# Are constant terms in second-quantization relevant?

I have a rather broad question and a specific problem. Let's take a orthonormal single-particle basis $\{ \vert i \rangle \}$, a simple single-particle Hamiltonian $$\tilde{H} = \sum_{i, j} h_{i j} \vert i \rangle \langle j \vert$$ and its second-quantized form $$H = \sum_{i, j} h_{ij} a^\dagger_i a_j ~.$$ Now I add a constant $C$, i.e. $H_C = \sum_{i, j} h_{ij} a^\dagger_i a_j + C ~.$ The broad question is, if this constant is relevant. Does it have any effect? How does one deal with a constant operator in Fock space? What is the single-particle version of $\tilde{H}_C$? I was unable to find literature about this. Such constants sometimes appear when one wants to define a Hamiltonian with a particular symmetry, e.g. particle-hole symmetry, and as far as I can tell, these constants are ignored or considered unimportant.

The specific problem, why I want to know about the consequences of these constants, is the following. (Don't worry about the details, just concentrate on the Hamiltonian.) The single-particle imaginary-time Green's function is defined as $$G_{k l}(\tau) = - \frac{1}{Z} \mathrm{Tr}(e^{-\beta H} \mathcal{T} a_k[\tau] a^\dagger_k) ~,$$ with $Z = \mathrm{Tr}\left( e^{-\beta H} \right)$, $\beta = 1/T$ and $a_k[\tau] = e^{\tau H} a_k e^{-\tau H}$. It's Fourier transformation is defined as $$G_{k l}(i \omega) = \int_0^\beta d\tau e^{i \omega \tau} G_{k l}(\tau) ~.$$ They fulfill the relations \begin{align} \partial_\tau G_{k l}(\tau) &= -\delta(\tau) \delta_{k l} - \sum_{m} h_{k m} G_{m l}(\tau) \\ \delta_{k l} &= \sum_{m} (i\omega \cdot \delta_{k m} - h_{k m}) G_{m l}(i\omega) ~. \tag{1} \end{align} We see that $G(i\omega)$ -- understood as a matrix -- is inverse to $Q$ with $Q_{k l} := (i\omega \cdot \delta_{k l} - h_{k l})$. Another approach is given by the resolvent $\mathcal{G}(z) = (z - \tilde{H})^{-1}$. Then $\mathcal{G}_{k l}(i \omega) = \langle k \vert (i \omega - \tilde{H})^{-1} \vert l \rangle$ which apparently satisfies Eq. (1) and therefore $\mathcal{G}_{k l}(i \omega) = G_{k l}(i\omega)$. This is a useful relation and used at several occasions in the previous research from students and my research.

Now what happens if we use $H_C$ instead of $H$ in the definitions of $G(\tau)$ and $\mathcal{G}$? Naively I would expect the constant to drop out from $G(\tau)$ as it appears in the enumerator and denominator and can be factored out. But I can't see something similar happen for $\mathcal{G}$. Does the equality not hold anymore?

• The identity operator $\mathrm{Id}$ of the Fock space acts like the identity on any subspace with fixed particles (this is very easy to prove), so $C=C\mathrm{Id}$ acts on the one particle subspace as the multiplication by $C$. As in QM, the constant is not so relevant since it does not change the dynamics, however it is sometimes useful: e.g. in the Nelson model, where subtracting a constant (that in the limit becomes infinite) is necessary to renormalize the Hamiltonian. – yuggib Aug 28 '14 at 23:30
• @yuggib Ok, yea, I guessed that it acts likes this but was not sure. Thank you. That means in this case that the constant induces an energy shift of the $h_{i i}$, and it probably does not simply cancel out of the imaginary-time Green's function. So the equality still holds. That's better than nothing, but I still need an argument why it was and is ignored in the hitherto research. They did not mention it. In my Hamiltonian there is an interaction $\propto (a^\dagger_i a_j - 1)(a^\dagger_k a_l -1)$ and we just ignore the resulting +1 even though it is an operator... hm. – hauntergeist Aug 29 '14 at 0:28

The constant $C$ is not a part of the single-particle Hamiltonian. It is what is called the vacuum energy, and has no observable effect unless we look at gravitation.

To be specific, adding this constant to the Hamiltonian makes $e^{-\beta H}\,\rightarrow e^{-\beta C}e^{-\beta H}$ and $Z=\rm{Tr}(e^{-\beta H})\, \rightarrow\, e^{-\beta C} Z$. That is, the same factor $e^{-\beta C}$ is introduced to both the numerator and denominator in the expression for the imaginary-time Green's function, and they cancel out.

Update: OP considered a Hamiltonian of the form $H = \sum_{ij} h_{ij} |i\rangle\langle j|$, which translates into $H = \sum_{ij} h_{ij} a_{i}^{\dagger} a_{j}$ in the second quantization. If we call this a single-particle term in the Hamiltonian, it is inescapable that an $n$-particle term should be what consists of $n$ creation and $n$ annihilation operators in its second quantized form.

According to this definition, an $n$-particle term acts on the subspace of the Fock space corresponding to (# of particles)$\ge n$. In this regard, an overall constant in the second-quantized Hamiltonian is a zero-particle term, or the vacuum energy.

As a concrete example, let's consider the following second quantized Hamiltonian:

$$H = C + \sum_{ij} h_{ij} a_{i}^{\dagger} a_{j} + \frac{1}{2}\sum_{ijkl} V_{ijkl} a_{i}^{\dagger} a_{j}^{\dagger}a_{l} a_{k}.$$ Then, its first-quantized form when there are $N$ particles should be $$H = C + \sum_{p=1}^{N}\sum_{ij} h_{ij} |i\rangle_{p}\langle j|_{p} + \sum_{p=1}^{N}\sum_{q=p+1}^{N} \sum_{ijkl}V_{ijkl} |i,j\rangle_{p,q}\langle k,l|_{p,q},$$ where $|i\rangle_{p}$ is a state in the single-particle Hilbert space of the $p$th particle, whereas $|i,j\rangle_{p,q}$ is a state in the two-particle Hilbert space of the $p$th and $q$th particles ($p<q$).

• To be acting only on the vacuum $\Omega$, it has to be of the form $C \lvert\Omega\rangle\langle\Omega\rvert$. Usually, when you write a number as an operator, you intend it multiplied by the identity operator, so it acts on any vector as the multiplication by $C$, the single particles ones as well. – yuggib Aug 29 '14 at 7:59
• @yuggib Vacuum energy is present regardless of the number of particles in the system. Hence it is indeed an operator that is proportional to the identity acting on the Fock space. – higgsss Aug 29 '14 at 8:01
• then it is also part of the single particle Hamiltonian, because it acts also on that subspace. – yuggib Aug 29 '14 at 8:08
• @yuggib I guess we should clarify what we mean by a single-particle Hamiltonian. To me it means a term of the form $H_{1} = \sum_{ij} h_{ij} a_{i}^{\dagger}a_{j}$. Similarly, a two-particle Hamiltonian would be of the form $H_{2} = \tfrac{1}{2}\sum_{i,j,k,l} h_{ijkl} a_{i}^{\dagger}a_{j}^{\dagger}a_{l}a_{k}$. On the other hand, the vacuum energy is just a constant. – higgsss Aug 29 '14 at 8:14
• @hauntergeist Shifting all interactions by $C$ is certainly unphysical. If your second-quatized Hamiltonian is given by $\sum_{ij} h_{ij} a_{i}^{\dagger}a_{j} + C$, this $C$ should not show up in its first-quantized form $\sum_{ij} h_{ij} |i\rangle\langle j|$. – higgsss Aug 29 '14 at 13:53