# Switching between Dirac notation and creation/annihilation operators?

I am wondering:

What is the explicit difference between the two Hamiltonians, $$H_1=\sum_{m=1}^Ntc_m^\dagger c_m+h.c.,$$ $$H_2=\sum_{m=1}^Nt |m\rangle\langle m | + h.c..$$ The Hamiltonians describe fermions restricted to live on a dicrete, one-dimensional lattice of length $N$. Here $|m \rangle$ is a position ket of site $m$.

What I think:

I think that $H_1$ is the so called second quantized Hamiltonian, which lives in a multi-particle Hilbert space. While the other Hamiltonian, $H_2$, is a single-particle Hamiltonian, living in a single-particle Hilbert space.

Is it true that these Hamiltonians live in different Hilbert spaces?

Going between the two Hamiltonians:

To what extent can one switch between the two Hamiltonians in the middle of a calculation? I believe that this often done every now and then, perhaps quite sloppily. What really happens as we switch between these formulations?

• So what's the status on this post? Do we need more information? – DanielSank Apr 9 '18 at 4:59
• @DanielSank I still haven't discussed this with my supervisor. But I am pretty satisfied with you answer. – Jens Roderus Apr 9 '18 at 6:36
• Hmm, I'll be interested to hear what your supervisor has to say. – DanielSank Apr 9 '18 at 6:43
• @DanielSank To provide some context: The reason why I asked is ebcause of this: physics.stackexchange.com/questions/381328/… The answer shows how there are differences for the symmetries of a first/second quantized Hamiltonian. Upon reading this I sorta wanted to take a step back and really understand what happens when we go between tehe first/second quantized Hamiltonian. hence this post. – Jens Roderus Apr 9 '18 at 6:46

I think that $H_1$ is the so called second quantized Hamiltonian, which lives in a multi-particle Hilbert space. While the other Hamiltonian, $H_2$, is a single-particle Hamiltonian, living in a single-particle Hilbert space.

If, as the question v4 says, the symbol $|m\rangle$ indicates a state of a particle localized to site $m$, then $H_2$ is indeed supposed to be viewed as a single particle Hamiltonian. In that case, the Hamiltonians are not necessarily the same. However, note that $H_1$ conserves the number of particles, so if you start with one particle, then practically speaking $H_1$ and $H_2$ are the same.

To what extent can one switch between the two Hamiltonians in the middle of a calculation? I believe that this often done every now and then, perhaps quite sloppily. What really happens as we switch between these formulations?

Often, that switch happens as follows:

• Identify non-interacting modes in the system.
• Treat each non-interacting mode separately as its own single-particle system where the energy level of this fake single particle actually corresponds to the number of quanta (Fock state) in that mode.

That's the only case I can think of where we regularly switch from second quantization to single-particle quantum mechanics.

In chat, OP asked what is "given up" by going from second quantization to single-particle. in the case described by the bullet points, nothing is sacrificed. We treat each mode separately because they're non-interacting. However, suppose we just have a box which may or may not contain electrons. In full generality, we should use second quantization and allow for an arbitrary number of particles. However, if we know that there's only one (or more) electrons and we know that we're working with low enough energy that spontaneous electron creation will not happen, then we can use single particle (or two-particle or three-particle... based on how many electrons there are) quantum mechanics to describe the system. We do this all the time when describing e.g. the electrons in a helium atom. We know there are two electrons and we know there isn't going to be any particle creation/annihilation (unless we're at high enough energy).

Note that the energy to create a massive particle is $E = m c^2$ were $c$ is the speed of light. It's because this energy cost is large that we know the number of electrons in a helium atom won't change under normal conditions. On the other hand, if we're working with photons, which have zero mass, or we're working in a thermodynamic situation where we have a reservoir of particles, then the particle number isn't fixed and we can't restrict to a specific particle number.

If we interpret $|m\rangle$ to indicate a Fock state with one occupation on site $m$, and if we're working with Fermions, then the Hamiltonians are the same, but written in two different notations. The rest of this post elaborates on the previous sentence.

## Representing linear transformations (operators) as sums of bras and kets

Given a vector space and a basis $e \equiv \{|e_1\rangle, |e_2\rangle,\ldots\}$, we can write any linear transformation $T$ on that vector space as a sum $$T = \sum_{ij} T_{ji} |e_j \rangle \langle e_i |$$ where $T_{ji}$ are literally the elements of the matrix representation of $T$ in the $e$ basis. We can show that trivially: $$\langle e_j | T | e_i \rangle = \sum_{kl} T_{lk} \underbrace{\langle e_j | e_l \rangle}_{\delta_{jl}} \langle \underbrace{e_i | e_ k \rangle}_{\delta_{ik}} = T_{ji} \, .$$ In other words, in the $e$ basis, the $j^\text{th}$ component of $T|e_i\rangle$ is $T_{ji}$.

An easy way to remember this is to use the fact that $\sum_j |e_j\rangle \langle e_j| = \text{identity}$, so $$T = \underbrace{\left( \sum_j |e_j \rangle \langle e_j| \right)}_\text{identity} T \underbrace{\left( \sum_i |e_i \rangle \langle e_i| \right)}_\text{identity} = \sum_{ij} \langle e_j | T | e_i \rangle |e_j\rangle\langle e_i| = \sum_{ij} T_{ji} |e_j \rangle \langle e_i| \, .$$

## Fermion creation/annihilation

In second quantization language, we represent quantum states etc. with creation annihilation operators $c_m$ and $c_m^\dagger$, as noted in the original post. Those operators are of course just linear transformations and can be represented in the bra/ket way, just the same as the arbitrary transformation $T$ in the discussion above.

Physically, the operator $c_m$ removes one unit of excitation from the $m^\text{th}$ mode of the system, which in this case is a spatially localized mode, i.e. occupation of a lattice site. The operator $c_m^\dagger c_m$ removes one excitation from the $m^\text{th}$ mode and then puts it back.

Suppose we work in a basis consisting of two elements for each lattice site: $\{|0\rangle, |1\rangle\}$ corresponding to zero or one occupation on the site (these are Fermions so we can't have more than one). We can then write $c^\dagger c$ for a particular lattice site as \begin{align} c^\dagger c &= \langle 0 | c^\dagger c | 0 \rangle |0 \rangle \langle 0 | + \langle 0 | c^\dagger c | 1 \rangle |0 \rangle \langle 1 | + \langle 1 | c^\dagger c | 0 \rangle |1 \rangle \langle 0 | + \langle 1 | c^\dagger c | 1 \rangle |1 \rangle \langle 1 | \\ &= \langle 1 | c^\dagger c | 1 \rangle |1 \rangle \langle 1 | \end{align} where in the second line we dropped terms that were zero. If a term in the Hamiltonian is $t (c^\dagger c)$, then we can write it as $t |1 \rangle \langle 1 |$.

All of that discussion was for a single mode. If we now go to all modes (sites on the lattice) and rename each $|1\rangle$ ket for that mode as $|m\rangle$, we can see that $H_1$ and $H_2$ are exactly the same thing.

• So are you saying that they are both multi or single particle Hamiltonians? – Jens Roderus Mar 28 '18 at 20:32
• If you have a reference to back up the fact that they are identical then that would be great. – Jens Roderus Mar 28 '18 at 20:33
• @JensRoderus They are both multi-particle hamiltonians. I can't think of a way to show a reference unless we can dig up a paper or textbook that happens to use both notations in the same context. The original question didn't ask for a reference though, and if it had, I wouldn't have posted this answer :-) – DanielSank Mar 28 '18 at 20:38
• @JensRoderus Actually I should say: it's possible to represent a single particle Hamiltonian with the expression you've given for $H_2$. The notation used in the question isn't specific enough on its own to know exactly what is meant by $H_2$. – DanielSank Mar 28 '18 at 20:39
• I don't see how $H_2$ could act on a multi-particle state. A two particle state would be written $| m_1\rangle \otimes | m_2\rangle \equiv | m_1,m_2\rangle$. Now how does $|m \rangle \langle m |$ act on that? – Jens Roderus Mar 28 '18 at 20:48