Timeline for Second quantization and Hamiltonian diagonalization
Current License: CC BY-SA 3.0
21 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Dec 16, 2015 at 1:22 | comment | added | Meng Cheng | 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. | |
| Dec 16, 2015 at 1:21 | comment | added | Meng Cheng | I'm afraid it is not me that lacks the full understanding of the theory of operators. | |
| Dec 16, 2015 at 1:20 | comment | added | gented | It seems you have not read any of my comments at all. | |
| Dec 16, 2015 at 1:19 | comment | added | Meng Cheng | 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. | |
| Dec 16, 2015 at 1:17 | comment | added | gented | 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. | |
| Dec 16, 2015 at 1:14 | comment | added | gented | 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. | |
| Dec 16, 2015 at 1:09 | comment | added | Meng Cheng | 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? | |
| Dec 16, 2015 at 1:04 | comment | added | gented | 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. | |
| Dec 16, 2015 at 1:00 | comment | added | Meng Cheng | 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. | |
| Dec 16, 2015 at 0:58 | comment | added | Sebastian Riese | $(a + a^\dagger)\left|n\right> = a\left|n\right> + a^\dagger\left|n\right>$. 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 $n > 0$, $a^\dagger\left|n \right> = \left|n + 1 \right>$. | |
| Dec 16, 2015 at 0:54 | comment | added | gented | Can you please write the formal definition of the operator $a+a^{\dagger}$ (domains, co-domains and actions)? | |
| Dec 16, 2015 at 0:51 | comment | added | Meng Cheng | 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. | |
| Dec 16, 2015 at 0:48 | comment | added | gented | 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. | |
| Dec 16, 2015 at 0:46 | comment | added | Sebastian Riese | @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). | |
| Dec 16, 2015 at 0:02 | comment | added | Caims | 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. | |
| Dec 15, 2015 at 23:59 | comment | added | gented | 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. | |
| Dec 15, 2015 at 23:52 | comment | added | Caims | In other problem that we will do in class in near future I have something like this: 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. | |
| Dec 15, 2015 at 23:39 | comment | added | gented | 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. | |
| Dec 15, 2015 at 23:33 | comment | added | Caims | 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 $H_n$ ? | |
| Dec 15, 2015 at 23:18 | comment | added | Caims | 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. | |
| Dec 15, 2015 at 23:13 | history | answered | gented | CC BY-SA 3.0 |
