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. ?

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Solution 1

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

    I1xI2x+I1yI2y

    Fig. 214 I1xI2x+I1yI2y

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

    2*I1z*I2y

    Fig. 215 2I1zI2y

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

    I1xI2y+I1yI2x

    Fig. 216 I1xI2y+I1yI2x

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

    2*(I1xI2xI3x-I1xI2yI3y-I1yI2xI3y-I1yI2yI3x)

    Fig. 217 2(I1xI2xI3xI1xI2yI3yI1yI2xI3yI1yI2yI3x)


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. ?

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Solution 2

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

    I1m

    Fig. 218 I1

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

    I1pI2p

    Fig. 219 I+1I+2

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

    2*I1mI2mI3m

    Fig. 220 2I1I2I3


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. ?

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Solution 3

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

    I1pI2m

    Fig. 221 I+1I2

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

    I1mI2p

    Fig. 222 I1I+2

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

    2I1pI2z

    Fig. 223 2I+1I2z


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.

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Solution 4

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

I1xI2x-I1yI2y

Fig. 224 ρ0=I1xI2xI1yI2y

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. 224.

Hence, 2Qx(I1,I2)=0.5(2I1xI2x)0.5(2I1yI2y)=I1xI2xI1yI2y.


Challenge 5: Explore how to interpret droplets

Using the menu Initial State ‣ MQ(+/− ops.) ‣ 3Q(I1,I2,I3), create the operator 2I+1I+2I+3 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 2I+1I+2I+3 corresponds to the following droplet:

2*I1pI2pI3p

Fig. 225 ρ0=2I+1I+2I+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. 225.

Hence,

2I+1I+2I+3=0.5(4I1xI2xI3x)+0.5i(4I1yI2xI3x)+0.5i(4I1xI2yI3x)0.5(4I1yI2yI3x)+0.5i(4I1xI2xI3y)0.5(4I1yI2xI3y)0.5(4I1xI2yI3y)0.5i(4I1yI2yI3y)=(2I1xI2xI3x2I1xI2yI3y2I1yI2xI3y2I1yI2yI3x2iI1yI2yI3y+2iI1yI2xI3y2iI1xI2yI3x+2iI1xI2xI3y


Challenge 6: Pulse sequence design challenge

Suppose in the course of a pulse sequence, you have created the state I1zI2zI1yI2y. How can you transfer this state into ±2 quantum coherence by a single pulse?

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Solution 6

Using the menu Initial State ‣ Edit Operator, create the operator I1zI2zI1yI2y. The corresponding DROPS representation should look like:

I1zI2z-I1yI2y

Fig. 226 I1zI2zI1yI2y

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. 227 ρ0=I1zI2zI1yI2y


Challenge 7: Define a new operator

Display the droplet corresponding to the operator 2I1xI2x+i2I1zI2x. (When the droplet is oriented properly, you will see the SpinDrops logo.)

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Solution 7

Using the menu Initial State ‣ Edit Operator, create the operator 2I1xI2x+i2I1zI2x. 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 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 2I1aI2b with ab and a,bx,y,z.

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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

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Solution 9

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

right tilt

2I1xI2z

Fig. 228 ±2I_{1a}I_{2z}

left tilt

2I1zI2x

Fig. 229 ±2I_{1z}I_{2a}

left tilt

-2I1zI2y

Fig. 230 ±2I_{1z}I_{2a}

right tilt

2I1yI2z

Fig. 231 ±2I_{1a}I_{2z}

left tilt

2I1zI2y

Fig. 232 ±2I_{1z}I_{2a}

right tilt

-2I1yI2z

Fig. 233 ±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

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Solution 10

2I1xI2z

Fig. 234 2I_{1x}I_{2z}

2I1zI2x

Fig. 235 2I_{1z}I_{2x}

-2I1zI2y

Fig. 236 -2I_{1z}I_{2y}

2I1yI2z

Fig. 237 2I_{1y}I_{2z}

2I1zI2y

Fig. 238 2I_{1z}I_{2y}

-2I1yI2z

Fig. 239 -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}

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Solution 11

2I1xI2z

Fig. 240 2I_{1x}I_{2z}

180°_x \rightarrow

-2I1xI2z

Fig. 241 -2I_{1x}I_{2z}

2I1xI2z

Fig. 242 2I_{1x}I_{2z}

180°_y \rightarrow

2I1xI2z

Fig. 243 2I_{1x}I_{2z}

2I1xI2z

Fig. 244 2I_{1x}I_{2z}

180°_z \rightarrow

-2I1xI2z

Fig. 245 -2I_{1x}I_{2z}

2I1yI2z

Fig. 246 2I_{1y}I_{2z}

180°_x \rightarrow

2I1yI2z

Fig. 247 2I_{1y}I_{2z}

2I1yI2z

Fig. 248 2I_{1y}I_{2z}

180°_y \rightarrow

-2I1yI2z

Fig. 249 -2I_{1y}I_{2z}

2I1yI2z

Fig. 250 2I_{1y}I_{2z}

180°_z \rightarrow

-2I1yI2z

Fig. 251 -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?

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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) (Fig. 252) and for isotropic mixing (Fig. 253). 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. 252 On-resonance Magnetization Transfer Sequence

I1x

Fig. 253 Isotropic Mixing Sequence


Challenge 13: Seeing coherence orders

Which coherence orders p are contained in the following operators?

a

2I1xI2z

Fig. 254 2I_{1x}I_{2z}

b

2I1xI2x

Fig. 255 2I_{1x}I_{2x}

c

I1xI2x-I1yI2y

Fig. 256 I_{1x}I_{2x}-I_{1y}I_{2y}

d

I1xI2x+I1yI2y

Fig. 257 I_{1x}I_{2x}+I_{1y}I_{2y}

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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. 258 2I_{1x}I_{2z}

2I1xI2x

Fig. 259 2I_{1x}I_{2x}

p = +1,-1

p = +2,0,-2

c

d

I1xI2x-I1yI2y

Fig. 260 I_{1x}I_{2x}-I_{1y}I_{2y}

I1xI2x+I1yI2y

Fig. 261 I_{1x}I_{2x}+I_{1y}I_{2y}

p = +2,-2

p = 0