Challenges

Challenge 1: Explore how to define desired states

Create the following droplets using the menu Initial State ‣ MQ(x/y ops.):

I1xI2x+I1yI2y
2*I1z*I2y
I1xI2y+I1yI2x
2*(I1xI2xI3x-I1xI2yI3y-I1yI2xI3y-I1yI2yI3x)
  1. ?
  2. ?
  3. ?
  4. ?

Scroll down for solution...

Solution 1

  1. MQ(x/y ops.) ‣ 0Q(I1,I2) ‣ 0Qx(I1,I2)

    I1xI2x+I1yI2y

    Fig. 205 \(I_{1x}I_{2x}+I_{1y}I_{2y}\)

  2. MQ(x/y ops.) ‣ 1Q(I1,I2) ‣ 2*I1z*I2y

    2*I1z*I2y

    Fig. 206 \(2 \cdot I_{1z} \cdot I_{2y}\)

  3. MQ(x/y ops.) ‣ 2Q(I1,I2) ‣ 2Qy(I1,I2)

    I1xI2y+I1yI2x

    Fig. 207 \(I_{1x}I_{2y}+I_{1y}I_{2x}\)

  4. MQ(x/y ops.) ‣ 3Q(I1,I2,I3) ‣ 2*3Qx(I1,I2,I3)

    2*(I1xI2xI3x-I1xI2yI3y-I1yI2xI3y-I1yI2yI3x)

    Fig. 208 \(2 \cdot (I_{1x}I_{2x}I_{3x}-I_{1x}I_{2y}I_{3y}-I_{1y}I_{2x}I_{3y}-I_{1y}I_{2y}I_{3x})\)


Challenge 2: Explore how to define desired states

Create the following droplets using the menu Initial State ‣ MQ(+/- ops.):

I1m
I1pI2p
2*I1mI2mI3m
  1. ?
  2. ?
  3. ?

Scroll down for solution...

Solution 2

  1. MQ(+/- ops.) ‣ 1Q(I1) ‣ I1(-) (turn off the preference Magnetization Vectors to get an opaque droplet).

    I1m

    Fig. 209 \(I_{1}^{-}\)

  2. MQ(+/- ops.) ‣ 2Q(I1,I2) ‣ I1(+)I2(+)

    I1pI2p

    Fig. 210 \(I_{1}^{+}I_{2}^{+}\)

  3. MQ(+/- ops.) ‣ 3Q(I1,I2,I3) ‣ 2*I1(-)*I2(-)*I3(-)

    2*I1mI2mI3m

    Fig. 211 \(2 \cdot I_{1}^{-}I_{2}^{-}I_{3}^{-}\)


Challenge 3: Explore how to define desired states

Create the following droplets using the menu Initial State ‣ MQ(+/- ops.):

I1pI2m
I1mI2p
2I1pI2z
  1. ?
  2. ?
  3. ?

Scroll down for solution...

Solution 3

  1. MQ(+/- ops.) ‣ 2Q(I1,I2) ‣ I1(+)I2(-)

    I1pI2m

    Fig. 212 \(I_{1}^{+}I_{2}^{-}\)

  2. MQ(+/- ops.) ‣ 2Q(I1,I2) ‣ I1(-)I2(+)

    I1mI2p

    Fig. 213 \(I_{1}^{-}I_{2}^{+}\)

  3. MQ(+/- ops.) ‣ 2Q(I1,I2) ‣ 2*I1(+)*I2z

    2I1pI2z

    Fig. 214 \(2I_{1}^{+}I_{2z}\)


Challenge 4: Explore how to interpret droplets

Using the menu 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 Cartesian product operators.

Scroll down for solution...

Solution 4

The double-quantum x operator involving spins I1 and I2 corresponds to the following droplet:

I1xI2x-I1yI2y

Fig. 215 \(\rho_0 = I_{1x}I_{2x}-I_{1y}I_{2y}\)

Select the menu 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 Fig. 215.

Hence, \(2Qx(I1,I2) = 0.5 (2I_{1x}I_{2x}) - 0.5 (2I_{1y}I_{2y}) = I_{1x}I_{2x} - I_{1y}I_{2y}\).


Challenge 5: Explore how to interpret droplets

Using the menu Initial State ‣ MQ(+/− ops.) ‣ 3Q(I1,I2,I3), create the operator \(2I_{1}^{+}I_{2}^{+}I_{3}^{+}\) and determine how it can be expressed as a linear combination of Cartesian product operators.

Scroll down for solution...

Solution 5

The +3-quantum operator \(2 \cdot I_{1}^{+}I_{2}^{+}I_{3}^{+}\) corresponds to the following droplet:

2*I1pI2pI3p

Fig. 216 \(\rho_0 = 2 \cdot I_{1}^{+}I_{2}^{+}I_{3}^{+}\)

Select the menu View ‣ List Prod. Ops…. The following terms will be displayed in the shorthand notation for Cartesian operators as shown in Fig. 216.

Hence,

\[\begin{split}2*I_1^+I_2^+I_3^+ & = 0.5 (4I_{1x}I_{2x}I_{3x}) \\ & + 0.5 i (4I_{1y}I_{2x}I_{3x}) \\ & + 0.5 i (4I_{1x}I_{2y}I_{3x}) \\ & - 0.5 (4I_{1y}I_{2y}I_{3x}) \\ & + 0.5 i (4I_{1x}I_{2x}I_{3y}) \\ & - 0.5 (4I_{1y}I_{2x}I_{3y}) \\ & - 0.5 (4I_{1x}I_{2y}I_{3y}) \\ & - 0.5 i (4I_{1y}I_{2y}I_{3y}) \\ \\ & = (2I_{1x}I_{2x}I_{3x} - 2I_{1x}I_{2y}I_{3y} - 2I_{1y}I_{2x}I_{3y} - 2I_{1y}I_{2y}I_{3x} \\ & - 2 i I_{1y}I_{2y}I_{3y} + 2 i I_{1y}I_{2x}I_{3y} - 2 i I_{1x}I_{2y}I_{3x} + 2 i I_{1x}I_{2x}I_{3y}\end{split}\]


Challenge 6: Pulse sequence design challenge

Suppose in the course of a pulse sequence, you have created the state \(I_{1z}I_{2z} - I_{1y}I_{2y}\). How can you transfer this state into ±2 quantum coherence by a single pulse?

Scroll down for solution...

Solution 6

Using the menu Initial State ‣ Edit Operator, create the operator \(I_{1z}I_{2z} - I_{1y}I_{2y}\). The corresponding DROPS representation should look like:

I1zI2z-I1yI2y

Fig. 217 \(I_{1z}I_{2z}-I_{1y}I_{2y}\)

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 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.)

I1zI2z-I1yI2y

Fig. 218 \(\rho_0 = I_{1z}I_{2z}-I_{1y}I_{2y}\)


Challenge 7: Define a new operator

Display the droplet corresponding to the operator \(2I_{1x}I_{2x} + i 2I_{1z}I_{2x}\). (When the droplet is oriented properly, you will see the SpinDrops logo.)

Scroll down for solution...

Solution 7

Using the menu Initial State ‣ Edit Operator, create the operator \(2I_{1x}I_{2x} + i 2I_{1z}I_{2x}\). Orienting the resulting droplet in the Drops display such that the view direction is along the y axis yields the SpinDrops logo :-)

2I1xI2x+i*2I1zI2x


Challenge 8: Rotations of operators and droplets

Starting from the term \(2I_{1x}I_{2y}\), 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 \(2I1_aI2_b\) with \(a≠b\) and \(a,b ∈ {x,y,z}\).

Scroll down for solution...

Solution 8

2I1xI2y \(90°_x \rightarrow\) 2I1xI2z \(90°_z \rightarrow\) 2I1yI2z
\(90°_y \uparrow\)       \(90°_y \downarrow\)
2I1zI2y \(\leftarrow 90°_x\) 2I1zI2x \(\leftarrow 90°_x\) 2I1yI2x


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 (\(±2I_{1z}I_{2a}\) with \(a ∈ \{x,y\}\) ) or with respect to the second spin (\(±2I_{1a}I_{2z}\) with \(a ∈ \{x,y\}\) )!

2I1xI2z 2I1zI2x -2I1zI2y
2I1yI2z 2I1zI2y -2I1yI2z

Scroll down for solution...

Solution 9

This challenge can be solved by simply observing the “kissing beans’ head tilt” (c.f. Head-tilt Rule).

right tilt

2I1xI2z

Fig. 219 \(±2I_{1a}I_{2z}\)

left tilt

2I1zI2x

Fig. 220 \(±2I_{1z}I_{2a}\)

left tilt

-2I1zI2y

Fig. 221 \(±2I_{1z}I_{2a}\)

right tilt

2I1yI2z

Fig. 222 \(±2I_{1a}I_{2z}\)

left tilt

2I1zI2y

Fig. 223 \(±2I_{1z}I_{2a}\)

right tilt

-2I1yI2z

Fig. 224 \(±2I_{1a}I_{2z}\)


Challenge 10: Recognizing droplets of antiphase operators (B)

Determine for each of the droplets the exact form of the corresponding antiphase operator!

2I1xI2z 2I1zI2x -2I1zI2y
2I1yI2z 2I1zI2y -2I1yI2z

Scroll down for solution...

Solution 10

2I1xI2z

Fig. 225 \(2I_{1x}I_{2z}\)

2I1zI2x

Fig. 226 \(2I_{1z}I_{2x}\)

-2I1zI2y

Fig. 227 \(-2I_{1z}I_{2y}\)

2I1yI2z

Fig. 228 \(2I_{1y}I_{2z}\)

2I1zI2y

Fig. 229 \(2I_{1z}I_{2y}\)

-2I1yI2z

Fig. 230 \(-2I_{1y}I_{2z}\)


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!

\(2I_{1x}I_{2z} \xrightarrow{180°x} -2I_{1x}I_{2z}\) \(2I_{1x}I_{2z} \xrightarrow{180°y} 2I_{1x}I_{2z}\) \(2I_{1x}I_{2z} \xrightarrow{180°z} -2I_{1x}I_{2z}\)
\(2I_{1y}I_{2z} \xrightarrow{180°x} 2I_{1y}I_{2z}\) \(2I_{1y}I_{2z} \xrightarrow{180°y} -2I_{1y}I_{2z}\) \(2I_{1y}I_{2z} \xrightarrow{180°z} -2I_{1y}I_{2z}\)

Scroll down for solution...

Solution 11

2I1xI2z

Fig. 231 \(2I_{1x}I_{2z}\)

\(180°_x \rightarrow\)
-2I1xI2z

Fig. 232 \(-2I_{1x}I_{2z}\)

2I1xI2z

Fig. 233 \(2I_{1x}I_{2z}\)

\(180°_y \rightarrow\)
2I1xI2z

Fig. 234 \(2I_{1x}I_{2z}\)

2I1xI2z

Fig. 235 \(2I_{1x}I_{2z}\)

\(180°_z \rightarrow\)
-2I1xI2z

Fig. 236 \(-2I_{1x}I_{2z}\)

2I1yI2z

Fig. 237 \(2I_{1y}I_{2z}\)

\(180°_x \rightarrow\)
2I1yI2z

Fig. 238 \(2I_{1y}I_{2z}\)

2I1yI2z

Fig. 239 \(2I_{1y}I_{2z}\)

\(180°_y \rightarrow\)
-2I1yI2z

Fig. 240 \(-2I_{1y}I_{2z}\)

2I1yI2z

Fig. 241 \(2I_{1y}I_{2z}\)

\(180°_z \rightarrow\)
-2I1yI2z

Fig. 242 \(-2I_{1y}I_{2z}\)


Challenge 12: Comparing the speed of coherence transfer

Compare the time required to transfer the initial state \(I_{1x}\) to \(I_{2x}\) in a homonuclear two-spin system when using either a sequence of delays and pulses (such as in a fully refocused INEPT) or an isotropic mixing sequence (TOCSY) for a given coupling constant J12. Which sequence is faster and by how much?

Scroll down for solution...

Solution 12

For J12 = 1 Hz (and v1 = v2 = 0 Hz), simulations are shown for the sequence 1/(2J12) - 90°x - 1/(2J12) (Fig. 243) and for isotropic mixing (Fig. 244). 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).

I1x

Fig. 243 On-resonance Magnetization Transfer Sequence

I1x

Fig. 244 Isotropic Mixing Sequence


Challenge 13: Seeing coherence orders

Which coherence orders \(p\) are contained in the following operators?

a

2I1xI2z

Fig. 245 \(2I_{1x}I_{2z}\)

b

2I1xI2x

Fig. 246 \(2I_{1x}I_{2x}\)

c

I1xI2x-I1yI2y

Fig. 247 \(I_{1x}I_{2x}-I_{1y}I_{2y}\)

d

I1xI2x+I1yI2y

Fig. 248 \(I_{1x}I_{2x}+I_{1y}I_{2y}\)

Scroll down for solution...

Solution 13

Create the terms using the Initial State ‣ Edit Operator menu. Choose View ‣ Separation ‣ Coh. Order p to see the corresponding droplet terms separated based on coherence order:

a b
2I1xI2z

Fig. 249 \(2I_{1x}I_{2z}\)

2I1xI2x

Fig. 250 \(2I_{1x}I_{2x}\)

\(p = +1,-1\) \(p = +2,0,-2\)
c d
I1xI2x-I1yI2y

Fig. 251 \(I_{1x}I_{2x}-I_{1y}I_{2y}\)

I1xI2x+I1yI2y

Fig. 252 \(I_{1x}I_{2x}+I_{1y}I_{2y}\)

\(p = +2,-2\) \(p = 0\)