogoliubov transformation and quite common and necessary for Hamiltonians that contain terms $aa$ and $a^\dagger a^\dagger$ (typically mean field Hamiltonians) such as the BCS Hamiltonian or Hamiltonians for exitations of a Bose-Einstein condensate. — Sebastian Riese, Dec 16 '15 at 00:43
The Bogoliubov transformation must be correctly defined tensoring with the appropriate identity, therefore its complete formulation is slightly more complicated than what showed. Moreover I was objecting on the expression given by Meng because that does not really find the eigenstates, rather it only re-writes the Hamiltonian in a more suitable way. — gented, Dec 16 '15 at 00:46
@GennaroTedesco I agree that the comment does not really help to find the solution (and is incomplete). Nevertheless, the way to solve this exercise is obviously by Bogoliubov transformation. But $[b, H] = Eb$ just is a consequence of $H = E b^\dagger b = En$. — Sebastian Riese, Dec 16 '15 at 00:48
Again, my only objection was on the formal writing of the Bogoliubov transformation, which ought to be correctly defined on common domains and so on and so forth. Writing $a+a^{\dagger}$ without any other prescription is incorrect (because the two operators have different domains and different co-domains, I wonder how you sum the results then). — gented, Dec 16 '15 at 00:52
I don't understand the objections. Of course $[b,H]=Eb$ is a consequence of $H=Eb^\dagger b$, that's exactly what you need to find $b$ and $E$ since we know $b$ must a linear combination of $a$ and $a^\dagger$ If you do not believe it, just use this relation and work out the resulting equations. — Meng Cheng, Dec 16 '15 at 00:56
@GennaroTedesco Maybe I am to tired, but I do not see the problem. Why do they have different domains? $a\left| 0 \right> = 0$. In fact, for the harmonic oscillator $x \propto a + a^\dagger$. (And a harmonic oscillator is nothing but a 0d, free bosonic field). (They do indeed have different co-domains, but so have $0$ and $1$). But the results are from a linear space, so you can always sum them (be it Fock space or whatever). — Sebastian Riese, Dec 16 '15 at 00:56
Over again, see my comments above and below the other answers. I am not objecting the results, I am objecting the definitions of $a+a^{\dagger}$ on the Fock space (please provide it with domains, actions and co-domains). Plus, the Fock space second quantisation is not equivalent to the harmonic oscillator exactly because of the issue with direct sums of Hilbert spaces (which does not appear in the latter). — gented, Dec 16 '15 at 01:02
There may be some subtleties I am ignoring, but I do not get what they are. $a$ and $a^\dagger$ are linear operators on the linear Fock space. Therefore I can add them. — Sebastian Riese, Dec 16 '15 at 01:05
The Fock space is not just one Hilbert space; it is the infinite direct sum of different Hilbert spaces (see all my comments and my answer). The action of the $a, a^{\dagger}$ is singularly defined on those Hilbert spaces only and needs to be glued in a suitable way to be applied to the entire infinite direct sum (i. e. the Fock space). — gented, Dec 16 '15 at 01:19

score 14 · Accepted Answer · answered Dec 16 '15 at 00:49

Diagonalizing the Hamiltonian means you want to bring it into the form $H=\omega b^\dagger b$, and it is pretty obvious that $b$ should be a linear combination of $a$ and $a^\dagger$, and $b$ should satisfy the canonical commutation of annihilation operators, namely $[b,b^\dagger]=1, [b,b]=0$.

Now let's write $b=ua+va^\dagger$ (this is called the Bogoliubov transformation, by the way). The condition $[b,b^\dagger]=1$ leads to $|u|^2-|v|^2=1$. Let us expand out $b^\dagger b$:

$$ b^\dagger b= |u|^2 a^\dagger a+ |v|^2 a a^\dagger + u^*v a^\dagger a^\dagger + uv^* aa. $$

Therefore

$$ \omega(|u|^2+|v|^2)=E+\Delta, \omega u^*v = \frac{1}{2}\Delta. $$

Together with $|u|^2-|v|^2=1$, we have three equations for three variables ($u, v, \omega$). In fact, in this case one can safely assume $u$ and $v$ are both real. The rest is just algebra.

That's what I asked for. The way to compute it. Thank you. – Caims Dec 16 '15 at 12:29 — Caims, Dec 16 '15 at 12:29

score 1 · Answer 2 · answered Dec 15 '15 at 23:13

1

Diagonalising an operator means finding its eigenstates.

Without loss of generality your Hamiltonian can be written as $$ H = c_1 a^{\dagger}a + c_2 a^{\dagger}a^{\dagger} + c_3 a a $$ with $a^{\dagger},a$ being operators of the type $a^{\dagger}\colon \mathcal{H}_n\mapsto \mathcal{H}_{n+1}$ (and conversely for $a$), where $\mathcal{H}_n$ is the $n$-particle Hilbert space contributing to the Fock space $\mathcal{F}= \oplus^{\infty}_n\mathcal{H}_n$.

There must be a few errors in your equation if you really mean that in a second quantisation procedure. First of all there is no general $a^{\dagger},a$ operator, rather you have one for each momentum $k$, that is $a^{\dagger}_k,a_k$ create and destroy (in quotation marks) particles with momentum $k$; there is no $k$ in your initial Hamiltonian, whereas the general form must be $\sum_k c_k\,a^{\dagger}_ka_k$.

Second of all: according to whether your particles are fermions or bosons the corresponding operators behave in a different way: for instance $a^{\dagger}_ka^{\dagger}_k=0$ for fermions.

If the Hamiltonian acts on a subspace of the Fock space with a certain number of particles $\mathcal{H}_n$, then the last two terms in your equation would bring the action onto $\mathcal{H}_{n\pm2}$, therefore the rhs will live in $\mathcal{H}_n +\mathcal{H}_{n+2} +\mathcal{H}_{n-2}$, which does not really make any sense since no prescription on how to sum elements in different Hilbert spaces is given (the last two pieces).

Either you assign a precise prescription to achieve the above, or there must be errors elsewhere in the formula, as pointed out; try giving more context so that one can work out what you mean. This said, suggested literature on how to write any Hamiltonian in second quantisation and find the corresponding solutions is, for example:

answered Dec 15 '15 at 23:13

gented

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Thank you so much. It's much clearer now. But there's no error in my hamiltonian (except that matrix which I wrote makes no sense). There's no summation, it is directly taken from my homework. – Caims Dec 15 '15 at 23:18
So basically my hamiltonian must keep me in the same Fock space and act on $H_n \rightarrow H_n$. Which means that my hamiltonian must have the same number of creation and annihilation operators in every coefficient to act from $H_n$ to $H_n$ ? – Caims Dec 15 '15 at 23:33
Yes, exactly. Plus, (I am not entirely sure) as far as I remember there is a general argument (see Schwabl) that shows that the most general Hamiltonian only contains products of one and two creation/annihiliation operators. – gented Dec 15 '15 at 23:39
In other problem that we will do in class in near future I have something like this: http://i.imgur.com/U0EAgfO.png . Is it legit? Is it ok with our definition? The hint is to solve it with Bogolyubov transformation. It's infite sum though, so I don't know. The only Hamiltonians that I found which acts on different Fock spaces is the one I posted in my first post and the one in the picture. – Caims Dec 15 '15 at 23:52
It does not make much sense to me as one would have to sum vectors in different Hilbert spaces, to be honest, unless some particular prescriptions are given. – gented Dec 15 '15 at 23:59
There's no prescriptions. There's just this hint and the goal is to find eigenstates. Well, I'm gonna ask my teacher then. Thank you for your time, your post really helped me to understand this better. – Caims Dec 16 '15 at 00:02
@GennaroTedesco Did you ever come across BCS theory or the Bogoliubov theory of the excitations of Bose-Einstein condensates? There exactly such terms $aa$ and $a^\dagger a^\dagger$ occur! This just means that the energy eigenstates do not have definite particle numbers. Summing vectors with different particle numbers is no problem at all in Fock space (which is a linear space). – Sebastian Riese Dec 16 '15 at 00:46
See my comments above. Bogoliubov transformation requires tensoring with the appropriate identity operators to be correctly defined, which do not appear in the above expression. Only then one may define sums among elements in different subspaces of the Fock space. – gented Dec 16 '15 at 00:48
2

What's so difficult about this Hamiltonian? Just consider $a, a^\dagger$ to be the ladder operators of the 1D harmonic oscillators. It's defined on the whole Fock space, and that is perfectly fine. – Meng Cheng Dec 16 '15 at 00:51
Can you please write the formal definition of the operator $a+a^{\dagger}$ (domains, co-domains and actions)? – gented Dec 16 '15 at 00:54
$(a + a^\dagger)\left|n\right> = a\left|n\right> + a^\dagger\left|n\right>$. Domain $H$, codomain some subspace of $H$. $a\left|0\right> = 0$. I don't see any problem. ($H = \text{lin} { \left|0\right>, \left|1\right>, \ldots }$). $a\left|n\right> = \left|n - 1 \right>$ for $n > 0$, $a^\dagger\left|n \right> = \left|n + 1 \right>$. – Sebastian Riese Dec 16 '15 at 00:58
Operators are defined on the Hilbert space. There is a well defined Fock space, spanned by $|n\rangle, n=0,1,\dots$. $a$ and $a^\dagger$ are both linear operators on this Hilbert space (so they map a state in this space to another state). What are their actions? You should know if you have taken any quantum mechanics course. – Meng Cheng Dec 16 '15 at 01:00
How do you sum $a|n\rangle + a^{\dagger}|n\rangle$, since the former lives in $\mathcal{H}{n-1}$ and the latter in $\mathcal{H}{n+1}$? You should see the inconsistency in the notation if you have taken any calculus course. – gented Dec 16 '15 at 01:04
The Hilbert space is spanned by the infinite number of states $|n\rangle$. They are all in the same Hilbert space. What prevents you from summing two states in the same Hilbert space up? It seems like you have never seen a Hilbert space whose dimension is larger than $1$, and states are labeled by some quantum numbers which can be changed by applying linear operators? – Meng Cheng Dec 16 '15 at 01:09
Just to give you some underlying basic definitions that you seem to overlook: $a\colon\mathcal{H}n\mapsto\mathcal{H}{n-1}$, whereas $a^{\dagger}\colon\mathcal{H}\mapsto\mathcal{H}_{n+1}$, and the Fock space is the infinite direct sum of all those different Hilbert spaces. As you see, they are not in the same Hilbert space. The Fock space is not the linear span of those, rather it is the tensor product span, that was my objection. If you don't see such difference I wonder how you deal with quantum field theories at all. – gented Dec 16 '15 at 01:14
If you correctly justify the procedure (and that was my initial question) then I totally agree with the final results (which are by the way of course correct); but if you sum things up without even seeing where the mistakes might be, well, that shows a very odd and lacking understanding of the theory of operators. – gented Dec 16 '15 at 01:17
2

You ask me about how I deal with quantum field theories at all? Here is how: quantum field theories deal with many-particle systems, and particles are allowed to be created and annihilated. So, the actual Hilbert space is spanned by the occupation number basis which specifies the occupation number in some single particle states. You can break this infinite dimensional Hilbert space into subspaces with different total number of occupations, that's your $\mathcal{H}_n$. So $a$ takes you between these subspaces, which is a linear operator in the actual infinite-dimensional Hilbert space. – Meng Cheng Dec 16 '15 at 01:19
It seems you have not read any of my comments at all. – gented Dec 16 '15 at 01:20
I'm afraid it is not me that lacks the full understanding of the theory of operators. – Meng Cheng Dec 16 '15 at 01:21
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I don't think there is any point in conutinuing this discussion. The answer is there, the OP seems to have no problem understanding $a$ and $a^\dagger$ (besides, the problem was probably assigned in the context of ladder operators for harmonic oscillators). We both think each other not reading the comments and confused about basic things, that's perfectly fine. I'll stop. – Meng Cheng Dec 16 '15 at 01:22

score -2 · Answer 3 · answered Dec 15 '15 at 22:51

How about just using Matrix representation.

https://en.wikipedia.org/wiki/Creation_and_annihilation_operators#Matrix_representation

You can have any number of Bosons from 0 to infinity. That will be your basis, and your wave function is represented as a vector in which element 0 gives probability amplitude of having 0 quanta, element 1 gives amplitude for 1 quantum, 2 for 2 quanta etc. in the system.

Calculating with matrices is easy:

N = 1000;
a = zeros(N);
for i=1:N-1
a(i,i+1) = sqrt(i);
end
H = 10*a*a' + 5 / 2 * (a*a+a'*a');
eig(N)

Disclaimer: I have worked with Fermions almost always, except for some quantum course 8 years ago.

Second quantization and Hamiltonian diagonalization

3 Answers3

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