# Product Fock spaces

1. Jun 8, 2015

### JorisL

Hi,

I'm having some issues with a piece of my notes. (relevant pages attached)
First we introduce an isomorphism $U = \oplus_n U_n$ from $\Gamma^{(a)s}\left(\mathcal{H}_1\oplus\mathcal{H}_2\right)$ to $\Gamma^{(a)s}\left(\mathcal{H}_1\right)\otimes\Gamma^{(a)s}\left(\mathcal{H}_2\right)$
With the $U_n$ mapping the n-particle space (layer if you like) of the Fock space to the product space.

So far I'm not seeing any real problems (this is what happens on the first page of the pdf).

The next page is where things get vague for me (lets ignore the paragraph about the Gibbs paradox until I get what's being defined)

We look at $n$ indistinguishable particles, each with the same single particle Hilbert space $\mathcal{H}$.
Now comes the first troubling part for me
Okay in a low density regime we can indeed look at such a system without too much loss of generality.

Now the big problem, equation (45) in the PDF says the following (for bosons)
$Ua^\dagger(\phi_1\oplus\phi_2)U^\dagger = a^\dagger(\phi_1)\otimes {1\!\!1} +{1\!\!1}\otimes a^\dagger(\phi_2)$

When the $U_n$ which build $U$ are defined in eq. (40-41) I see
$U_1(\phi_1\oplus\phi_2) = \phi_1\oplus\phi_2$ is this shorthand for $\phi_1 \otimes {1\!\!1}\oplus {1\!\!1}\otimes\phi_2$?
If so how do they look at $a^\dagger(\phi_1\oplus\phi_2)U^\dagger$?

If I can get my head around this part I can finish the rest but it's just not coming to me.
Any recommended books on the topic?

Thanks,

Joris

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2. Jun 8, 2015

### micromass

Staff Emeritus
Remember that $\mathbb{C}\otimes \mathcal{H} \cong \mathcal{H}$. Thus what he does in the text is identifying
$$\mathcal{H}\oplus \mathcal{K}\cong (\mathbb{C}\otimes \mathcal{K})\oplus (\mathcal{H}\otimes \mathbb{C})$$
through the isomorphism $\varphi\oplus \psi\rightarrow (1\otimes \psi)\oplus (\varphi\otimes 1)$.

3. Jun 8, 2015

### micromass

Staff Emeritus
If you want me to answer the second part, you will have to tell me what $a$ and $a^\dagger$ are. I don't know much physics, but this is basically math.

4. Jun 8, 2015

### JorisL

Okay, thanks for the first clarification.

The operator a isn't defined thus far actually, in the next section we assume that $a = (a^\dagger)^\dagger$.

But let me give the definition we used.
$\Phi\in\Gamma^{(a)s}$ we have from the construction that it's of the form $\Phi = \phi^{(0)}\oplus \phi^{(1)}\oplus\phi^{(2)}\oplus \ldots$.
The creation operator $a^\dagger(\psi)$ then maps $\Phi$ to
$a^\dagger(\psi)\Phi = 0\oplus\phi^{(0)}\psi\oplus\psi\wedge\phi^{(1)}\oplus\psi\wedge\phi^{(2)}\oplus\ldots$ for fermions

Here the wedge denotes an anti-symmetrized tensorproduct. ($\phi_1\wedge\phi_2 = -\phi_2\wedge\phi_1$) This is suitable for fermions.
For bosons we replace it by a symmetrized product which we denote by $\odot$ in the text (not standard by any means)

That's the definition we used.

5. Jun 8, 2015

### micromass

Staff Emeritus
It suffices to show that $U a^\dagger (\varphi_1\oplus \varphi_2) = (a^\dagger(x)\otimes \mathbb{1} + \mathbb{1}\otimes a^\dagger(y)) U$.

It suffices to do this on the various levels. So let me do this on the first level. So if $\Phi\in \Gamma^{(a)s}(\mathcal{H}_1\oplus \mathcal{H}_2)$ has the form
$$\Phi = (0,(\varphi\oplus \psi),0,0,...)$$
Then $$U\Phi = (0,\varphi\otimes 1 + 1\otimes \psi,0,0,...) = (0,\varphi,0,0,...)\otimes (1,0,0,0,...) + (1,0,0,0,...)\otimes (0,\psi,0,0,...)$$
Applying $(a^\dagger(x)\otimes \mathbb{1} + \mathbb{1}\otimes a^\dagger(y))$ gives us
$$(0,0,x\otimes \varphi,0,...)\otimes (1,0,0,...) + (0,\varphi,0,0,...)\otimes (0,y,0,0,...) + (1,0,0,0,...)\otimes (0,0,y\otimes \psi,0,...) + (0,x,0,0,...)\otimes (0,\psi,0,0,...)$$
Notice that this corresponds to
$$(0,0,x\otimes \varphi,0,...) + (0,0,\varphi\otimes y,0,...) + (0,0,y\otimes \psi,0,...) + (0,0,x\otimes \psi,0,...)~~~(*)$$
On the other hand
$$a^\dagger(x\oplus y)\Phi = (0,0,(x\oplus y)\otimes (\varphi\oplus \psi),0,0,...)$$
Applying $U$ to this gives
$$(0,0,(x\otimes \varphi)\oplus ((x\otimes \psi) + (\varphi\otimes y) \oplus (y\otimes \psi),0,...)$$
This corresponds to $(*)$.

6. Jun 8, 2015

### JorisL

Great stuff!
This was exactly what I needed.
You know of any good (mathematical) references for this kind of results?
The physics text I've found kind of glance over it and rush to apply it. (not good enough for my inner mathematician but it takes time I don't have at the moment)

Thanks,

Joris

7. Jun 8, 2015

### micromass

Staff Emeritus
I'm afraid I don't have anything precise for you. What I did comes mainly from my knowledge of abstract algebra and a bit of functional analysis. So I can't give you one reference for the entire thing. But if you are asking for details on only one specific step, then I probably have references for that. Nevertheless, the math behind this can be found in Roman's advanced linear algebra, but this doesn't mention Fock spaces at all, so it's only for the deeper math.

8. Jun 8, 2015

### JorisL

Thanks, I'll see if I can get any more out of it using that resource.
But I believe my understanding is deep enough for what is expected. It really was a crash course in 4 two hour lectures so he'll not be too strict about these things.

Thanks,

Joris