I'm looking to compute the tensor product of photon number states. I suspect this is a fairly simple quantum optics problem, but am having the following problem.

Consider a qubit which is in the following state:

$|\psi> = |1>_x$

where x is the mode containing 1 photon. Similarly, $|2>_x$ is assumed to be the $x$ mode which contains 2 photons, and $|0>_x$ is the vacuum.

How can I compute the tensor product of 2 of these qubits? i.e. $|\psi_\mathrm{out}> = |\psi> \otimes|\psi> = |1>_x \otimes |1>_x = ?????$

I'm confused as the tensor product typically increases the size of the subspace, but here, the tensor product is between 2 of the same quantity.

Is the answer simply, $|\psi_\mathrm{out}> = |2>_x ~~~ $?

I tried to formulate a solution in terms of the creation operator, $\hat{a}^\dagger_{x}$, since we know: $\hat{a}^\dagger_{x} |n>_x = (n+1) |n+1>_x$ and thus $|1>_x = \hat{a}^\dagger_{x} |0>_x $.

Therefore, I wondered if we could write:

$|\psi_\mathrm{out}> ~ = ~ |1>_x \otimes |1>_x ~ = ~ |1>_x |1>_x ~ = ~ (\hat{a}^\dagger_{x})^2 |0>_x ~ = ~ \sqrt{2} |2>_x$

Are any of the above correct?


EDIT, following useful comments below from user3502079, I've added the following:

For clarity (to avoid lots of 0s and 1s), I'm going to use $|H\rangle$ as the $|0\rangle$ basis, and $|V\rangle$ in place of $|1\rangle$ basis. From your answer, is it correct to write the following, for the case of 2 input qubits in V state (i.e. photons that are vertically polarized):

$|\psi_\mathrm{out}\rangle = |V\rangle \otimes |V\rangle = a^\dagger_V |vac\rangle \otimes a^\dagger_V |vac\rangle = (a^\dagger_V \otimes a^\dagger_V)\cdot(|vac\rangle \otimes |vac\rangle)~~~~~~~~$ ?

I note that the creation operator $a^\dagger_V$ is the same whether it operates on either vacuum state. My problem is then, how do I evaluate $(a^\dagger_V \otimes a^\dagger_V)$?

Given that the definition of creation operators are: $a^\dagger_V |vac\rangle = |1\rangle_V $ and $a^\dagger_V |1\rangle_V = \sqrt{2}|2\rangle_V $, can I therefore say that:

$(a^\dagger_V \otimes a^\dagger_V)\cdot(|vac\rangle \otimes |vac\rangle) = a^\dagger_V a^\dagger_V |vac\rangle = a^\dagger_V |1\rangle_V = \sqrt{2}|2\rangle_V$

i.e. the answer is two photons occpying the V state, with coefficient $\sqrt{2}$.

Is this valid? Or am I confusing tensor representation and occupation number expressions? Thanks!


3 Answers 3


Usually the tensor product of photon number states are used when these states are composed of different Hilbert spaces. In other words, if ${\cal H}={\cal H}_A\otimes {\cal H}_B$, then we expect to find states like $|n\rangle_A\otimes|m\rangle_B$ in ${\cal H}$, where $n$ and $m$ can represent different occupation numbers.

If, on the other hand, one wants to consider only one Hilbert space associated with only one mode (in other words, we only look at the particle-number degree of freedom and ignore all other degrees of freedom), then what would it mean to form tensor products, such as $|\psi'\rangle=|1\rangle\otimes|1\rangle$?

Well, it would not be normalized: $\langle\psi'|\psi'\rangle=2$. (One can see this from the fact that, since both states live in the same Hilbert space, there are two ways to pair off the individual states in the tensor product.) However, one can normalize the state by an appropriate normalization constant: $$|2\rangle=\frac{1}{\sqrt{2}}|1\rangle\otimes|1\rangle .$$ For the general case, this becomes $$|n\rangle=\frac{1}{\sqrt{n!}}|1\rangle^{\otimes n} .$$ Such a definition satisfies all the relationships and requirements for well-defined photon number states.

  • $\begingroup$ Thanks, this is the relationship I was looking for. Out of interest, is there a mathematical way to prove $\langle\psi'|\psi'\rangle=2$ as you mention in your post? Or is it simply logic that says: since we have 2 photons (called X and Y, say), the product could be XY or YX, but doesn't matter since they're indistinguishable? Thanks. $\endgroup$
    – SLhark
    Apr 25, 2019 at 15:54
  • $\begingroup$ Yes, it is the mathematics of combinatorics. It comes down to how one can form all possible sequences of elements from finite sets. Anyway, I'm glad the answer helped. $\endgroup$ Apr 26, 2019 at 4:22

I think you're getting confused between occupation number representation and the regular way of representing states (using the word regular for the lack of a better word: I mean the symmetrized or antisymmetrized state depending on whether you're talking about bosons or fermions). Occupation number representation: $| n_1,n_2,n_3....n_i.....\rangle$ This means in the $i$th state there are $n_i$ particles. The symmetrized state $|1,0\rangle + |0,1\rangle$ is written in the occupation number representation as $|1,1,0,0,0...\rangle$ It means you have one particle in the state represented by $0$ and one in the state represented by $1$.


I don't know much about quantum optics but I can help you with tensor products. Define the following basis for the qubits: $$|0\rangle=\begin{pmatrix}1\\0\end{pmatrix};\quad|1\rangle=\begin{pmatrix}0\\1\end{pmatrix};$$ In this basis you can explicitly compute the tensor product: $$\begin{pmatrix}a\\b\end{pmatrix}\otimes\begin{pmatrix}c\\d\end{pmatrix}=\begin{pmatrix}a\begin{pmatrix}c\\d\end{pmatrix}\\b\begin{pmatrix}c\\d\end{pmatrix}\end{pmatrix}=\begin{pmatrix}ac\\ad\\bc\\bd\end{pmatrix}$$

You can now calculate the new basis by taking tensor products of the old basis vectors, like I did in the line above. Using the shorthand $|1\rangle\otimes|0\rangle=|10\rangle$ we get \begin{align} &|00\rangle=\begin{pmatrix}1&0&0&0\end{pmatrix}^T\\ &|01\rangle=\begin{pmatrix}0&1&0&0\end{pmatrix}^T\\ &|10\rangle=\begin{pmatrix}0&0&1&0\end{pmatrix}^T\\ &|11\rangle=\begin{pmatrix}0&0&0&1\end{pmatrix}^T \end{align} The tensor product has increased the size of the Hilbert space: instead of two dimensional we now have four dimensional vectors. These new vectors don't correspond to photon states yet. Since photons are bosons we must have that exchanging the first photon with the second photon leaves the states unchanged.

There are three states (I think) that fulfill this last constraint. These correspond to the numbered states you mention.

\begin{align} |0\rangle_x&=|00\rangle\\ |1\rangle_x&=\tfrac{1}{\sqrt{2}}\left(|10\rangle+|01\rangle\right)\\ |2\rangle_x&=|11\rangle \end{align}

Finally you mention the creation operator. When using operators in conjunction with tensor product you usually first define operators that work on only one of the states. When you take a tensor product of two operators and apply it to a product state it is the same as applying the operators on each of the states. $$(\hat A\otimes\hat B)\cdot(|\psi_1\rangle\otimes|\psi_2\rangle)=(\hat A|\psi_1\rangle)\otimes(\hat B|\psi_2\rangle)$$

Verify for yourself that the following operators only act on only one of the states: $$a^\dagger_1=a^\dagger\otimes I;\quad a^\dagger_2=I\otimes a^\dagger$$ with I the identity operator. You can now construct the raising operators from your question using these one-state operators. So you can write down the raising operators, it's just a little more work to do it explicitly for multiple states.

  • $\begingroup$ Thanks for the clear explanation. I took this advice and tried to apply it to the problem, but still have concerns that my final solution looks incorrect (shouldn't wavefunction coefficients squared always be less than 1 if they are to represent probabilities). Please see edited OP question. Thanks for your help (will up vote when I can (need 15 rep)). $\endgroup$
    – SLhark
    Apr 25, 2019 at 9:45

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