# Projection of Quantum State onto Bell State

I am very interested by this paper on entanglement swapping and timelike entanglement.

The one thing I get really tripped up with is the whole idea of a projection onto a bell basis. I understand the whole idea of projecting into different basises, but I am not sure what the authors are doing. From the article:

When the two photons of time  (photons 2 and 3) are
projected onto any Bell state, the first and last photons
(1 and 4) collapse also into the same state and entanglement
is swapped. The first and last photons, that did not share
between them any correlations, become entangled.


I am trying to see how that is the case from Equation 3. I am missing the way to mathematically take the projection and then see what is happening with photons 1 and 4. Any help in understanding how to project onto the bell basis would be very much appreciated!

• Please make a reference in your question to the free access paper in arxiv Commented Jul 25, 2014 at 9:03

Firstly the link you have provided for the paper is only accessible for those who have a subscription to the APS journal. So maybe instead you could just give a preview of the equation you are referring to.

But I can give you a general overview on projections in quantum mechanics and Bell states. I will use Dirac's Bra-Ket notation for the rest of this post:

Mathematically speaking, a projection of e.g. one state $\left|v\right>$ onto another state $\left|u\right>$, is just written as follows: $$P_{u} := \left\langle u\right| v\rangle \left|u\right>$$ Where $P_u$ stands for projection onto $\left|u\right>$ and not a probability.

The $\left\langle u\right| v\rangle$ term is just the inner product between the two kets (states) in Dirac's notation.

Now in a physical sense, projection of states in quantum mechanics results from simple measurement of a system's observables, e.g. a photon's polarization, in a superposition of states $\left|\psi\right>= \alpha \left|H\right> + \beta \left|V\right>$ once measured, can collapse (be projected) onto one of its eigenstates, being horizontal or vertical polarization in this case.

As for Bell states, each Bell state describes a unique maximally entangled state of two qubits(photons e.g.). Meaning all 4 Bell states together: (A and B to distinguish between the photons) $$\left|\Phi_{\pm}^{AB}\right> = \frac{1}{\sqrt{2}}\left(\left|H^{A},H^{B}\right> \pm \left|V^{A},V^{B}\right>\right)$$ $$\left|\Psi_{\pm}^{AB}\right> = \frac{1}{\sqrt{2}}\left(\left|H^{A},V^{B}\right> \pm \left|V^{A},H^{B}\right>\right)$$ form a basis in the Hilbert space $H_4^{AB}$ of 2-qubit entangled states. So as shown before, here one can also talk about the projection of two qubit states onto a given Bell state, using the same formula. An example to showcase: If the Bell states are said to form a complete basis for 2-qubit states, then for any given 2-qubit state, one should be able to decompose and express it in terms of superpositon of Bell states. Simply meaning that you project the 2-qubit state on each one of the Bell states. For example lets see whether the state $\left|v\right>=\left|H^{A},V^{B}\right>$ has a non-vanishing probability of collapsing (being projected after measurement) onto the first Bell state: \begin{align*} \left\langle v\right| \Phi_+^{AB}\rangle &= \frac{1}{\sqrt{2}} \left\langle V^{B},H^{A}\right| \left(\left|H^{A},H^{B}\right> + \left|V^{A},V^{B}\right>\right) \\ &=\frac{1}{\sqrt{2}}\left(\left\langle V^{B},H^{A}\right| \left|H^{A},H^{B}\right> + \left\langle V^{B},H^{A}\right| \left|V^{A},V^{B}\right>\right)\\ &=\frac{1}{\sqrt{2}}\left(\left\langle H^{A}\right| H^{A}\rangle \left\langle V^{B}\right| H^{B}\rangle + \left\langle H^{A}\right| V^{A}\rangle \left\langle V^{B}\right| V^{B}\rangle\right)\\ &= 0 \end{align*} So we see that such projection is impossible, a result which is expected as $\left|H^{A},V^{B}\right>$ is not an entangled state since it can be written down as a single outer product of the individual kets.

I hope this overview gives you enough insight to be able to follow the mathematical steps involved in the paper you're reading.

• So am I correct in saying that the Bell States form a basis of an 'entangled' Hilbert space. What I mean is any entangled state can be represented by a projection onto these states? Commented Jul 25, 2014 at 13:17
• Yes, but note that it forms a complete basis only for two-qubit systems. So yes if you have any pair of entangled qubits, you can represent them in a superposition of Bell states. Commented Jul 26, 2014 at 14:33
• Wouldn't this mean the photon with the entangled state already collapsed (and therefore correlated to the first photon) is just impacting the reading on the second entangled pair, as opposed to some time-traversing effect? Commented Oct 3, 2018 at 16:32

In the context of your paper, the time-like Bell states are :

$|\psi^\pm\rangle^{\tau, \tau'}_{ab} = \frac{1}{\sqrt{2}}(|h_a^\tau v_b^{\tau'}\rangle \pm |h_a^\tau v_b^{\tau'}\rangle)$

$|\phi^\pm\rangle^{\tau, \tau'}_{ab} = \frac{1}{\sqrt{2}}(|h_a^\tau h_b^{\tau'}\rangle \pm |v_a^\tau v_b^{\tau'}\rangle)$

So, equation $(3)$ just reexpress $|S\rangle = |\psi^-\rangle^{0, \tau}_{ab} \otimes |\psi^-\rangle^{\tau, 2\tau}_{ab}$, which represents a tensor product of a particles $1(0),2(\tau)$ states times a particles $3(\tau),4(2\tau)$ state, in a decomposition on Bell states at equal times Bell states at different times: $|S\rangle = \frac{1}{2}(|\psi^-\rangle^{0, 2\tau}_{ab} \otimes |\psi^-\rangle^{\tau, \tau}_{ab} + ...)$ (just chek with the definition of the $\phi, \psi$ above that the decomposition is correct)

So, if you project, for instance, the $2,3$ particles on a equal times Bell state as $|\psi^-\rangle^{\tau, \tau}_{ab}$, then the $1,4$ particles are entangled because they are projected in the Bell state $|\psi^-\rangle^{0, 2\tau}_{ab}$

• So I have that state |S>. If I want to project it onto $|\psi^{\tau,\tau}_{ab}$, then I take $\langle \psi|S\rangle|\psi\rangle$? Commented Jul 25, 2014 at 14:17
• And then this should result in 1 and 4 being projected into that other bell state? Commented Jul 25, 2014 at 14:18
• Yes to your second question. For your first question, by measurement of a $2,3$ particle state ($|\psi^-\rangle^{\tau, \tau}_{ab}$), $|S\rangle$ is projected on the state $|\psi^-\rangle^{0, 2\tau}_{ab} \otimes |\psi^-\rangle^{\tau, \tau}_{ab}$, so you have : $\langle\psi^-|^{0, 2\tau}_{ab} \otimes \langle\psi^-|^{\tau, \tau}_{ab} |S\rangle = \frac{1}{2}$, and finally the projection of $|S\rangle$ is $\langle\psi^-|^{0, 2\tau}_{ab} \otimes \langle\psi^-|^{\tau, \tau}_{ab} |S\rangle |S\rangle = \frac{1}{2}|\psi^-\rangle^{0, 2\tau}_{ab} \otimes |\psi^-\rangle^{\tau, \tau}_{ab}$ Commented Jul 26, 2014 at 9:31

In the article - "Entanglement Swapping between Photons that Have Never Coexisted", the authors consider two pairs of photons generated with a time-gap of $$\tau$$. First, the first pair of two Photons(1,2) is created at time $$t=0$$, and the first photon is measured immediately after it is created. After this, the second photon is delayed until the other pair of Photons (3,4) is created at time $$t = \tau$$. After this, the second photon of the first pair and the first photon of the second pair, Photon 2 and Photon 3, are projected onto the bell basis. The second photon of the second pair is also delayed by $$\tau$$ in a delay line.

The combined state of these four photons can be represented as:

$$|\Psi^-\rangle_{1,2}^{0,0} \otimes |\Psi^-\rangle_{3,4}^{\tau,\tau}$$

Where we are using the definition of time-like bell states: $$|\Psi^\pm \rangle_{a,b}^{\tau,\tau'} = \frac{1}{\sqrt{2}} (|h_a^{\tau} v_b^{\tau'} \rangle \pm |v_a^{\tau} h_b^{\tau'} \rangle )$$

$$|\Phi^\pm \rangle_{a,b}^{\tau,\tau'} = \frac{1}{\sqrt{2}} (|h_a^{\tau} h_b^{\tau} \rangle \pm |v_a^{\tau} v_b^{\tau'} \rangle )$$

After applying the given procedure, the final combined state is denoted as:

$$|\Psi^-\rangle_{1,2}^{0,\tau} |\Psi^-\rangle_{3,4}^{\tau,2\tau} = \frac{1}{2} [ |\Psi^+\rangle_{1,4}^{0,2\tau} |\Psi^+\rangle_{2,3}^{\tau,\tau} - |\Psi^-\rangle_{1,4}^{0,2\tau} |\Psi^-\rangle_{2,3}^{\tau,\tau} -|\Phi^+\rangle_{1,4}^{0,2\tau} |\Phi^+\rangle_{2,3}^{\tau,\tau} + |\Phi^-\rangle_{1,4}^{0,2\tau} |\Phi^-\rangle_{2,3}^{\tau,\tau} ]$$

We can do a bit of mathematics to see how the above state can be written in the given form. From the definition of time-like bell states, we can write:

$$|\Psi^-\rangle_{1,2}^{0,\tau} |\Psi^-\rangle_{3,4}^{\tau,2\tau} = \frac{1}{2} (| h_1^0 v_2^\tau \rangle - | v_1^0 h_2^\tau \rangle) \otimes (| h_3^\tau v_4^{2\tau} \rangle - | v_3^\tau h_4^{2\tau} \rangle)$$

If we project Photon 2 and Photon 3 in the above state onto a bell state ( | \Psi^{+}_{2,3} \rangle ), then I can be expressed in the following way:

$$\langle \Psi^{+}|_{2,3}^{\tau,\tau} [|\Psi^-\rangle_{1,2}^{0,\tau} |\Psi^-\rangle_{3,4}^{\tau,2 \tau}] | \Psi^{+} \rangle_{2,3}^{\tau,\tau} = \frac{1}{2} | \Psi^+ \rangle_{1,4}^{0,2 \tau} | \Psi^{+} \rangle_{2,3}^{\tau,\tau}$$

Similarly, other terms can be calculated.

$$\langle \Psi^{-}|_{2,3}^{\tau,\tau} [|\Psi^-\rangle_{1,2}^{0,\tau} |\Psi^-\rangle_{3,4}^{\tau,2 \tau}] | \Psi^{-} \rangle_{2,3}^{\tau,\tau} = -\frac{1}{2} | \Psi^- \rangle_{1,4}^{0,2 \tau} | \Psi^{-} \rangle_{2,3}^{\tau,\tau}$$

$$\langle \Phi^{-}|_{2,3}^{\tau,\tau} [|\Psi^-\rangle_{1,2}^{0,\tau} |\Psi^-\rangle_{3,4}^{\tau,2 \tau}] | \Phi^{-} \rangle_{2,3}^{\tau,\tau} = \frac{1}{2} | \Phi^- \rangle_{1,4}^{0,2 \tau} | \Phi^{-} \rangle_{2,3}^{\tau,\tau}$$

$$\langle \Phi^{+}|_{2,3}^{\tau,\tau} [|\Psi^-\rangle_{1,2}^{0,\tau} |\Psi^-\rangle_{3,4}^{\tau,2 \tau}] | \Phi^{+} \rangle_{2,3}^{\tau,\tau} = -\frac{1}{2} | \Phi^+ \rangle_{1,4}^{0,2 \tau} | \Phi^{+} \rangle_{2,3}^{\tau,\tau}$$

From the above term, we can see that if Photon 2 and Photon 3 are projected on state $$| \Psi^{+}_{2,3} \rangle$$, then the resulting state of Photon 1 and 4 is:

$$| \Psi^+ \rangle_{1,4}^{0,2 \tau} = \frac{1}{\sqrt{2}} ( | h_1^{0} v_4^{2\tau} \rangle + | v_1^{0} h_4^{2\tau} \rangle )$$

We can see that after performing bell projection on Photons 2 and 3, Photons 1,4, which did not share any correlations between them, became entangled. This entanglement can be seen as entanglement between photons that do not coexist, and it is referred to as Quantum entanglement in time.

I hope it will be helpful. The full details are provided here for more information - https://quantumthermodynamic.com/quantum-entanglement-in-time/