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Symmetry operator 2 -Symmetry operator 4

Symmetry operator 2 -Symmetry operator 4. - x - y ½+z - ( ½+y ½-x ¼+z) -½-x-y -½+x-y ¼. Symmetry operator 2 -Symmetry operator 4. - x - y ½+z - ( ½+y ½-x ¼+z) -½-x-y -½+x-y ¼. Plug in u. u=-½-x-y

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Symmetry operator 2 -Symmetry operator 4

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  1. Symmetry operator 2-Symmetry operator 4 -x - y ½+z - ( ½+y ½-x ¼+z) -½-x-y -½+x-y ¼

  2. Symmetry operator 2-Symmetry operator 4 -x - y ½+z - ( ½+y ½-x ¼+z) -½-x-y -½+x-y ¼ Plug in u. u=-½-x-y 0.18=-½-x-y 0.68=-x-y Plug in v. v=-½+x-y 0.22=-½+x-y 0.72=x-y Add two equations and solve for y. 0.68=-x-y +(0.72= x-y) 1.40=-2y -0.70=y Plug y into first equation and solve for x. 0.68=-x-y 0.68=-x-(-0.70) 0.02=x

  3. Symmetry operator 2-Symmetry operator 4 -x - y ½+z - ( ½+y ½-x ¼+z) -½-x-y -½+x-y ¼ Plug in u. u=-½-x-y 0.18=-½-x-y 0.68=-x-y Plug in v. v=-½+x-y 0.22=-½+x-y 0.72=x-y Add two equations and solve for y. 0.68=-x-y +(0.72= x-y) 1.40=-2y -0.70=y Plug y into first equation and solve for x. 0.68=-x-y 0.68=-x-(-0.70) 0.02=x

  4. Symmetry operator 2-Symmetry operator 4 -x - y ½+z - ( ½+y ½-x ¼+z) -½-x-y -½+x-y ¼ Plug in u. u=-½-x-y 0.18=-½-x-y 0.68=-x-y Plug in v. v=-½+x-y 0.22=-½+x-y 0.72=x-y Add two equations and solve for y. 0.68=-x-y +(0.72= x-y) 1.40=-2y -0.70=y Plug y into first equation and solve for x. 0.68=-x-y 0.68=-x-(-0.70) 0.02=x

  5. Symmetry operator 2-Symmetry operator 4 -x - y ½+z - ( ½+y ½-x ¼+z) -½-x-y -½+x-y ¼ Plug in u. u=-½-x-y 0.18=-½-x-y 0.68=-x-y Plug in v. v=-½+x-y 0.22=-½+x-y 0.72=x-y Add two equations and solve for y. 0.68=-x-y +(0.72= x-y) 1.40=-2y -0.70=y Plug y into first equation and solve for x. 0.68=-x-y 0.68=-x-(-0.70) 0.02=x

  6. Symmetry operator 3 -Symmetry operator 6 ½-y ½+x ¾+z - ( -y -x ½-z) ½ ½+2x ¼+2z Plug in v. v= ½+2x 0.48= ½+2x -0.02=2x -0.01=x Plug in w. w= ¼+2z 0.24= ¼+2z -0.01=2z -0.005=z

  7. Symmetry operator 3 -Symmetry operator 6 ½-y ½+x ¾+z - ( -y -x ½-z) ½ ½+2x ¼+2z Plug in v. v= ½+2x 0.46= ½+2x -0.04=2x -0.02=x Plug in w. w= ¼+2z 0.24= ¼+2z -0.01=2z -0.005=z

  8. Symmetry operator 3 -Symmetry operator 6 ½-y ½+x ¾+z - ( -y -x ½-z) ½ ½+2x ¼+2z Plug in v. v= ½+2x 0.46= ½+2x -0.04=2x -0.02=x Plug in w. w= ¼+2z 0.24= ¼+2z -0.01=2z -0.005=z

  9. From step 3 Xstep3= 0.02 ystep3=-0.70 zstep3=?.??? From step 4 Xstep4=-0.02 ystep4= ?.?? zstep4=-0.005 Clearly, Xstep3does not equalXstep4 . Use a Cheshire symmetry operator that transforms xstep3= 0.02 into xstep4=- 0.02. For example, let’s use: -x, -y, z And apply it to all coordinates in step 3. xstep3-transformed = - (+0.02) = -0.02 ystep3-transformed = - (- 0.70)= +0.70 Now xstep3-transformed = xstep4 And ystep3 has been transformed to a reference frame consistent with x and z from step 4. So we arrive at the following self-consistent x,y,z: Xstep4=-0.02, ystep3-transformed=0.70, zstep4=-0.005 Or simply, x=-0.02, y=0.70, z=-0.005 The x, y coordinate in step 3 describes one of the heavy atom positions in the unit cell. The x, z coordinate in step 4 describes a symmetry related copy. We can’t combine these coordinates directly. They don’t describe the same atom. Perhaps they even referred to different origins. How can we transform x, y from step 3 so it describes the same atom as x and z in step 4?

  10. From step 3 Xstep3= 0.02 ystep3=-0.70 zstep3=?.??? Cheshire Symmetry Operators for space group P43212 X, Y, Z -X, -Y, Z -Y, X, 1/4+Z Y, -X, 1/4+Z Y, X, -Z -Y, -X, -Z X, -Y, 1/4-Z -X, Y, 1/4-Z 1/2+X, 1/2+Y, Z 1/2-X, 1/2-Y, Z 1/2-Y, 1/2+X, 1/4+Z 1/2+Y, 1/2-X, 1/4+Z 1/2+Y, 1/2+X, -Z 1/2-Y, 1/2-X, -Z 1/2+X, 1/2-Y, 1/4-Z 1/2-X, 1/2+Y, 1/4-Z X, Y, 1/2+Z -X, -Y, 1/2+Z -Y, X, 3/4+Z Y, -X, 3/4+Z Y, X, 1/2-Z -Y, -X, 1/2-Z X, -Y, 3/4-Z -X, Y, 3/4-Z 1/2+X, 1/2+Y, 1/2+Z 1/2-X, 1/2-Y, 1/2+Z 1/2-Y, 1/2+X, 3/4+Z 1/2+Y, 1/2-X, 3/4+Z 1/2+Y, 1/2+X, 1/2-Z 1/2-Y, 1/2-X, 1/2-Z 1/2+X, 1/2-Y, 3/4-Z 1/2-X, 1/2+Y, 3/4-Z From step 4 Xstep4=-0.02 ystep4= ?.?? zstep4=-0.005 Clearly, Xstep3does not equalXstep4 . Use a Cheshire symmetry operator that transforms xstep3= 0.02 into xstep4=- 0.02. For example, let’s use: -x, -y, z And apply it to all coordinates in step 3. xstep3-transformed = - (+0.02) = -0.02 ystep3-transformed = - (- 0.70)= +0.70 Now xstep3-transformed = xstep4 And ystep3 has been transformed to a reference frame consistent with x and z from step 4. So we arrive at the following self-consistent x,y,z: Xstep4=-0.02, ystep3-transformed=0.70, zstep4=-0.005 Or simply, x=-0.02, y=0.70, z=-0.005

  11. From step 3 Xstep3= 0.02 ystep3=-0.70 zstep3=?.??? Cheshire Symmetry Operators for space group P43212 X, Y, Z -X, -Y, Z -Y, X, 1/4+Z Y, -X, 1/4+Z Y, X, -Z -Y, -X, -Z X, -Y, 1/4-Z -X, Y, 1/4-Z 1/2+X, 1/2+Y, Z 1/2-X, 1/2-Y, Z 1/2-Y, 1/2+X, 1/4+Z 1/2+Y, 1/2-X, 1/4+Z 1/2+Y, 1/2+X, -Z 1/2-Y, 1/2-X, -Z 1/2+X, 1/2-Y, 1/4-Z 1/2-X, 1/2+Y, 1/4-Z X, Y, 1/2+Z -X, -Y, 1/2+Z -Y, X, 3/4+Z Y, -X, 3/4+Z Y, X, 1/2-Z -Y, -X, 1/2-Z X, -Y, 3/4-Z -X, Y, 3/4-Z 1/2+X, 1/2+Y, 1/2+Z 1/2-X, 1/2-Y, 1/2+Z 1/2-Y, 1/2+X, 3/4+Z 1/2+Y, 1/2-X, 3/4+Z 1/2+Y, 1/2+X, 1/2-Z 1/2-Y, 1/2-X, 1/2-Z 1/2+X, 1/2-Y, 3/4-Z 1/2-X, 1/2+Y, 3/4-Z From step 4 Xstep4=-0.02 ystep4= ?.?? zstep4=-0.005 Clearly, Xstep3does not equalXstep4 . Use a Cheshire symmetry operator that transforms xstep3= 0.02 into xstep4=- 0.02. For example, let’s use: -x, -y, z And apply it to all coordinates in step 3. xstep3-transformed = - (+0.02) = -0.02 ystep3-transformed = - (- 0.70)= +0.70 Now xstep3-transformed = xstep4 And ystep3 has been transformed to a reference frame consistent with x and z from step 4. So we arrive at the following self-consistent x,y,z: Xstep4=-0.02, ystep3-transformed=0.70, zstep4=-0.005 Or simply, x=-0.02, y=0.70, z=-0.005

  12. From step 3 Xstep3= 0.02 ystep3=-0.70 zstep3=?.??? Cheshire Symmetry Operators for space group P43212 X, Y, Z -X, -Y, Z -Y, X, 1/4+Z Y, -X, 1/4+Z Y, X, -Z -Y, -X, -Z X, -Y, 1/4-Z -X, Y, 1/4-Z 1/2+X, 1/2+Y, Z 1/2-X, 1/2-Y, Z 1/2-Y, 1/2+X, 1/4+Z 1/2+Y, 1/2-X, 1/4+Z 1/2+Y, 1/2+X, -Z 1/2-Y, 1/2-X, -Z 1/2+X, 1/2-Y, 1/4-Z 1/2-X, 1/2+Y, 1/4-Z X, Y, 1/2+Z -X, -Y, 1/2+Z -Y, X, 3/4+Z Y, -X, 3/4+Z Y, X, 1/2-Z -Y, -X, 1/2-Z X, -Y, 3/4-Z -X, Y, 3/4-Z 1/2+X, 1/2+Y, 1/2+Z 1/2-X, 1/2-Y, 1/2+Z 1/2-Y, 1/2+X, 3/4+Z 1/2+Y, 1/2-X, 3/4+Z 1/2+Y, 1/2+X, 1/2-Z 1/2-Y, 1/2-X, 1/2-Z 1/2+X, 1/2-Y, 3/4-Z 1/2-X, 1/2+Y, 3/4-Z From step 4 Xstep4=-0.02 ystep4= ?.?? zstep4=-0.005 Clearly, Xstep3does not equalXstep4 . Use a Cheshire symmetry operator that transforms xstep3= 0.02 into xstep4=- 0.02. For example, let’s use: -x, -y, z And apply it to all coordinates in step 3. xstep3-transformed = - (+0.02) = -0.02 ystep3-transformed = - (- 0.70)= +0.70 Now xstep3-transformed = xstep4 And ystep3 has been transformed to a reference frame consistent with x and z from step 4. So we arrive at the following self-consistent x,y,z: Xstep4=-0.02, ystep3-transformed=0.70, zstep4=-0.005 Or simply, x=-0.02, y=0.70, z=-0.005

  13. Use x,y,z to predict the position of a non-Harker Patterson peak • x,y,z vs. –x,y,z ambiguity remains In other words x=-0.02, y=0.70, z=-0.005 or x=+0.02, y=0.70, z=-0.005 could be correct. • Both satisfy the difference vector equations for Harker sections • Only one is correct. 50/50 chance • Predict the position of a non Harker peak. • Use symop1-symop5 • Plug in x,y,z solve for u,v,w. • Plug in –x,y,z solve for u,v,w • I have a non-Harker peak at u=0.28 v=0.28, w=0.0 • The position of the non-Harker peak will be predicted by the correct heavy atom coordinate.

  14. x y z -( y x -z) x-y -x+y 2z symmetry operator 1 -symmetry operator 5 u v w First, plug in x=-0.02, y=0.70, z=-0.005 u=x-y = -0.02-0.70 =-0.72 v=-x+y= +0.02+0.70= 0.72 w=2z=2*(-0.005)=-0.01 The numerical value of these co-ordinates falls outside the section we have drawn. Lets transform this uvw by Patterson symmetry v,-u,-w. -0.72,0.72,-0.01 becomes -0.72,-0.72,0.01 then add 1 to u and v 0.28, 0.28, 0.01 This corresponds to the peak shown u=0.28, v=0.28, w=0.01 Thus, x=-0.02, y=0.70, z=-0.005 is correct. Hurray! We are finished! In the case that the above test failed, we would change the sign of x.

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