==========
Challenges
==========
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Scroll down for solution...
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Challenge 1: Explore how to define desired states
-------------------------------------------------
Create the following droplets using the menu :menuselection:`Initial
State --> MQ(x/y ops.)`:
.. list-table::
:class: no-cap-num
* - a.
.. drops:: I1xI2x + I1yI2y
:width: 50%
:caption:
:view: p170
- b.
.. drops:: 2*I1z*I2y
:width: 50%
:caption:
:view: p170
* - c.
.. drops:: I1xI2y + I1yI2x
:width: 50%
:caption:
:view: p170
- d.
.. drops:: 2*(I1xI2xI3x -I1xI2yI3y - I1yI2xI3y - I1yI2yI3x)
:width: 50%
:caption:
:view: p170
a. ?
b. ?
c. ?
d. ?
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Solution 1
==========
a) :menuselection:`MQ(x/y ops.) --> 0Q(I1,I2) --> 0Qx(I1,I2)`
.. drops:: I1xI2x + I1yI2y
:width: 50%
:view: p170
:class: no-cap-num
b) :menuselection:`MQ(x/y ops.) --> 1Q(I1,I2) --> 2*I1z*I2y`
.. drops:: 2*I1z*I2y
:width: 50%
:view: p170
:class: no-cap-num
c) :menuselection:`MQ(x/y ops.) --> 2Q(I1,I2) --> 2Qy(I1,I2)`
.. drops:: I1xI2y + I1yI2x
:width: 50%
:view: p170
:class: no-cap-num
d) :menuselection:`MQ(x/y ops.) --> 3Q(I1,I2,I3) --> 2*3Qx(I1,I2,I3)`
.. drops:: 2*(I1xI2xI3x -I1xI2yI3y - I1yI2xI3y - I1yI2yI3x)
:width: 50%
:view: p170
:class: no-cap-num
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Challenge 2: Explore how to define desired states
-------------------------------------------------
Create the following droplets using the menu :menuselection:`Initial
State --> MQ(+/- ops.)`:
.. list-table::
:class: no-cap-num
* - a.
.. drops:: I1m
:width: 40%
:caption:
:view: p170
:prefs: Magnetization Vectors=false
- b.
.. drops:: I1pI2p
:width: 40%
:caption:
:view: p170
- c.
.. drops:: 2*I1mI2mI3m
:width: 40%
:caption:
:view: p170
a) ?
b) ?
c) ?
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Solution 2
==========
a) :menuselection:`MQ(+/- ops.) --> 1Q(I1) --> I1(-)` (turn off the
preference :ref:`Magnetization Vectors` to get an opaque droplet).
.. drops:: I1m
:width: 40%
:view: Bp170
:class: no-cap-num
:prefs: Magnetization Vectors=false
b) :menuselection:`MQ(+/- ops.) --> 2Q(I1,I2) --> I1(+)I2(+)`
.. drops:: I1pI2p
:width: 40%
:view: p170
:class: no-cap-num
c) :menuselection:`MQ(+/- ops.) --> 3Q(I1,I2,I3) --> 2*I1(-)*I2(-)*I3(-)`
.. drops:: 2*I1mI2mI3m
:width: 40%
:view: p170
:class: no-cap-num
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Challenge 3: Explore how to define desired states
-------------------------------------------------
Create the following droplets using the menu :menuselection:`Initial
State --> MQ(+/- ops.)`:
.. list-table::
:class: no-cap-num
* - a.
.. drops:: I1pI2m
:width: 40%
:caption:
:view: Cp170
- b.
.. drops:: I1mI2p
:width: 40%
:caption:
:view: Cp170
- c.
.. drops:: 2I1pI2z
:width: 40%
:caption:
:view: p170
a) ?
b) ?
c) ?
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Solution 3
==========
a) :menuselection:`MQ(+/- ops.) --> 2Q(I1,I2) --> I1(+)I2(-)`
.. drops:: I1pI2m
:width: 40%
:view: Cp170
:class: no-cap-num
b) :menuselection:`MQ(+/- ops.) --> 2Q(I1,I2) --> I1(-)I2(+)`
.. drops:: I1mI2p
:width: 40%
:view: Cp170
:class: no-cap-num
c) :menuselection:`MQ(+/- ops.) --> 2Q(I1,I2) --> 2*I1(+)*I2z`
.. drops:: 2I1pI2z
:width: 40%
:view: p170
:class: no-cap-num
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Challenge 4: Explore how to interpret droplets
----------------------------------------------
Using the menu :menuselection:`Initial State --> MQ(x/y ops.) -->
2Q(I1,I2)`, create the operator `2Qx(I1,I2)` and determine how it can
be expressed as a linear combination of :term:`Cartesian product
operators `.
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Solution 4
==========
The double-quantum x operator involving spins **I1** and **I2**
corresponds to the following droplet:
.. drops:: I1xI2x - I1yI2y
:sequence: Standard/FreeEvo1sec
:window: List Basis Elems
:aspect: 200
:name: challenges_4
Select the menu :menuselection:`View --> List Prod. Ops...` to see the
operator decomposition. The Cartesian terms will be displayed in the
shorthand notation for Cartesian operators as shown in
:numref:`challenges_4`.
Hence, :pton:`2Qx(I1,I2) = 0.5 (2I1xI2x) - 0.5 (2I1yI2y) = I1xI2x -
I1yI2y`.
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Challenge 5: Explore how to interpret droplets
----------------------------------------------
Using the menu :menuselection:`Initial State --> MQ(+/− ops.) -->
3Q(I1,I2,I3)`, create the operator :pton:`2I1pI2pI3p` and determine
how it can be expressed as a linear combination of Cartesian product
operators.
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Solution 5
==========
The +3-quantum operator :pton:`2*I1pI2pI3p` corresponds to the
following droplet:
.. drops:: 2*I1pI2pI3p
:sequence: Standard/FreeEvo1sec
:window: List Basis Elems
:name: challenges_5
Select the menu :menuselection:`View --> List Prod. Ops...`. The
following terms will be displayed in the shorthand notation for
Cartesian operators as shown in :numref:`challenges_5`.
Hence,
.. math::
2*I_1^+I_2^+I_3^+ & = 0.5 (4I_{1x}I_{2x}I_{3x}) \\
& + 0.5 i (4I_{1y}I_{2x}I_{3x}) \\
& + 0.5 i (4I_{1x}I_{2y}I_{3x}) \\
& - 0.5 (4I_{1y}I_{2y}I_{3x}) \\
& + 0.5 i (4I_{1x}I_{2x}I_{3y}) \\
& - 0.5 (4I_{1y}I_{2x}I_{3y}) \\
& - 0.5 (4I_{1x}I_{2y}I_{3y}) \\
& - 0.5 i (4I_{1y}I_{2y}I_{3y}) \\
\\
& = (2I_{1x}I_{2x}I_{3x} - 2I_{1x}I_{2y}I_{3y} - 2I_{1y}I_{2x}I_{3y} - 2I_{1y}I_{2y}I_{3x} \\
& - 2 i I_{1y}I_{2y}I_{3y} + 2 i I_{1y}I_{2x}I_{3y} - 2 i I_{1x}I_{2y}I_{3x} + 2 i I_{1x}I_{2x}I_{3y}
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Challenge 6: Pulse sequence design challenge
--------------------------------------------
Suppose in the course of a pulse sequence, you have created the state
:pton:`I1zI2z - I1yI2y`. How can you transfer this state into ±2 quantum
coherence by a single pulse?
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Solution 6
==========
Using the menu :menuselection:`Initial State --> Edit Operator`,
create the operator :pton:`I1zI2z - I1yI2y`. The corresponding DROPS
representation should look like:
.. drops:: I1zI2z - I1yI2y
:nspin: 3; v1=0; v2=0; v3=0; J13=.1; J23=.1; J12=.1
:view: Cp160
:aspect: 200%
The droplet representing the bilinear terms involving spins **I1** and
**I2** already has the desired shape and color of ±2 quantum
coherence, except that it is not correctly oriented! Rotating it by
90° around the y (or −y) axis will yield the correct orientation,
i.e. a non-selective 90°y (or a 90°−y) pulse yields the desired ±2
quantum coherence. This can be tested by selecting
:menuselection:`Pulse Sequence --> Rotation --> 90° Pulse --> 90°(-y)`
and the resulting DROPS representation at the end of the pulse is
shown on the right. (Note the rectangle at the bottom, which
represents the pulse, where the blue color represents the pulse phase
−y.)
.. drops:: I1zI2z - I1yI2y
:nspin: 3; v1=0; v2=0; v3=0; J13=.1; J23=.1; J12=.1; Homonuclear=1
:sequence: Standard/Pulse90degmy
:window: List Basis Elems
:view: Cp160
:time: 0.025
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Challenge 7: Define a new operator
----------------------------------
Display the droplet corresponding to the operator :pton:`2I1xI2x + i
2I1zI2x`. (When the droplet is oriented properly, you will see the
SpinDrops logo.)
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Solution 7
==========
Using the menu :menuselection:`Initial State --> Edit Operator`,
create the operator :pton:`2I1xI2x + i 2I1zI2x`. Orienting the
resulting droplet in the Drops display such that the view direction is
along the y axis yields the SpinDrops logo :code:`:-)`
.. drops:: 2I1xI2x + i*2I1zI2x
:caption:
:width: 80%
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Challenge 8: Rotations of operators and droplets
------------------------------------------------
Starting from the term :pton:`2I1xI2y`, find a sequence of
non-selective rotations (90±x, 90±y, or 90±z) to incrementally
create all of the bilinear operators of the form :math:`2I1_aI2_b`
with :math:`a≠b` and :math:`a,b ∈ {x,y,z}`.
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Solution 8
==========
.. list-table::
:class: no-cap-num nocol norow
* - .. drops:: 2I1xI2y
:width: 40%
:caption:
:view: p170
- :math:`90°_x \rightarrow`
- .. drops:: 2I1xI2z
:width: 40%
:caption:
:view: p170
- :math:`90°_z \rightarrow`
- .. drops:: 2I1yI2z
:width: 40%
:caption:
:view: p170
* - :math:`90°_y \uparrow`
-
-
-
- :math:`90°_y \downarrow`
* - .. drops:: 2I1zI2y
:width: 40%
:caption:
:view: p170
- :math:`\leftarrow 90°_x`
- .. drops:: 2I1zI2x
:width: 40%
:caption:
:view: p170
- :math:`\leftarrow 90°_x`
- .. drops:: 2I1yI2x
:width: 40%
:caption:
:view: p170
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Challenge 9: Recognizing droplets of antiphase operators (A)
------------------------------------------------------------
Determine for each of the following droplets whether it represents
antiphase coherence with respect to the `first` spin
(:math:`±2I_{1z}I_{2a}` with :math:`a ∈ \{x,y\}` ) or with respect to
the `second` spin (:math:`±2I_{1a}I_{2z}` with :math:`a ∈ \{x,y\}` )!
.. list-table::
:class: no-cap-num
* - .. drops:: 2I1xI2z
:width: 40%
:caption:
:view: p170
- .. drops:: 2I1zI2x
:width: 40%
:caption:
:view: p170
- .. drops:: -2I1zI2y
:width: 40%
:caption:
:view: p170
* - .. drops:: 2I1yI2z
:width: 40%
:caption:
:view: p170
- .. drops:: 2I1zI2y
:width: 40%
:caption:
:view: p170
- .. drops:: -2I1yI2z
:width: 40%
:caption:
:view: p170
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Solution 9
==========
This challenge can be solved by simply observing the “kissing beans’
head tilt” (c.f. :ref:`drops_props_kb_summary`).
.. list-table::
:class: no-cap-num
* - right tilt
.. drops:: 2I1xI2z
:width: 40%
:view: p170
:caption: :math:`±2I_{1a}I_{2z}`
- left tilt
.. drops:: 2I1zI2x
:width: 40%
:view: p170
:caption: :math:`±2I_{1z}I_{2a}`
- left tilt
.. drops:: -2I1zI2y
:width: 40%
:view: p170
:caption: :math:`±2I_{1z}I_{2a}`
* - right tilt
.. drops:: 2I1yI2z
:width: 40%
:view: p170
:caption: :math:`±2I_{1a}I_{2z}`
- left tilt
.. drops:: 2I1zI2y
:width: 40%
:view: p170
:caption: :math:`±2I_{1z}I_{2a}`
- right tilt
.. drops:: -2I1yI2z
:width: 40%
:view: p170
:caption: :math:`±2I_{1a}I_{2z}`
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Challenge 10: Recognizing droplets of antiphase operators (B)
-------------------------------------------------------------
Determine for each of the droplets the **exact** form of the
corresponding antiphase operator!
.. list-table::
:class: no-cap-num
* - .. drops:: 2I1xI2z
:width: 40%
:caption:
:view: p170
- .. drops:: 2I1zI2x
:width: 40%
:caption:
:view: p170
- .. drops:: -2I1zI2y
:width: 40%
:caption:
:view: p170
* - .. drops:: 2I1yI2z
:width: 40%
:caption:
:view: p170
- .. drops:: 2I1zI2y
:width: 40%
:caption:
:view: p170
- .. drops:: -2I1yI2z
:width: 40%
:caption:
:view: p170
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Solution 10
===========
.. list-table::
:class: no-cap-num
* - .. drops:: 2I1xI2z
:width: 40%
:view: p170
- .. drops:: 2I1zI2x
:width: 40%
:view: p170
- .. drops:: -2I1zI2y
:width: 40%
:view: p170
* - .. drops:: 2I1yI2z
:width: 40%
:view: p170
- .. drops:: 2I1zI2y
:width: 40%
:view: p170
- .. drops:: -2I1yI2z
:width: 40%
:view: p170
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Challenge 11: Symmetry of antiphase operators under 180° rotations (A)
----------------------------------------------------------------------
Verify that the following symmetry relations of anti-phase operators
under 180° rotations are faithfully represented by the corresponding
antiphase droplets!
.. list-table::
:class: no-cap-num
:align: center
* - :pton:`2I1xI2z \xrightarrow{180°x} -2I1xI2z`
- :pton:`2I1xI2z \xrightarrow{180°y} 2I1xI2z`
- :pton:`2I1xI2z \xrightarrow{180°z} -2I1xI2z`
* - :pton:`2I1yI2z \xrightarrow{180°x} 2I1yI2z`
- :pton:`2I1yI2z \xrightarrow{180°y} -2I1yI2z`
- :pton:`2I1yI2z \xrightarrow{180°z} -2I1yI2z`
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Solution 11
===========
.. list-table::
:class: no-cap-num nocol
* - .. drops:: 2I1xI2z
:width: 40%
:view: p170
- :math:`180°_x \rightarrow`
- .. drops:: -2I1xI2z
:width: 40%
:view: p170
* - .. drops:: 2I1xI2z
:width: 40%
:view: p170
- :math:`180°_y \rightarrow`
- .. drops:: 2I1xI2z
:width: 40%
:view: p170
* - .. drops:: 2I1xI2z
:width: 40%
:view: p170
- :math:`180°_z \rightarrow`
- .. drops:: -2I1xI2z
:width: 40%
:view: p170
* - .. drops:: 2I1yI2z
:width: 40%
:view: p170
- :math:`180°_x \rightarrow`
- .. drops:: 2I1yI2z
:width: 40%
:view: p170
* - .. drops:: 2I1yI2z
:width: 40%
:view: p170
- :math:`180°_y \rightarrow`
- .. drops:: -2I1yI2z
:width: 40%
:view: p170
* - .. drops:: 2I1yI2z
:width: 40%
:view: p170
- :math:`180°_z \rightarrow`
- .. drops:: -2I1yI2z
:width: 40%
:view: p170
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Challenge 12: Comparing the speed of coherence transfer
-------------------------------------------------------
Compare the time required to transfer the initial state :pton:`I1x` to
:pton:`I2x` in a homonuclear two-spin system when using either a
sequence of delays and pulses (such as in a fully refocused
:term:`INEPT`) or an isotropic mixing sequence (:term:`TOCSY`) for a
given coupling constant J12. Which sequence is faster and by how much?
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Solution 12
===========
For J12 = 1 Hz (and v1 = v2 = 0 Hz), simulations are shown for the
sequence `1/(2J12) - 90°x - 1/(2J12)` (:numref:`challenges_12a`) and
for isotropic mixing (:numref:`challenges_12b`). The isotropic mixing
sequence only requires half the amount of time compared to the
INEPT-type transfer, i.e. it is twice as fast. The times shown in
these examples are not related exactly by a factor of two (1.025 s and
0.5 s) because the 1.025 s includes the 0.025 second 90°x pulse
duration, from a pulse amplitude of 10 Hz, but in a real experiment
would be extremely short due to much higher actual pulse amplitudes
(typically on the order of 1 kHz - 20 kHz).
.. drops:: I1x
:nspin: 3; Homonuclear=1; v1=0; v2=0; v3=0; J13=0; J23=0; J12=1
:aspect: 200
:sequence: { "Spin System": "Homonuclear",
"channels" : {"f1": ["Ixee", "Iyee", "Izee"]},
"name": "DelPulDel",
"desc": "Mag Transfer",
"table":[{"delay":{"duration":0.5,"unit":3}},
{"f1":{"angle":90.0,"phase":0.0}},
{"delay":{"duration":0.5,"unit":3}}] }
:view: Cp110
:class: border
:time: 1.025
:name: challenges_12a
:caption: On-resonance Magnetization Transfer Sequence
:link:
.. drops:: I1x
:nspin: 3; v1=0; v2=0; v3=0; J13=0; J23=0; J12=1
:aspect: 200
:sequence: Standard/IM12
:view: Cp110
:link:
:class: border
:time: 0.5
:name: challenges_12b
:caption: Isotropic Mixing Sequence
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Challenge 13: Seeing coherence orders
-------------------------------------
Which coherence orders :math:`p` are contained in the following
operators?
.. list-table::
:class: no-cap-num
* - a
.. drops:: 2I1xI2z
:width: 50%
:view: p170
- b
.. drops:: 2I1xI2x
:width: 50%
:view: p170
* - c
.. drops:: I1xI2x - I1yI2y
:width: 50%
:view: p170
- d
.. drops:: I1xI2x + I1yI2y
:width: 50%
:view: p170
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Solution 13
===========
Create the terms using the :menuselection:`Initial State --> Edit
Operator` menu. Choose :menuselection:`View --> Separation -->
Coh. Order p` to see the corresponding droplet terms separated based
on coherence order:
.. list-table::
:class: no-cap-num norow
* - a
- b
* - .. drops:: 2I1xI2z
:width: 50%
:view: p170
:prefs: Current Separation=3
- .. drops:: 2I1xI2x
:width: 50%
:view: p170
:prefs: Current Separation=3
* - :math:`p = +1,-1`
- :math:`p = +2,0,-2`
.. list-table::
:class: no-cap-num norow
* - c
- d
* - .. drops:: I1xI2x - I1yI2y
:width: 50%
:view: p170
:prefs: Current Separation=3
- .. drops:: I1xI2x + I1yI2y
:width: 50%
:view: p170
:prefs: Current Separation=3
* - :math:`p = +2,-2`
- :math:`p = 0`
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