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Consider the following state for some bosons represented in Fock space:

$$|2\rangle_{k_1}|1\rangle_{k_2}$$

where $k_i$ is some distinguishing index. You may think of these as the two different wavevectors for photons.

Now, if we use the Hilbert space representation of each individual boson, the same normalized state is written as

$$\frac{1}{\sqrt{3}} \left( |k_1,k_1,k_2\rangle+|k_1,k_2,k_1\rangle+|k_2,k_1,k_1\rangle \right)$$

Is this correct? Are the two expressions above equivalent? Now, consider that the particle in mode $k_2$ is transferred to the mode $k_1$ (photon addition from one mode into another, for example). This would give the state

$$\frac{3}{\sqrt{3}} |k_1,k_1,k_1\rangle$$

which is clearly not normalized. How is this possible? Does it mean Fock space is more fundamental? However, the Hilbert space picture appears more intuitive at first glance.

In any case, how do we make sense of the prefactor of $\sqrt{3}$ intuitively?

I am trying to somehow relate it to the prefactors that appear due to the action of creation operator in mode $k_1$:

$\hat{a}^{\dagger}|2\rangle=\sqrt{3}|3\rangle$

Saurabh Shringarpure
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    "However, the Hilbert space picture appears more intuitive at first glance." That's odd. The (anti)symmetrized wave function picture should seem considerably less intuitive. See this post. – DanielSank May 25 '20 at 23:12
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    Let me pose a counter-question that may be enlightening: Suppose we have the spin state $\frac{1}{\sqrt{2}}(|\uparrow\rangle+|\downarrow\rangle)$. Now, if we let $|\uparrow\rangle=|\downarrow\rangle$, our state becomes $\frac{2}{\sqrt{2}}|\uparrow\rangle$, which is not normalized. How is this possible? – probably_someone May 25 '20 at 23:17
  • Or, for another, even simpler counter-question: Suppose we have the vector $\frac{1}{\sqrt{2}}\hat{x}+\frac{1}{\sqrt{2}}\hat{y}$. If we let $\hat{x}=\hat{y}$, the vector then becomes $\frac{2}{\sqrt{2}}\hat{x}$, which is no longer of length 1. How is this possible? – probably_someone May 25 '20 at 23:19
  • @probably_someone I don't know. The reason I used $k_2$ goes to $k_1$ was that I was picturing photon addition where we add a photon from one mode into another. – Saurabh Shringarpure May 26 '20 at 00:12
  • @DanielSank I was visualizing each boson as a separate particle, like a billiard ball, and so I liked the representation where we see the state of each individual particle separately. – Saurabh Shringarpure May 26 '20 at 00:20
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    The reason that $k_2 \to k_1$ breaks the normalization is that the inner product is not continuous as a function of $k$ (because it is a delta funciton). – Javier May 26 '20 at 00:20
  • @DanielSank Thanks for the link. – Saurabh Shringarpure May 26 '20 at 00:36
  • @Javier Isn't the issue more that we can't relabel basis states that way? In particular, when you relabel basis states, you have to do in a unitary fashion. For example, the issue will persist if you do something like what @probablysomeone suggested which doesn't involve delta functions at all. –  May 26 '20 at 00:43
  • @DvijD.C. That was my point, actually - the action "let $k_2=k_1$" is as illegal as declaring that two orthogonal basis vectors are actually the same vector. You're destroying the structure of the vector space you're working with, so you shouldn't expect anything about that vector space to behave well afterward. – probably_someone May 26 '20 at 00:45
  • @probablysomeone Yes, I was just echoing your point in order to convey that delta functions don't seem to be the issue, the issue is what you point out, you're essentially reducing the dimensionality of your Hilbert space when you name two orthogonal basis states the same, so the obvious consequence would be messed up normalization (among other things). –  May 26 '20 at 00:49
  • @probably_someone and Dvij D.C. I was wondering how to think about photon addition then, where a creation operator acts on number state to give a higher number state. The operation is non-unitary but it can still be achieved. – Saurabh Shringarpure May 26 '20 at 01:41
  • @DvijD.C. You're not turning one basis vector into another, you're defining a function $f(k_1, k_2) = |k_1, k_2\rangle$ (or something like that) and then looking at its norm. We don't think of this function as having an analog in the finite-dimensional case because we can't label the basis vectors with a continuous variable. – Javier May 26 '20 at 01:53
  • @SaurabhU.Shringarpure There is a separate creation operator for every value of $k$. Acting with the creation operator on a particular state will increase the occupancy of that value of $k$ only, leaving the occupancies for other values of $k$ unaffected. – probably_someone May 26 '20 at 01:53

1 Answers1

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Fock space is the Hilbert space of multiparticle systems.

The difference you're alluding to is a difference of representation. The one you're naturally associating with Fock space is the occupation number representation, which is obviously more natural to describe multiparticle systems due to the indistinguishability of particles.

However, notice that the basis states of the occupation number representation are the same basis states that you get when you construct the full Hilbert space as a product of single-particle Hilbert spaces and write down the full basis as a product of bases of single-particle Hilbert spaces (see, for example, $|1\rangle_k|1\rangle_K$ is the same as $\frac{1}{\sqrt{2}}(|k\rangle |K\rangle + |K\rangle |k\rangle)$.

Hence, the occupation number representation just writes all of the same basis states in a new notation, a much helpful one, of course). So, not only is Fock space the same Hilbert space, the occupation number basis is not even some new basis used to span the same space.

The $k_1$ going to $k_2$ trick is misleading for the reason that we're talking about $\underline{\text{basis}}$ and you're mixing them with $\textit{states}$. One can perform some unitary transformations on the basis to get to a new basis but keeping everything else the same and just letting $k_1$ go to $k_2$ is not such a transformation.

If you want to talk about a physical state labeled by arbitrary $k_1$ and $k_2$, then you can do such a trick, and say that when this happens (in the expansion of the state into basis vectors), the coefficient in front of the basis vector with different $k_1$ and $k_2$ become zero and those coefficient which have all three particles having the same momentum i.e. ($k_1$ = $k_2$) goes to one.

  • What about the case of fermions and the following reason given as an explanation for the Pauli exclusion principle: We have two identical fermions in a completely asymmetric state $\frac{|k\rangle |K\rangle -|K\rangle |k\rangle}{\sqrt 2}$. Now if we imagine that we put both fermions in the same state $|k\rangle $, then this expression tells that it disappears and hence cannot exist. – Saurabh Shringarpure May 26 '20 at 22:37
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    @SaurabhUShringarpure The state you wrote is explicitly normalized for different $k$ and $K$ so you cannot let $k$ go to $K$ in that expression. The idea is that to construct the basis for the full Hilbert space, you need to consider anti-symmetric superpositions of product of basis. And the anti-symmetric version of $|k\rangle |k\rangle$ vanishes so you simply don't have such basis in a Fermionic Hilbert space. –  May 26 '20 at 22:46
  • Ok, that makes sense. But, it is ok to say that the boson creation operator, that is applied to the Fock space representations, would act in the manner I described in the question in the Hilbert space representation. The creation operator is nonunitary and so I might be safe? – Saurabh Shringarpure May 26 '20 at 22:55
  • @SaurabhU.Shringarpure Yes, that's right. A creation operator is under no obligation to preserve the norm. –  May 26 '20 at 23:00