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Spacetime Path Integrals for Entangled States

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Abstract

Although the path-integral formalism is known to be equivalent to conventional quantum mechanics, it is not generally obvious how to implement path-based calculations for multi-qubit entangled states. Whether one takes the formal view of entangled states as entities in a high-dimensional Hilbert space, or the intuitive view of these states as a connection between distant spatial configurations, it may not even be obvious that a path-based calculation can be achieved using only paths in ordinary space and time. Previous work has shown how to do this for certain special states; this paper extends those results to all pure two-qubit states, where each qubit can be measured in an arbitrary basis. Certain three-qubit states are also developed, and path integrals again reproduce the usual correlations. These results should allow for a substantial amount of conventional quantum analysis to be translated over into a path-integral perspective, simplifying certain calculations, and more generally informing research in quantum foundations.

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Notes

  1. There are reasonable arguments as to why the approach here could also be extended to other types of entanglement, but for the purposes of this paper, we shall focus on path integral accounts of which-way entanglement.

  2. Granted, many physicists commonly think of entangled states as “superpositions” of configurations in ordinary space, which is arguably comparable to a set of possible paths. But apart from the disconnect between this viewpoint and the formal mathematics, this model of entangled states further requires the visualization of a mysterious direct connection between spacelike-separated events, a connection not required in the path integral.

  3. While there are various physical mechanisms that emit photons in opposite directions, further details of this source are not important for any of the calculations below, except we will need certain directions to be more probable than others.

  4. If the photons are emitted in any other direction, they will not be detected, and will not enter into to the measurement probabilities.

  5. These measurement devices are slightly different than those used in the previous sections, but here this allows for a smaller total number of phase plates, simplifying the system.

  6. Here we are using the word ”classical” to indicate that only straight-line paths are required. If a photon is thought of as an electromagnetic wave, it is not literally classical for that wave to entirely pass through a beamsplitter (or to entirely reflect), but the path integral formalism does not evidently allow for including both beamsplitter outputs into a single history if only one detector fires.

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Acknowledgements

The authors gratefully thank Emily Adlam, Hilary Hurst, and several anonymous referees for helpful comments and suggestions.

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Appendices

Appendix 1: The Schmidt Basis

For some applications of the two-qubit path integral, one may want to start with a completely general two-qubit pure state

$$\begin{aligned} \left| \psi \right\rangle =a\left| {\bar{0}}{\bar{0}} \right\rangle +b\left| {\bar{0}}{\bar{1}} \right\rangle +c\left| {\bar{1}}{\bar{0}} \right\rangle +d\left| {\bar{1}}{\bar{1}} \right\rangle , \end{aligned}$$
(30)

where abc and d are complex and \(|a|^2 + |b|^2 + |c|^2 + |d|^2 = 1\). Here the overbars on \(\left| {\bar{0}} \right\rangle\) and \(\left| {\bar{1}} \right\rangle\) are to distinguish these states from the Schmidt basis \(\left| {0} \right\rangle\) and \(\left| {1} \right\rangle\) used in Sect. 3.

Each of the two qubits in \(\left| \psi \right\rangle\) can then be measured in some arbitrary direction on the Bloch sphere, corresponding to the angles \(({\bar{\theta }}_1,{\bar{\phi }}_1)\) for the first qubit and \(({\bar{\theta }}_2,{\bar{\phi }}_2)\) for the second. These angles are defined relative to the corresponding basis on the right side of (30), and will generally be different than the angles \((\theta _1,\phi _1)\) and \((\theta _2,\phi _2)\) used in Sect. 3. (These latter angles are defined relative to the Schmidt basis states \(\left| {0} \right\rangle\) and \(\left| {1} \right\rangle\).)

However, it is a relatively straightforward manner to relate these various coordinate systems, if one can solve for the direction \((\alpha ,\beta )\) on the given Bloch sphere which lines up with the z-axis of the Schmidt basis Bloch sphere. For the first qubit, this relationship is defined by

$$\begin{aligned} \left| 0 \right\rangle _1= & {} \cos \frac{\alpha _1}{2} \left| {\bar{0}} \right\rangle _1 + \sin \frac{\alpha _1}{2} e^{i\beta _1} \left| {\bar{1}} \right\rangle _1 \end{aligned}$$
(31)
$$\begin{aligned} \left| 1 \right\rangle _1= & {} \sin \frac{\alpha _1}{2} e^{-i\beta _1} \left| {\bar{0}} \right\rangle _1 - \cos \frac{\alpha _1}{2} \left| {\bar{1}} \right\rangle _1. \end{aligned}$$
(32)

And a similar (but reversed) expression for the second qubit is

$$\begin{aligned} \left| 0 \right\rangle _2= & {} \sin \frac{\alpha _2}{2} e^{-i\beta _2} \left| {\bar{0}} \right\rangle _2 - \cos \frac{\alpha _2}{2} \left| {\bar{1}} \right\rangle _2 \end{aligned}$$
(33)
$$\begin{aligned} \left| 1 \right\rangle _2= & {} \cos \frac{\alpha _2}{2} \left| {\bar{0}} \right\rangle _2 + \sin \frac{\alpha _2}{2} e^{i\beta _2} \left| {\bar{1}} \right\rangle _2. \end{aligned}$$
(34)

Recall, that if written in the Schmidt basis, the state \(\left| \psi \right\rangle\) takes the much simpler form given by Eq. (9). If one knows the angles \((\alpha ,\beta )\), then one knows precisely how the measurement angles \(({\bar{\theta }},{\bar{\phi }})\) in the given basis are related to the measurement angles \((\theta ,\phi )\) in the Schmidt basis. Geometrically, on the Bloch sphere, they are related by a rotation which takes \((\alpha ,\beta )\) to the z-axis, but any careful analysis would certainly want to use the precise above equations.

All that remains is to find the four angles \(\alpha _1\), \(\beta _1\), \(\alpha _2\) and \(\beta _2\). We have not been able to find a derivation of these angles in the published literature, but they can essentially be recovered from the results in [25], except for the special case of maximally-entangled states which will be worked out separately. In terms of the four complex parameters in Eq. (30), these angles are

$$\begin{aligned} \cos \alpha _1= & {} \frac{|a|^2+|b|^2-|c|^2-|d|^2}{\sqrt{1-4|ad-bc|^2}}, \end{aligned}$$
(35)
$$\begin{aligned} e^{-i\beta _1}= & {} \frac{ac^*+bd^*}{|ac^*+bd^*|}, \end{aligned}$$
(36)
$$\begin{aligned} \cos \alpha _2= & {} \frac{|a|^2+|c|^2-|b|^2-|d|^2}{\sqrt{1-4|ad-bc|^2}}, \end{aligned}$$
(37)
$$\begin{aligned} e^{-i\beta _2}= & {} \frac{ab^*+cd^*}{|ab^*+cd^*|}. \end{aligned}$$
(38)

Via tedious algebra, all of the above equations in this appendix can be combined to ascertain that both \(\left\langle \psi |00 \right\rangle\) and \(\left\langle \psi |11 \right\rangle\) are exactly zero. (We have also checked this numerically.) This algebra is much easier if one uses the normalization of \(\left| \psi \right\rangle\) to prove the relationships

$$\begin{aligned} \sin \alpha _1= & {} \frac{2|ac^*+bd^*|^2}{\sqrt{1-4|ad-bc|^2}}, \end{aligned}$$
(39)
$$\begin{aligned} \sin \alpha _2= & {} \frac{2|ab^*+cd^*|^2}{\sqrt{1-4|ad-bc|^2}}. \end{aligned}$$
(40)

These equations also indicate what is going on when one of the \(\beta\) terms become undefined – that is, when either \(|ac^*+bd^*|\) or \(|ab^*+cd^*|\) is zero. In these cases, the corresponding \(\alpha\) also goes to zero or \(\pi\), such that the two bases are aligned for that qubit. (For \(\alpha =0\) it is already in the Schmidt basis, and for \(\alpha =\pi\) one need merely set \(\left| 0 \right\rangle =\left| {\bar{1}} \right\rangle\) and \(\left| 1 \right\rangle =\left| {\bar{0}} \right\rangle\).) The fact that \(\beta\) is undefined is therefore just the usual spherical-coordinate ambiguity for \(\phi\) on the z-axis; in this case \(\beta\) can take any value, such as \(\beta =0\).

The only remaining problem is the special case of maximally-entangled states, where \({{1-4|ad-bc|^2}}=0\); evidently in this case the above angles \(\alpha\) are undefined. In that case, it is always possible to leave one of the qubits unchanged, and then find a Schmidt basis using only the other qubit. These maximally entangled states can always [26] be written in the form

$$\begin{aligned} {\left| \psi \right\rangle }=\frac{1}{\sqrt{2}}\left[ e^{i\chi _1}cos\gamma \!\left| {\bar{0}}{\bar{0}} \right\rangle +e^{i\chi _2}sin\gamma \!\left| {\bar{0}}{\bar{1}} \right\rangle +e^{-i\chi _2}sin\gamma \!\left| {\bar{1}}{\bar{0}} \right\rangle -e^{-i\chi _1}cos\gamma \!\left| {\bar{1}}{\bar{1}} \right\rangle \right] . \end{aligned}$$
(41)

For this general maximally-entangled state, if one leaves the first qubit unchanged, \(\bar{\left| 0 \right\rangle }_1=\left| 0 \right\rangle _1\), the second qubit can be written as a superposition of a Schmidt basis \(\left| 0 \right\rangle _2\) and \(\left| 1 \right\rangle _2\)

$$\begin{aligned} \bar{\left| 0 \right\rangle }_2= & {} f\left| 0 \right\rangle _2+g\left| 1 \right\rangle _2, \end{aligned}$$
(42)
$$\begin{aligned} \bar{\left| 1 \right\rangle }_2= & {} g^*\left| 0 \right\rangle _2-f^*\left| 1 \right\rangle _2. \end{aligned}$$
(43)

For a Schmidt basis of the form Eq. (9), we require \(\left\langle \psi |00 \right\rangle =0\) and \(\left\langle \psi |11 \right\rangle =0\). Therefore

$$\begin{aligned}&(e^{i\chi _1}cos\gamma )^*f+(e^{i\chi _2}sin\gamma )^*g=0, \end{aligned}$$
(44)
$$\begin{aligned}&(e^{-i\chi _2}sin\gamma )^*g^*-(-e^{-i\chi _1}cos\gamma )^*f^*=0. \end{aligned}$$
(45)

As the coefficients f and g are normalized,we get:

$$\begin{aligned} f= & {} sin\gamma , \end{aligned}$$
(46)
$$\begin{aligned} g= & {} -e^{-i(\chi _1-\chi _2)}\,cos\gamma . \end{aligned}$$
(47)

The above results can be combined to indicate precisely how to rotate the coordinate systems of the measurement devices, such that the results from Sect. 3 can directly apply to any possible pure two qubit state.

Appendix 2: GHZ Details

In this Appendix, we provide the details of the GHZ path integral calculations for the other possible outcomes. The QM probabilities can be calculated from (25) as:

$$\begin{aligned} \begin{aligned} \left| \left\langle \psi |{++-} \right\rangle \right| ^2\! = & {} \frac{1}{2} \left| cos \frac{\theta _1}{2}cos\frac{\theta _2}{2}sin\frac{\theta _3}{2}\,e^{-i\phi _3} - sin \frac{\theta _1}{2}sin\frac{\theta _2}{2}cos\frac{\theta _3}{2}\,e^{i(\phi _1 + \phi _2)}\right| ^2 ,\\ \left| \left\langle \psi |{+-+} \right\rangle \right| ^2= & {} \frac{1}{2} \left| cos \frac{\theta _1}{2}sin\frac{\theta _2}{2}cos\frac{\theta _3}{2}\,e^{-i\phi _2} - sin \frac{\theta _1}{2}cos\frac{\theta _2}{2}sin\frac{\theta _3}{2}\,e^{i(\phi _1 + \phi _3)}\right| ^2 ,\\ \left| \left\langle \psi |{+--} \right\rangle \right| ^2= & {} \frac{1}{2} \left| cos \frac{\theta _1}{2}sin\frac{\theta _2}{2}sin\frac{\theta _3}{2}\,e^{-i(\phi _2 + \phi _3)} + sin \frac{\theta _1}{2}cos\frac{\theta _2}{2}cos\frac{\theta _3}{2}\,e^{i\phi _1}\right| ^2 ,\\ \left| \left\langle \psi |{-++} \right\rangle \right| ^2= & {} \frac{1}{2} \left| sin \frac{\theta _1}{2}cos\frac{\theta _2}{2}cos\frac{\theta _3}{2}\,e^{-i\phi _1} - cos \frac{\theta _1}{2}sin\frac{\theta _2}{2}sin\frac{\theta _3}{2}\,e^{i(\phi _2 + \phi _3)}\right| ^2 ,\\ \left| \left\langle \psi |{-+-} \right\rangle \right| ^2= & {} \frac{1}{2} \left| sin \frac{\theta _1}{2}cos\frac{\theta _2}{2}sin\frac{\theta _3}{2}\,e^{-i(\phi _1 + \phi _3)} + cos \frac{\theta _1}{2}sin\frac{\theta _2}{2}cos\frac{\theta _3}{2}\,e^{i\phi _2}\right| ^2 ,\\ \left| \left\langle \psi |{--+} \right\rangle \right| ^2= & {} \frac{1}{2} \left| sin \frac{\theta _1}{2}sin\frac{\theta _2}{2}cos\frac{\theta _3}{2}\,e^{-i(\phi _1 + \phi _2)} + cos \frac{\theta _1}{2}cos\frac{\theta _2}{2}sin\frac{\theta _3}{2}\,e^{i\phi _3}\right| ^2 ,\\ \left| \left\langle \psi |{---} \right\rangle \right| ^2= & {} \frac{1}{2} \left| sin \frac{\theta _1}{2}sin\frac{\theta _2}{2}sin\frac{\theta _3}{2}\,e^{-i(\phi _1 + \phi _2+ \phi _3)} - cos \frac{\theta _1}{2}cos\frac{\theta _2}{2}sin\frac{\theta _3}{2}\right| ^2 . \end{aligned} \end{aligned}$$
(48)

Going to the path integral formalism, each outcome has two histories that lead to it; the amplitudes for these histories are listed below and have been calculated using the procedure outlined by Sect. 3.3 for Fig. 3. In each of these cases the first parentheses corresponds to particle 1, the second corresponds to particle 2 and the third corresponds to particle 3.

$$\begin{aligned} \begin{aligned} {\mathcal {E}}^{solid}_{++-} =&\frac{1}{\sqrt{2}}\left( e^{i\phi _1}sin\frac{\theta _1}{2}\right) \left( e^{i\phi _2}sin\frac{\theta _2}{2}\right) \left( ie^{i\phi _3}cos\frac{\theta _3}{2}\right) , \\ {\mathcal {E}}^{dashed}_{++-} =&\frac{1}{\sqrt{2}}\left( i\,cos\frac{\theta _1}{2}\right) \left( i\,cos\frac{\theta _2}{2}\right) \left( i\,sin\frac{\theta _3}{2}\right) , \\ {\mathcal {E}}^{solid}_{+-+} =&\frac{1}{\sqrt{2}}\left( e^{i\phi _1}\,sin\frac{\theta _1}{2}\right) \left( e^{i\phi _2}\,i\,cos\frac{\theta _2}{2}\right) \left( e^{i\phi _3}\,sin\frac{\theta _3}{2}\right) , \\ {\mathcal {E}}^{dashed}_{+-+} =&\frac{1}{\sqrt{2}}\left( i\,cos\frac{\theta _1}{2}\right) \left( sin\frac{\theta _2}{2})\right) \left( i^2\,cos\frac{\theta _3}{2}\right) ,\\ {\mathcal {E}}^{solid}_{+--} =&\frac{1}{\sqrt{2}}\left( e^{i\phi _1}\,sin\frac{\theta _1}{2}\right) \left( e^{i\phi _2}\,i\,cos\frac{\theta _2}{2}\right) \left( e^{i\phi _3}\,i\,cos\frac{\theta _3}{2}\right) ,\\ {\mathcal {E}}^{dashed}_{+--} =&\frac{1}{\sqrt{2}}\left( i\,cos\frac{\theta _1}{2}\right) \left( sin\frac{\theta _2}{2}\right) \left( i\,sin\frac{\theta _3}{2}\right) , \\ {\mathcal {E}}^{solid}_{-++} =&\frac{1}{\sqrt{2}}\left( e^{i\phi _1}\,i\,cos\frac{\theta _1}{2}\right) \left( e^{i\phi _2}\,sin\frac{\theta _2}{2}\right) \left( e^{i\phi _3}\,sin\frac{\theta _3}{2}\right) , \\ {\mathcal {E}}^{dashed}_{-++} =&\frac{1}{\sqrt{2}}\left( sin\frac{\theta _1}{2}\right) \left( i\,cos\frac{\theta _2}{2}\right) \left( i^2\,cos\frac{\theta _3}{2}\right) , \\ {\mathcal {E}}^{solid}_{-+-} =&\frac{1}{\sqrt{2}}\left( e^{i\phi _1}\,i\,cos\frac{\theta _1}{2}\right) \left( e^{i\phi _2}\,sin\frac{\theta _2}{2}\right) \left( e^{i\phi _3}\,i\,cos\frac{\theta _3}{2}\right) ,\\ {\mathcal {E}}^{dashed}_{-+-} =&\frac{1}{\sqrt{2}}\left( sin\frac{\theta _1}{2}\right) \left( i\,cos\frac{\theta _2}{2}\right) \left( i\,sin\frac{\theta _3}{2}\right) ,\\ {\mathcal {E}}^{solid}_{--+} =&\frac{1}{\sqrt{2}}\left( e^{i\phi _1}\,i\,cos\frac{\theta _1}{2}\right) \left( e^{i\phi _2}\,i\,cos\frac{\theta _2}{2}\right) \left( e^{i\phi _3}\,sin\frac{\theta _3}{2}\right) , \\ {\mathcal {E}}^{dashed}_{--+} =&\frac{1}{\sqrt{2}}\left( sin\frac{\theta _1}{2}\right) \left( i\,cos\frac{\theta _2}{2}\right) \left( i\,sin\frac{\theta _3}{2}\right) , \\ {\mathcal {E}}^{solid}_{---} =&\frac{1}{\sqrt{2}}\left( e^{i\phi _1}\,i\,cos\frac{\theta _1}{2}\right) \left( e^{i\phi _2}\,i\,cos\frac{\theta _2}{2}\right) \left( e^{i\phi _3}i\,cos\frac{\theta _3}{2}\right) ,\\ {\mathcal {E}}^{dashed}_{---} =&\frac{1}{\sqrt{2}}\left( sin\frac{\theta _1}{2}\right) \left( sin\frac{\theta _2}{2}\right) \left( i\,sin\frac{\theta _3}{2}\right) . \end{aligned} \end{aligned}$$
(49)

Plugging these amplitudes into Eq. (18) yields the probabilities given by Eq. (48).

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Tyagi, N., Wharton, K. Spacetime Path Integrals for Entangled States. Found Phys 52, 9 (2022). https://doi.org/10.1007/s10701-021-00520-2

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