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We have two places, call them Here and There where a particle could be. When the particle isn’t spatially superposed we write its state as $ |1\rangle_{Here}\otimes|0\rangle_{There}$ zero being the vacuum mode.

My question is why when the particle is superposed, for example in the state $$\frac{1}{\sqrt{2}}(|Here\rangle + |There\rangle)$$ if we want to write its state including the vacuum modes we write it like this $$\frac{1}{\sqrt{2}}(|1\rangle_{Here}\otimes|0\rangle_{There} + |0\rangle_{Here}\otimes|1\rangle_{There})$$ which looks like entangled form. And we don’t write it in some separable form in analogy with the case in first equation: $$\frac{|1\rangle_{Here}+|1\rangle_{There}}{\sqrt{2}}\otimes\frac{|0\rangle_{Here}-|0\rangle_{There}}{\sqrt{2}}.$$ Or even $$|1\rangle_{\frac{H+T}{\sqrt{2}}}\otimes|0\rangle_{\frac{H-T}{\sqrt{2}}}.$$

Superpositional +/- states are as legitimate basis as localized ones, but vacuum modes notation seems to suggest there is difference - one state is separable, the other - entangled. I was thinking that this happens because the creation and annihilation operators are defined for local points like Here, but then again, why not define them for $\frac{H+T}{\sqrt{2}}$.

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Both expressions are legitimate: you have identified the difference between first quantization and second quantization. The specific state you mention was subject to a lot of debate as to whether or not there was Single-particle entanglement; ultimately the real question is whether the entanglement you've found is useful for your particular purpose.

The best way I can think of for drawing out your question is to look a single photon that is horizontally polarized. If you literally rotate your head, all of a sudden the particle looks diagonally polarized (superposition of horizontal and vertical polarizations). In the first quantized notation, this means $$|H\rangle\to|D\rangle=\frac{|H\rangle+|V\rangle}{\sqrt{2}},$$ while in the second-quantized notation it means $$|1\rangle_H\otimes |0\rangle_V\to \frac{|1\rangle_H\otimes |0\rangle_V+|0\rangle_H\otimes |1\rangle_V}{\sqrt{2}}.$$ This is exactly the scenario described in the question, but it is an explicit physical example where literally tilting your head generates entanglement. That seems preposterous! But yet it is entanglement in the mode basis or the second-quantized notation. You'll probably be happy to know that this kind of entanglement on its own is seldom useful, so people sometimes get defensive about what is "true" or "useful" entanglement, but one should realize that this mathematical structure is indeed entanglement and that entanglement always depends on your decomposition of Hilbert space (and thus on your mode structure). Thus entanglement on its own is not the most important thing in quantum theory; entanglement in a particular decomposition of Hilbert space tends to be more important.

What is really happening here? We need a Hilbert space decomposition to agree upon if we want to discuss entanglement. Our quantum state always belongs to some Hilbert space $\mathcal{H}$. In some scenarios, this space can be factorized into the tensor product of two Hilbert spaces $\mathcal{H}=\mathcal{H}_A\otimes\mathcal{H}_B$, which means that the basis vectors of the full Hilbert space can be written as tensor products of the basis vectors of the constituent Hilbert spaces. Then, we can discuss entanglement between these two constituent Hilbert spaces. But it turns out the decomposition is seldom unique; there might be another decomposition $\mathcal{H}=\mathcal{H}_{A^\prime}\otimes\mathcal{H}_{B^\prime}$. And a state that is separable (not entangled) when written in the first decomposition might look entangled when written in the second ($^\prime$) decomposition. The other typical example of this is solving the Schrodinger equation for energy states of the hydrogen atom: when we decompose the system into centre of mass coordinates and relative coordinates, we can solve for each independently. The centre of mass's position is separable from (not entangled with) the relative position of the proton and the electron. But if you look in the mode decomposition where you treat the proton and electron's positions as the two relevant degrees of freedom, then they look entangled! In the examples here, one treatment only looks at one Hilbert space $\mathcal{H}$ for a single particle, and uses various basis states for that particle like $|H\rangle$ and $|V\rangle$ or $|Here\rangle$ and $|There\rangle$. The other treatment considers the factorization into H and V or Here and There subspaces such that the joint basis states are $|i\rangle_{H}\otimes |j\rangle_V$ for various non-negative integers $i$ and $j$.

These distinctions become more rich when there is more than one particle. If one has exactly two particles, I can think of decomposing the Hilbert space in terms of the basis states of each particle (like $|H\rangle\otimes |V\rangle$ or $|Here\rangle\otimes|Here\rangle$), or I can decompose it in terms of basis states that count the number of H and V particles like $|i\rangle_H\otimes |j\rangle_V$ with $i+j=2$. It happens that both of these treatments have only two subspaces, so much confusion may arise. Maybe it is better to already go to the three or four or five-particle spaces: then the first-quantized decomposition has basis states that look like $|H\rangle\otimes |H\rangle\otimes|V\rangle\otimes|H\rangle\otimes|V\rangle$ while the second-quantized decomposition still only has two subspaces being tensor product-ed for the basis states like $|3\rangle_H\otimes|2\rangle_V$. These look manifestly different but can describe the same states. Now, if the particles are identical, then often one requires superpositions of first-quantized basis states to describe a single second-quantized state (because $|3\rangle_H\otimes|2\rangle_V$ does not tell you which of the particles is H versus V).

You can also think of questions like this in the reverse direction: if I have a state that is separable in the second-quantized notation, it might look entangled in the first-quantized one, due to symmetry arguments! E.g., if you have indistinguishable particles like bosons then $$|1\rangle_H\otimes |1\rangle_V=\frac{|H\rangle\otimes |V\rangle+|V\rangle\otimes |H\rangle}{\sqrt{2}}.$$ A recent result in this domain is that Entanglement between Identical Particles Is a Useful and Consistent Resource and there is lots of ongoing work on figuring out what exactly are the most important resources for quantum information protocols.

Now one important point of second quantization is that it allows us to describe situations where the number of particles is not fixed. You can do what you've written because you only have one particle, and indeed you could do something like that for two or three or $N$ particles, but it would be hard to use the $|Here\rangle$ and $|There\rangle$ states (first quantization) if you had a superposition of one and two particles. Yet the basis with the vacuum modes does just fine (second quantization); e.g., $$(|2\rangle_{Here}\otimes |0\rangle_{There}+ |1\rangle_{Here}\otimes |1\rangle_{There}+|0\rangle_{Here}\otimes |1\rangle_{There})/\sqrt{3}.$$

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    – Buzz
    Commented Mar 3 at 23:06
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In quantum mechanics, the state of a particle can be described using a vector in a complex vector space known as a Hilbert space. The tensor product is used to represent composite systems, such as when considering the state of a particle in two different locations. Let's delve into the mathematical details to understand why the notation changes when dealing with superpositions involving vacuum modes.

First, consider the state of the particle in a non-superposed state:

$|\psi_1⟩ = |1⟩_{\text{Here}} ⊗ |0⟩_{\text{There}}$.

Here, $|1⟩_{\text{Here}}$ represents the particle being present at the "Here" location, and $|0⟩_{\text{There}}$ represents the absence of the particle at the "There" location.

Now, let's examine the superposed state:

$|\psi_2⟩ = \frac{1}{\sqrt{2}} (|1⟩_{\text{Here}} + |1⟩_{\text{There}})$.

In this state, the particle is in a superposition of being present at both "Here" and "There" locations simultaneously.

To represent this superposition including vacuum modes, we use the tensor product to account for all possible combinations of states:

$|\psi_2⟩ = \frac{1}{\sqrt{2}} (|1⟩_{\text{Here}} ⊗ |0⟩_{\text{There}} + |0⟩_{\text{Here}} ⊗ |1⟩_{\text{There}})$.

Here, the first term represents the particle being at "Here" and the vacuum mode at "There," while the second term represents the particle being at "There" and the vacuum mode at "Here." This is not an entangled state; rather, it represents a superposition of states where the particle can be found in either location.

The reason we don't represent the superposition as $|1⟩_{\text{Here}} + |1⟩_{\text{There}} ⊗ |0⟩_{\text{Here}} - |0⟩_{\text{There}}$ or using other notations is because they don't accurately represent the physical situation. The vacuum mode is not just a separate state but a part of the composite system. Defining creation and annihilation operators for combined states like $|1⟩_{\text{Here}} + |1⟩_{\text{There}}$ isn't as straightforward as for individual locations.

Thus, the notation involving the tensor product accurately captures the superposition of states in both locations, including vacuum modes. While the representation may resemble entanglement, it's important to distinguish between entanglement and superposition in this context. In entanglement, the states of composite systems cannot be factored into individual states, which is not the case here.

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  • $\begingroup$ why don't the states where particle is in superposed state and vacuum too and they are overall in separable state don't represent accurately the situation? What do you mean by "not straightforward" - is it impossible or counterintuitive? And why isn't this entanglement - as far as I understand, we can't factor the states of the particle and vacuum. $\endgroup$
    – Sutasu
    Commented Mar 3 at 19:40
  • $\begingroup$ I agree with @Sutasu - mathematically this is entanglement, you may just argue that physically it is not the most useful entanglement. That's why I linked an old paper that discusses this exact debate in my answer (Single-particle entanglement, doi.org/10.1103/PhysRevA.72.064306) $\endgroup$ Commented Mar 4 at 14:44

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