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Updated Slack URL invite link, fixed typos in documentation
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josh146 committed May 9, 2018
1 parent d8e1d43 commit d35cc22
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2 changes: 1 addition & 1 deletion doc/algorithms/hamiltonian_simulation.rst
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Expand Up @@ -38,7 +38,7 @@ The `Bose-Hubbard Hamiltonian <https://en.wikipedia.org/wiki/Bose%E2%80%93Hubbar

.. math:: H = J\sum_{i}\sum_j A_{ij} \ad_i\a_j + \frac{1}{2}U\sum_i \hat{n}_i(\hat{n}_i-1)= J(\ad_1 \a_2 + \ad_2\a_1) + \frac{1}{2}U ( \hat{n}_1^2 - \hat{n}_1 + \hat{n}_2^2 - \hat{n}_2)

where :math:`J` represents the transfer integral or hopping term of the boson between nodes, and :math:`U` is the on-site interaction potential. Here, :math:`\ad_1 \a_2` represents a boson transitioning from node 1 to node 2, while :math:`\ad_2\a_1` represents a boson transitioning from node 2 to node 1, and :math:`\hat{n}_i=\ad\a` is the number operator applied to mode :math:`i`. Applying the Lie-product formula, we find that
where :math:`J` represents the transfer integral or hopping term of the boson between nodes, and :math:`U` is the on-site interaction potential. Here, :math:`\ad_1 \a_2` represents a boson transitioning from node 1 to node 2, while :math:`\ad_2\a_1` represents a boson transitioning from node 2 to node 1, and :math:`\hat{n}_i=\ad_i\a_i` is the number operator applied to mode :math:`i`. Applying the Lie-product formula, we find that

.. math:: e^{-iHt} = \left[\exp\left({-i\frac{ J t}{k}(\ad_1 \a_2 + \ad_2\a_1)}\right)\exp\left(-i\frac{Ut}{2k}\hat{n}_1^2\right)\exp\left(-i\frac{Ut}{2k}\hat{n}_2^2\right)\exp\left(i\frac{Ut}{2k}\hat{n}_1\right)\exp\left(i\frac{Ut}{2k}\hat{n}_2\right)\right]^k+\mathcal{O}\left(t^2/k\right),

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2 changes: 1 addition & 1 deletion doc/conventions/gates.rst
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Expand Up @@ -157,7 +157,7 @@ Rotation
We write the phase space rotation operator as

.. math::
R(\phi) = \exp\left(i \phi \ad \a\right)=\exp\left(i \frac{\phi}{2} \left(\frac{\x^2+ \p^2}{\hbar}-\frac{1}{2}\right)\right), \quad R^\dagger(\phi) \a R(\phi) = \a e^{i \phi}
R(\phi) = \exp\left(i \phi \ad \a\right)=\exp\left(i \frac{\phi}{2} \left(\frac{\x^2+ \p^2}{\hbar}-\I\right)\right), \quad R^\dagger(\phi) \a R(\phi) = \a e^{i \phi}
.. tip::

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2 changes: 1 addition & 1 deletion doc/index.rst
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Expand Up @@ -64,7 +64,7 @@ Support
- **Source Code:** https://github.com/XanaduAI/strawberryfields
- **Issue Tracker:** https://github.com/XanaduAI/strawberryfields/issues

If you are having issues, please let us know by posting the issue on our Github issue tracker, or by joining our `Strawberry Fields Slack channel <https://join.slack.com/t/sf-xanadu/shared_invite/enQtMzQ0NzA0NTczNTkxLTJhOWM0MzY4N2U1ODc5NWZmY2JlN2FmOTY3OTE3ZmVjMWFhZjMwZWY0ODRkZDFmMTA5NjA1YjgyYzA3NDg3NGM>`_.
If you are having issues, please let us know by posting the issue on our Github issue tracker, or by joining our `Strawberry Fields Slack channel <https://u.strawberryfields.ai/slack>`_.

For more details on contributing or performing research with Strawberry Fields, please see
:ref:`research`.
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