Is the Fock state a superposition of product states?

Question

I know that in principle, the Fock state is not the same as a product state because if it were, the photons would be uncorrelated. What I don't understand is, can it be expanded in product states?

For example, if two photons share the same mode, is the state $|2,0\rangle$ the same as the product $|1,0\rangle|1,0\rangle$?

If they are in different modes, is the Fock state $|1,1\rangle$ the same as superposition of two Fock states, $\frac{1}{\sqrt{2}}\left(|1_a,1_b\rangle+|1_b,1_a\rangle\right)$ (with $a$ and $b$ labeling the photons)? If yes, then is this superposition, in turn, the same as $\frac{1}{\sqrt{2}}\left(|1_a,1_b\rangle+|1_b,1_a\rangle\right)$ ?

Answer 1

There are many things to be clarified here.

First, you probably mean "tensor products". If you do, then it's important to distinguish "tensor product of states" and "tensor product of spaces". The Fock space is an infinite-dimensional (or at least multi-dimensional) harmonic oscillator and it is isomorphic to a tensor product of one-dimensional Fock spaces or harmonic oscillators.

However, the tensor product of states is not quite the same thing in the sense that the tensor product space is not composed of purely states that are tensor product states, products of elements from the original spaces. Instead, the tensor product space contains all conceivable linear combinations of the tensor products of the original states.

Now, $|2,0\rangle$ in which two particles are in the same mode is in no way equivalent to a tensor product of $|1,0\rangle$ with itself. By computing the tensor product, we are "extending the number of modes", so a tensor product $|1,0\rangle \otimes |1,0\rangle$ is always something of the form $|1,0,1,0\rangle$. In this 4-mode Hilbert space, the first and third mode may have the same properties, there may even be a symmetry between them, but they are not the same mode.

Concerning the last two questions, there is nothing that could be meaningfully called $$ \frac{1}{\sqrt{2}} (|1a,1b\rangle + |1b,1a\rangle) $$ whether or not you replace the commas by $\rangle |$ – which clearly shouldn't matter (it's just a different typographical convention). This expression of yours uses some completely inconsistent notation. You must first decide what properties $x,y$ of the state the symbol $|x,y\rangle$ denotes and you can't ever permute them. You failed to do so because $x$ sometimes refers to "the first photon" and sometimes to "the second photon", and so on. It just makes no sense.

You could choose an alternative multi-body notation in which the wave function may fail to be symmetric or antisymmetric. If you did so, $x$ would always correspond to the first photon and $y$ would always correspond to the second photon. In this notation, the state $|2,0\rangle$ in the occupation number basis would be written as $|\alpha,\alpha\rangle$ where $\alpha,\beta$ are the modes that had the occupation numbers $2,0$, respectively. No nontrivial superposition has to be constructed in this case to symmetrize the state because the state is already symmetric: the wave functions of both photons are $\alpha$, they are the same.

Answer 2

You are right, and here is why. $|0,1\rangle$ and $|1,0\rangle$ form a basis in the Hilbert space $H$, which has two complex dimensions. It can be the space of one boson which may be in two quantum states. The state space for two bosons will be not merely the tensor product of the space $H$ with itself, $H\otimes H$. It will be a subspace of it, which is denoted sometimes by $H\odot H$, and is the symmetric tensor product. That is, in general, while the tensor product of two vectors $u$ and $v$ is $u\otimes v$, the symmetric tensor product is $u\odot v=\frac{1}{\sqrt {2}}\left(u\otimes v + v \otimes u\right)$. The factor $\frac{1}{\sqrt {2}}$ is chosen for normalization. The complex dimension of $H$ is $2$, and that of $H\otimes H$ is $4$. But the complex dimension of $H\odot H$ is $3$. The basis of $H\otimes H$ is given by $|0,1\rangle|0,1\rangle$ ($ :=|0,1\rangle\otimes|0,1\rangle$, but it is customary to omit $\otimes$), $|0,1\rangle|1,0\rangle$, $|1,0\rangle|0,1\rangle$, $|1,0\rangle|1,0\rangle$. But the basis of $H\odot H$ is $|0,2\rangle:=|0,1\rangle|0,1\rangle$, $|1,1\rangle:=\frac{1}{\sqrt {2}}\left(|0,1\rangle|1,0\rangle+|1,0\rangle|0,1\rangle\right)$, and $|2,0\rangle:=|1,0\rangle|1,0\rangle$. So you are right that $|0,2\rangle=|0,1\rangle|0,1\rangle$, and that $|1,1\rangle=\frac{1}{\sqrt {2}}\left(|0,1\rangle|1,0\rangle+|1,0\rangle|0,1\rangle\right)$, and that the state $|1,1\rangle$ is independent on the chosen basis.

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I add this to explain in more detail some points which were raised. The Fock space for a boson whose one-particle Hilbert space is $H$ is $$F=\bigoplus_{k=0}^{\infty}\odot^kH=\mathbb C\oplus H \oplus \left(H\odot H\right)\oplus\ldots\oplus\odot^kH\oplus\ldots.$$

Now we see that $H$ is a subspace of $F$, or $H\lt F$. Tensor products between $H$ and $H$ don't stay in general in $F$, because $H\otimes H\nleq F$. But $F$ is closed to symmetric tensor products, that is $H\odot H<F$.

So, it is perfectly legit to take tensor products inside the Fock space, provided that they are symmetric tensor products.

The full basis of the Fock space, in the full notation involving tensor products and direct sums of vector spaces, is complicated to write. So, rewrite it like this:

$$|n_1,\ldots,n_m\rangle:=|1,0,\ldots,0\rangle^{n_1}\odot|0,1,\ldots,0\rangle^{n_2}\odot\ldots\odot|0,0,\ldots,1\rangle^{n_m}$$

where $m=\dim H$, and the exponent in $|1,0,\ldots,0\rangle^{n_1}$ means $n_1$ times tensor product with itself (it is automatically symmetric).

Now go back to $\dim H=m=2$. The basis will be

$$|n_1,n_2\rangle=|1,0\rangle^{n_1}\odot |0,1\rangle^{n_2}.$$

If you check this for $n_1+n_2=2$, you will find exactly what I said in the answer.

Answer 3

I know that in principle, the Fock state is not the same as a product state because if it were, the photons would be uncorrelated.

While product states form a basis of the tensor product space, the full vector space is given by linear combination of these, which in general can't be decomposed into a single product vector. This fact is known as entanglement.

Now, the Fock spaces are made up (as the sum of vector spaces) of symmetrizations (in case of Bosons) or antisymmetrizations (in case of Fermions) of such product spaces (of arbitrary power). This automatically leads to entanglement except if all particles occupy the same state.

What I don't understand is, can it be expanded in product states? For example, if two photons share the same mode, is the state |2,0> the same as the product |1,0>|1,0>?

Yes, you can expand Fock states in terms of product states. In your example $|2,0\rangle_{F}$ only a single state is occupied, so no entanglement will be present: $$ \begin{align*} |2,0\rangle_{F}&=|2,0\rangle_{H\vee H}\\ &=|1,0\rangle_H\vee|1,0\rangle_H\\ &=\mathrm{Sym}(|1,0\rangle_H\otimes|1,0\rangle_H)\\ &=\frac12(|1,0\rangle_H\otimes|1,0\rangle_H+|1,0\rangle_H\otimes|1,0\rangle_H)\\ &=|1,0\rangle_H\otimes|1,0\rangle_H \end{align*} $$

Here, $F$ denotes the Fock space over the vector space $H$ and $\vee$ the symmetrized tensor product. All states are labeled by occupation numbers.

If they are in different modes, is the Fock state |1,1> the same as superposition of two Fock states, (1/sqrt 2) ( |1a,1b> + |1b,1a> ) (with a and b labeling the photons)? If yes, then is this superposition, in turn, the same as (1/sqrt 2) ( |1a>|1b> + |1b>|1a> ) ?

The correct expansion for $|1,1\rangle_F$ is $$ \begin{align*} |1,1\rangle_F&=|1,1\rangle_{H\vee H}\\ &=|1,0\rangle_H\vee|0,1\rangle_H\\ &=\mathrm{Sym}(|1,0\rangle_H\otimes|0,1\rangle_H)\\ &=\frac12(|1,0\rangle_H\otimes|0,1\rangle_H+|0,1\rangle_H\otimes|1,0\rangle_H) \end{align*} $$ an entangled state, whereas the expansion you give is just $$ \begin{align*} &\frac1{\sqrt2}(|1,0\rangle_H\otimes|1,0\rangle_H+|1,0\rangle_H\otimes|1,0\rangle_H)\\ =&\sqrt2(|1,0\rangle_H\otimes|1,0\rangle_H)\\ =&\sqrt2|2,0\rangle_F \end{align*} $$ assuming $|1\rangle\equiv|1,0\rangle_H$. Note that a label for the particle is superfluous as it is implied by order of the kets.

Be careful with your notation as $|1,1\rangle$ can also be understood as shorthand for $$ |1,1\rangle\equiv|1\rangle\otimes|1\rangle $$ which would again correspond to $|2,0\rangle_F$ under the assumption from above.