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 $\frac{1}{\sqrt{2}}\left(|1_a,1_b\rangle+|1_b,1_a\rangle\right)$ ?


3 Answers 3


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.

  • $\begingroup$ Did I understand you correctly that states such as 12√(|1a,1b⟩+|1b,1a⟩) can only be used to construct the Hilbert space for photon pairs, but the photon pair itself will never exist in that state? $\endgroup$
    – ph_14
    Sep 30, 2012 at 5:53
  • $\begingroup$ No, what I say is that your expression has no meaning to start with, and I am confident that I stated this rather clearly. There is no sense in which states $|1a,1b\rangle$ and $|1b,1a\rangle$ if $a,b$ label "two different photons" could differ, so it's a misleading notation that pretends that things that are the same are different by chaotically reordering the labels. And if you say that these two states are the same, then the right normalization factor is $1/2$ rather than $1/\sqrt{2}$ and you may write them simply as a single state. $\endgroup$ Sep 30, 2012 at 6:03
  • $\begingroup$ It's like when you discuss objects in the state $(|\text{cat alive}\rangle + |\text{alive cat}\rangle )/ \sqrt{2}$. The ordering of the words may be different and you want to create the impression that they're different, orthogonal states, but any conceivable interpretation of these states implies that they're really the same states in the Hilbert space, so if you want to add "them" (it, twice) up and normalize it, the factor should be $1/2$ and not $1/\sqrt{2}$ and you may just write $|\text{alive cat}\rangle$. BTW: All my comments were about two-photon or photon pair states. $\endgroup$ Sep 30, 2012 at 6:24
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    $\begingroup$ Dear ph_14, right, but notice that the state Psi_s you wrote down has two different particles and two different states (a,b; and 1,2, whichever is which), so in the occupation basis, it would be a $|1,1\rangle$ state. You only had one one-particle state/mode i.e. $|2,0\rangle$ diphoton state but you pretended that the states with the two orderings of the labels 1a,1b are different. $\endgroup$ Sep 30, 2012 at 7:02
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    $\begingroup$ $|0,1\rangle$ and $|1,0\rangle$ are not tensor products, they represent two orthonormal states in one particle Hilbert space. $\endgroup$ Sep 30, 2012 at 7:26

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.

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:


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.

  • $\begingroup$ I got a minus in less than $5$ minutes. I'd like to know the reason. Is there a mistake in what I wrote? $\endgroup$ Sep 30, 2012 at 7:28
  • $\begingroup$ Dear Cristi, I got a -1 quickly, too. The reason why I gave you a downvote is that you claim that the OP is right which he or she is surely not. You also write several meaningless formulae in which states in occupation bases are written as tensor products of other states in the occupation basis on the same Hilbert space which is surely inconsistent. If you write states like $|0,1\rangle$ which is everywhere on the RHS of your equations, it's an element of a 2-mode Fock space, and by tensoring two such spaces/states, you get a 4-mode Hilbert space, not a 2-mode one like $0,2\rangle$. $\endgroup$ Sep 30, 2012 at 7:49
  • $\begingroup$ @Lubos Motl: I gave you a -1 not too quickly, but after having enough time to read what you wrote, and the comments. Not from the first sentence, as you did. You are wrong in your answer, you make confusion between two different notations. On the other hand, my answer is correct, it's only problem is that it contradicts yours. I did not to make my answer about yours, but anyone who knows how the Fock state is constructed will agree with me. In your opinion, $|0,0,0,1,0,0,0,0\rangle$ (in the Fock space notation) is a state of how many particles? One or eight? $\endgroup$ Sep 30, 2012 at 7:57
  • $\begingroup$ @Lubos Motl: I didn't have a reference right now to prove what I said, but I googled a bit and found this mathematik.hu-berlin.de/~maphy/Skript1.pdf. Please check section $1.2$. $\endgroup$ Sep 30, 2012 at 8:07
  • $\begingroup$ I believe this answer is correct, but it might be a good idea to add some index denoting the underlying vector space to the kets so it's more obvious in which cases we're dealing with the occupation number representation; also, I've seen the symbol $\vee$ used for the symmetrized tensor product as analogue to the anti-symmetrized exterior product $\wedge$ $\endgroup$
    – Christoph
    Sep 30, 2012 at 10:11

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.

  • $\begingroup$ How is your answer different from mine? $1.$ You replaced $\odot$, which is more standard, with $\vee$, which can be confounded with the Hodge dual of $\wedge$. BTW, most textbooks use for the symmetric tensor product the same $\otimes$ as for the usual tensor product, but I think this leads to confusions. $2.$ And you added the index indicating the space. This is correct, but given that the Fock space is a direct sum of these spaces, the index is redundant. $\endgroup$ Sep 30, 2012 at 11:57
  • $\begingroup$ @CristiStoica: I believe my answer is sufficiently different: it mentions entanglement, uses the correct normalization and the presentation is quite different; not liking $\vee$ for symmetrized tensor products is fine, but I don't really buy your argument about Hodge duals - most symbols in mathematics have different meanings depending on context, including $\odot$, which is a pretty generic symbol used for arbitrary group operations; of course the main contents of both our answers is the same as they answer the same question, which is why I did upvote yours in addition to wrinting my own $\endgroup$
    – Christoph
    Sep 30, 2012 at 16:02
  • $\begingroup$ @Christoph: Your point is well taken. I did try to "symmetrize" the state |1,1> itself, instead of constructing the symmetric basis vector |1,0> V |0,1> of the new 3D subspace. But I have a new question. If we say that |2,0⟩F=|1,0⟩H⊗|1,0⟩H, as it appears from Linear algebra, then should not all N photns in state |N,0,0,0> behave totally independently of each other in all possible experiments? $\endgroup$
    – ph_14
    Sep 30, 2012 at 17:47
  • $\begingroup$ @ph_14: There's no $|N,0,0,0\rangle$, you probably mean $|N,0\rangle$. Anyway, if you can make two identical photons behave differently, then you evolved $|2,0\rangle$ into something different, like $\alpha|2,0\rangle+\beta|1,1\rangle+\gamma|0,2\rangle$. $\endgroup$ Sep 30, 2012 at 19:54
  • $\begingroup$ @Christoph:|N,0⟩ if you only have two polarization modes. I really meant a "generic" boson (maybe a spin-2 boson, or maybe spatial modes are also included). And yes, logically it makes sense that bosons in |N,0> or |0,N> state should behave independently. Isn't this what happens in the HOM experiment when both photons enter the same port? $\endgroup$
    – ph_14
    Sep 30, 2012 at 20:55

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