# Entangled wave function and polarisation operator

I was working on the following problem from Quantum Chemistry and Spectroscopy by T. Engel (3rd Edition), and was stumped in a few places. I wish get some feedback on my solution

The problem is the following one from chapter 6 of the book:

Consider the entangled wave function for two photons:

$$\Psi_{12} = \frac{1}{\sqrt{2}}[\Psi_{1}(H)\Psi_{2}(V)+\Psi_{1}(V)\Psi_{2}(H)]$$

Assume that the polarization operator $\hat{P_i}$ has the properties $\hat{P_i}\Psi_{i}(H) = - \Psi_{i}(H)$ and $\hat{P_i}\Psi_{i}(V)= \Psi_{i}(V)$ where $i \in$ {1,2}

a. Show that $\Psi_{12}$ is not an eigenfunction $\hat{P_1}$ and $\hat{P_2}$ .

b. Show that each of the two terms in $\Psi_{12}$ is an eigenfunction of the polarization $\hat{P_1}$

c. What is the average value of the $\hat{P_1}$that you will measure on identically prepared systems?

Solution

$\hat{P_1}\Psi_{12} = \hat{P_1}(\frac{1}{\sqrt{2}}[\Psi_{1}(H)\Psi_{2}(V)+\Psi_{1}(V)\Psi_{2}(H)])$

or, (using linearity)

$\hat{P_1}\Psi_{12} = \frac{1}{\sqrt{2}}(\hat{P_1}[\Psi_{1}(H)\Psi_{2}(V)]+\hat{P_1}[\Psi_{1}(V)\Psi_{2}(H)])$

Here, I did the following: For $\hat{P_i}\Psi_{i}\Psi_{j}$ (for $j\neq i$), I wrote $\hat{P_i}\Psi_{i}\Psi_{j} = \Psi_{j}\hat{P_i}\Psi_{i}$ i.e, treated $\Psi_{j}$ as a constant and pulled it out. question: Is this treatment correct? the question doesn't specify any other properties besides the two listed for the polarisation operator, so I am not so sure about what I did.

This gives me:

$\hat{P_1}\Psi_{12} = \frac{1}{\sqrt{2}}[-\Psi_{1}(H)\Psi_{2}(V)+\Psi_{1}(V)\Psi_{2}(H)]\neq (\textrm{constant})\Psi_{12}$

Similarly,

$\hat{P_2}\Psi_{12} = \frac{1}{\sqrt{2}}[\Psi_{1}(H)\Psi_{2}(V)-\Psi_{1}(V)\Psi_{2}(H)]\neq (\textrm{constant})\Psi_{12}$

Thus, $\Psi_{12}$ is not an eigenfunction of $\hat{P_i}$

b. Proceeding in a similar fashion, I computed

$\hat{P_1}[\Psi_{1}(H)\Psi_{2}(V)] = -\Psi_{1}(H)\Psi_{2}(V)$ Thus, this is an eigenfunction with an eigenvalue -1 and, $\hat{P_1}[\Psi_{1}(V)\Psi_{2}(H)] = \Psi_{1}(V)\Psi_{2}(H)$ This is also an eigenfunction with eigenvalue +1

c. Since we have identified 2 eigenfunctions and their corresponding eigenvalues we can write an arbitrary state as a superposition: $C_1\phi_1+ C_2\phi_2$ and $|C_1|^2+|C_2|^2= 1$ (square of the expansion coefficients gives the probability of measuring a particular eigenvalue).

and, the expectation values is like a weighted average given by $\langle\hat{P_1}\rangle = |C_1|^2\cdot \lambda_1+ |C_2|^2\cdot \lambda_2$ where, $\lambda_1 ~\textrm{and}~ \lambda_2$ are eigenvalues of $\phi_1, \phi_2$ respectively.

For, the given state $\Psi_{12}$ , $\phi_1 = \Psi_{1}(H)\Psi_{2}(V)$ and $\phi_2 = \Psi_{1}(V)\Psi_{2}(H)$ and the eigenvalues are -1 and +1

The expansion coefficients $C_1 = C_2 = \frac{1}{\sqrt2}$ so,

$$\langle\hat{P_1}\rangle = 0$$

question: is this line of reasoning correct?

• The first question should be correct. On part (c), what does it mean to the state of the systems by "identically prepared systems"? Think about it. – Xiaodong Qi Jun 4 '16 at 3:57
• @XiaodongQi So here's what I thought--we duplicate the system several times and in each case the photon is described by the superposition Ψ12. Then when we make measurements on each system, we either get +1 or -1, each with some probability that is given by the expansion coefficients and thus on average it is zero since both the expansion coefficients are equal and thus either value is equally likely. – getafix Jun 4 '16 at 4:17

The solution on part (a) should be correct, and it is solved correctly to my knowledge.

However, on part (c), there may be something wrong on your understanding. Here is what I think is the correct way to solve it.

What you want to calculate is the expectation value of $\hat{P}_1$, which can be expressed as

\begin{align} \langle \hat{P}_1\rangle &= \Psi^*_{12}\hat{P}_1\Psi_{12}\\ &=\frac{1}{2}[\phi_1^*+\phi_2^*]\hat{P}_1[\phi_1+\phi_2]\\ &=\frac{1}{2}[\phi_1^*+\phi_2^*]\cdot [-\phi_1+\phi_2]\\ &=\frac{1}{2}[-\phi_1^*\phi_1+\phi_2^*\phi_2+\phi_1^*\phi_2-\phi_2^*\phi_1]\\ &=\frac{1}{2}[|\phi_2|^2-|\phi_1|^2 + 2i\mathrm{Im}(\phi_1^*\phi_2)] \end{align} where I have defined $\phi_1=\Psi_1(H)\Psi_2(V)$ and $\phi_2=\Psi_1(V)\Psi_2(H)$. From the question, we may be able to use the fact that $|\phi_1|^2=|\Psi_1(H)|^2|\Psi_2(V)|^2$ and $|\phi_2|^2=|\Psi_1(V)|^2|\Psi_2(H)|^2$ to show \begin{align} |\phi_2|^2-|\phi_1|^2=|\Psi_1(V)|^2|\Psi_2(H)|^2-|\Psi_1(H)|^2|\Psi_2(V)|^2 \end{align}

while $\phi_1^*\phi_2=\Psi_1^*(H)\Psi_2(H)\Psi_1(V)\Psi_2^*(V)$ should be real to make the measurement real, and hence $2i\mathrm{Im}(\phi_1^*\phi_2)=0$. As one can assume photon functions are exchangeable in the amplitude at least, that is $|\Psi_1(x)|=|\Psi_2(x)|$, one may obtain $|\phi_2|^2-|\phi_1|^2=0$.

Under our assumption above, one may find $\langle\hat{P}_1\rangle=0$.

Although you got the same final result, I feel you may have missed the interference terms (like $\phi_1^*\phi_2$) in processing the calculations.