# Normalization of photon wavefunction and beam splitters

I am having trouble with the normalization of photon wave functions, when passed through a beam splitter.

Let me define the single photon state as $$|1\rangle = \int \text{d}\omega \phi(\omega) a^\dagger(\omega)|0\rangle$$ and the two photon state as $$|2\rangle = \int \text{d}\omega_1\text{d}\omega_2\phi(\omega_1)\phi(\omega_2) a^\dagger(\omega_1)a^\dagger(\omega_2)|0\rangle$$ These states are normalized if $\int \text{d}\omega |\phi(\omega)|^2 = 1$.

I have used the operator $a^\dagger$ to denote the creation operator in the first of my two modes. The operator for the second mode is $b^\dagger$. Let me insert the two photon state into the first port of a beam splitter which takes $a \rightarrow \frac{a-b}{\sqrt{2}}$. My output state is then $$|\psi\rangle = \frac{1}{2}\int \text{d}\omega_1\text{d}\omega_2 \phi(\omega_1)\phi(\omega_2) [a^\dagger(\omega_1)-b^\dagger(\omega_1)][a^\dagger(\omega_2)-b^\dagger(\omega_2)]|0,0\rangle$$ I can compute the amplitude of the various output events. The amplitude for both $|2,0\rangle$ and $|0,2\rangle$ is $\frac{1}{2}$. The amplitude for $|1,1\rangle$ comes out to be $-1$. The state does not seem to be normalized. Why not?

### Further details

I compute amplitudes in the following way. $$\langle 2,0|\psi\rangle = \frac{1}{2}\int \text{d}\omega_1^\prime\text{d}\omega_2^\prime\phi^*(\omega_1^\prime)\phi^*(\omega_2^\prime) \int \text{d}\omega_1\text{d}\omega_2 \phi(\omega_1)\phi(\omega_2)\langle 0,0 | a(\omega_1^\prime)a(\omega_2^\prime) [a^\dagger(\omega_1)-b^\dagger(\omega_1)][a^\dagger(\omega_2)-b^\dagger(\omega_2)]|0,0\rangle$$

Only the term corresponding to two photons in the first mode survives. $$\langle 0,0 |a(\omega_1^\prime)a(\omega_2^\prime) a^\dagger(\omega_1)a^\dagger(\omega_2)|0,0\rangle = \delta(\omega_1^\prime-\omega_1)\delta(\omega_2^\prime-\omega_2)$$ Therefore the above integrals collapse to $$\frac{1}{2}\int \text{d}\omega_1\text{d}\omega_2 |\phi(\omega_1)|^2|\phi(\omega_2)|^2 = \frac{1}{2}$$ Similarly $$\langle 1,1|\psi\rangle = \frac{1}{2}\int \text{d}\omega_1^\prime\text{d}\omega_2^\prime\phi^*(\omega_1^\prime)\phi^*(\omega_2^\prime) \int \text{d}\omega_1\text{d}\omega_2 \phi(\omega_1)\phi(\omega_2)\langle 0,0 | a(\omega_1^\prime)b(\omega_2^\prime) [a^\dagger(\omega_1)-b^\dagger(\omega_1)][a^\dagger(\omega_2)-b^\dagger(\omega_2)]|0,0\rangle$$ This time two different terms survive, one corresponding to the $\omega_1$ photon jumping to the second channel, the other corresponding to the $\omega_2$ photon jumping to the second channel. $$\langle 0,0 |a(\omega_1^\prime)b(\omega_2^\prime) a^\dagger(\omega_1)b^\dagger(\omega_2)|0,0\rangle = \delta(\omega_1^\prime-\omega_1)\delta(\omega_2^\prime-\omega_2)$$ $$\langle 0,0 |a(\omega_1^\prime)b(\omega_2^\prime) b^\dagger(\omega_1)a^\dagger(\omega_2)|0,0\rangle = \delta(\omega_1^\prime-\omega_2)\delta(\omega_2^\prime-\omega_1)$$ Therefore, the integrals collapse and we get $$2\times \frac{-1}{2}\int \text{d}\omega_1\text{d}\omega_2 |\phi(\omega_1)|^2|\phi(\omega_2)|^2 = -1$$

I have added the amplitudes of these two terms rather than the probabilities because the two photons have an identical spectrum. Is this incorrect for some reason?

### Further edited in details

I have used the following unitary for the transformation of the field operators. $$\frac{1}{\sqrt{2}}\begin{bmatrix} 1 & -1 \\ 1 & \phantom{-}1 \end{bmatrix}$$ This transformation can be found in any quantum optics book such as Scully and Zubairy.

The normalization of the $|2\rangle$ state follows $$\langle 2|2\rangle = \int \text{d}\omega_1^\prime\text{d}\omega_2^\prime\phi^*(\omega_1^\prime)\phi^*(\omega_2^\prime) \int \text{d}\omega_1\text{d}\omega_2\phi(\omega_1)\phi(\omega_2) \langle 0| a(\omega_1^\prime)a(\omega_2^\prime) a^\dagger(\omega_1)a^\dagger(\omega_2)|0\rangle$$

Now note that there are four operators, each corresponding to a different frequency. The crucial commutation relation between them is $[a(\omega_1),a^\dagger(\omega_2)] = \delta(\omega_1-\omega_2)$. The braket is non-zero if I equate $\omega_1^\prime$ and $\omega_1$ and similarly $\omega_2^\prime$ and $\omega_2$. The braket reduces to $$\langle 0| a(\omega_1) a^\dagger(\omega_1) a(\omega_2) a^\dagger(\omega_2)|0\rangle \delta(\omega_1^\prime-\omega_1) \delta(\omega_2^\prime-\omega_2) = \delta(\omega_1^\prime-\omega_1) \delta(\omega_2^\prime-\omega_2)$$ Then the integral above collapses to $$\int \text{d}\omega_1\text{d}\omega_2|\phi(\omega_1)|^2|\phi(\omega_2)|^2 = 1$$

The two-photon input state as you have defined

$$|\tilde{2}\rangle = \int \text{d}\omega_1\text{d}\omega_2\phi(\omega_1)\phi(\omega_2) a^\dagger(\omega_1)a^\dagger(\omega_2)|0\rangle$$

is not-normalized. Consider

$$\langle \tilde{2}|\tilde{2}\rangle = \int\int \text{d}\omega_1 \text{d}\omega_2 \text{d}\omega_1^\prime \text{d}\omega_2^\prime \phi(\omega_1)\phi(\omega_2) \phi(\omega_1^\prime)\phi(\omega_2^\prime) \langle 0| a(\omega_1^\prime)a(\omega_2^\prime) a^\dagger(\omega_1)a^\dagger(\omega_2)|0\rangle\\ \phantom{\langle 2|2\rangle}= \int\int \text{d}\omega_1 \text{d}\omega_2 \text{d}\omega_1^\prime \text{d}\omega_2^\prime \phi(\omega_1)\phi(\omega_2) \phi(\omega_1^\prime)\phi(\omega_2^\prime)\phantom{\langle 0| a(\omega_1^\prime)a(\omega_2^\prime) a^\dagger(\omega_1)a^\dagger(\omega_2)|0\rangle} \\\times\left(\delta(\omega_1^\prime-\omega_1)\delta(\omega_2^\prime-\omega_2)+ \delta(\omega_2^\prime-\omega_1) \delta(\omega_1^\prime-\omega_2)\right)\phantom{\langle 0| a(\omega_1^\prime)|0\rangle}\\ \phantom{\langle 2|2\rangle} = 2, \phantom{\int+\text{d}\omega_1 \text{d}\omega_2 \text{d}\omega_1^\prime \text{d}\omega_2^\prime \phi(\omega_1)\phi(\omega_2) \phi(\omega_1^\prime)\phi(\omega_2^\prime) \langle 0| a(\omega_1^\prime)a(\omega_2^\prime) a^\dagger(\omega_1)a^\dagger(\omega_2)|0\rangle}$$ where I have repeatedly used the Bosonic commutation relations to simplify (see calculations appended) the expression above, but the same can be accomplished using Wick's theorem .

The correctly normalized state is thus $$|2\rangle = \frac{1}{\sqrt{2}}\int \text{d}\omega_1\text{d}\omega_2\phi(\omega_1)\phi(\omega_2) a^\dagger(\omega_1)a^\dagger(\omega_2)|0\rangle.$$

Using the correctly normalized state, we obtain the three amplitudes $$\langle 20|U_{\text{BS}}|20\rangle = \frac{1}{2} \\ \langle 11|U_{\text{BS}}|20\rangle = -\frac{1}{\sqrt{2}} \\ \langle 02|U_{\text{BS}}|20\rangle = \frac{1}{2},$$ the sum of whose squares is $1$. The beamsplitter is safe again for quantum mechanics.

Appendix: Calculations for $\langle \tilde{2}|\tilde{2}\rangle = 2$ $$\langle \tilde{2}|\tilde{2}\rangle = \int\int \text{d}\omega_1 \text{d}\omega_2 \text{d}\omega_1^\prime \text{d}\omega_2^\prime \phi(\omega_1)\phi(\omega_2) \phi(\omega_1^\prime)\phi(\omega_2^\prime) \langle 0| a(\omega_1^\prime)a(\omega_2^\prime) a^\dagger(\omega_1)a^\dagger(\omega_2)|0\rangle\\ = \int\int \text{d}\omega_1 \text{d}\omega_2 \text{d}\omega_1^\prime \text{d}\omega_2^\prime \phi(\omega_1)\phi(\omega_2) \phi(\omega_1^\prime)\phi(\omega_2^\prime) \\\times \left(\langle 0| a(\omega_1^\prime) a^\dagger(\omega_1)a(\omega_2^\prime)a^\dagger(\omega_2)|0\rangle +\\ \langle 0| a(\omega_1^\prime) \delta(\omega_2^\prime-\omega_1)a^\dagger(\omega_2)|0\rangle \right), \\ = \int\int \text{d}\omega_1 \text{d}\omega_2 \text{d}\omega_1^\prime \text{d}\omega_2^\prime \phi(\omega_1)\phi(\omega_2) \phi(\omega_1^\prime)\phi(\omega_2^\prime) \\\times \left(\langle 0| (a^\dagger(\omega_1)a(\omega_1^\prime)+ \delta(\omega_1^\prime-\omega_1) (a^\dagger(\omega_2)a(\omega_2^\prime)+\delta(\omega_2^\prime-\omega_2))|0\rangle +\\ \langle 0| \delta(\omega_2^\prime-\omega_1) (a^\dagger(\omega_2)a(\omega_1^\prime)+\delta(\omega_1^\prime-\omega_2)|0\rangle \right)\\= \int\int \text{d}\omega_1 \text{d}\omega_2 \text{d}\omega_1^\prime \text{d}\omega_2^\prime \phi(\omega_1)\phi(\omega_2) \phi(\omega_1^\prime)\phi(\omega_2^\prime) \\\times\left(\delta(\omega_1^\prime-\omega_1)\delta(\omega_2^\prime-\omega_2)+ \delta(\omega_2^\prime-\omega_1) \delta(\omega_1^\prime-\omega_2)\right)\\ = 2.$$ The correctly normalized state is thus: $$|2\rangle = \frac{1}{\sqrt{2}}|\tilde{2}\rangle = \frac{1}{\sqrt{2}}\int \text{d}\omega_1\text{d}\omega_2\phi(\omega_1)\phi(\omega_2) a^\dagger(\omega_1)a^\dagger(\omega_2)|0\rangle.$$ NB: The correctly normalized $n$-photon state is $$|n\rangle = \frac{1}{\sqrt{n!}}\int \text{d}\omega_1\text{d}\omega_2\ldots\omega_n \phi(\omega_1)\phi(\omega_2) \ldots \phi(\omega_n) a^\dagger(\omega_1)a^\dagger(\omega_2)\ldots a^\dagger(\omega_n)|0\rangle.$$

I am not sure what you mean by "normalized" here. Usually it is a state with norm 1, i.e. $$\langle\psi|\psi\rangle = 1$$ A multi-particle state as you defined is not normalized: $$\langle2|2\rangle = \langle0|aaa^\dagger a^\dagger|0\rangle = \langle0|a(a^\dagger a+1)a^\dagger|0\rangle = \langle0|(aa^\dagger + aa^\dagger)|0\rangle = 2\langle1|1\rangle$$

To normalize, one divides a state by the square root of number of particles.

Now, lets have a look at the transformation you called $a\rightarrow\frac{a-b}{\sqrt{2}}$. This transformation may be written out as

$$\hat{A} = \int d\omega \frac{a^\dagger(\omega)-b^\dagger(\omega)}{\sqrt{2}}a(\omega)$$ i.e. it annihilates a particle in state $a$ and creates a new particle in the antisymmetric superposition of states $a$ and $b$. This operator is not unitary:

$$\hat{A}^\dagger\hat{A} \neq Id$$

Non-unitary operators are not norm-conserving, so if you let them act on a state, the result will have a different norm.

• Your beam splitter transformation is not correct. An ideal beam splitter obviously implements a unitary (and thus norm-conserving) operation since the output intensity is the same as the input intensity. The beam splitter Hamiltonian looks like (schematically) $ab^{\dagger} + ba^{\dagger}$, if you exponentiate this with the right prefactor you will get the desired transformation. – Mark Mitchison Jun 17 '14 at 14:49
• Exactly. My point was that the state in which OP arrived was not normalized because he got there via non-unitary operation. – M.Herzkamp Jun 17 '14 at 16:48
• But the transformation $a \to \frac{a-b}{\sqrt{2}}$, $b\to \frac{a+b}{\sqrt{2}}$ is obviously unitary (in fact orthogonal). – Mark Mitchison Jun 17 '14 at 18:27
• I have added further details on the normalization of the $|2\rangle$ state. My transformation is unitary as confirmed by @MarkMitchison. By normalization I used the commonly accepted meaning, probabilities sum to 1. – Abdullah Khalid Jun 17 '14 at 19:06