INADEQUATE scraps

The INADEQUATE Analysis PhL 1.25.08 Part I. Effect of page 405 lower pulse sequence on two spins 1. First pulse. We start with Iz and get quickly to +Iy = I1y+I2y with the first (/2)-x pulse. 2. The free run. If we use the lower picture on page 405, the free run uses U = exp(-iJ2I1zI2zt) because this is what is meant by "only J-couplings". We go at once to sandwich rules (6) which says: exp(- i [2jzJz]) Jy exp(+ i [2jzJz]) = Jy cos [2jzJx ] sin() j,k,l anti-cyclic and set = J. So let's apply this like so: exp(- i [2I1zI2z]) I1y exp(+ i [2I1zI2z]) = I1y cos [2I1x I2z] sin() However, we are supposed to assume that we have the "official" SES value 2J = 1, meaning that = J = /2 so only the second term survives. We then end up with = 2I1xI2z (12) 3. The second pulse. Our usual rules are Rx pulse are and we want /2 so exp(- i Jx) Jy exp(+ i Jx) = Jy cos + Jz sin = + Jz exp(- i Jx) Jz exp(+ i Jx) = Jz cos Jy sin = Jy so we have = 2I1xI2z + (12) + 2I1xI2y + (12) and we comment that this includes orders 2 and 0. 4. The third pulse. I already verified these so I won't do them again. The main conclusion is this: you end up with, averaged over the cycle, a- = +1 - = 01 + J a- = -1 - = 01 - J and the same for the two which have the - first, but they are at 01. Thus, we get in the single spectrum two pairs of anti-phase lines, and we have blocked the string peaks from the single-C13 molecules. Part II. Verify the coherence flow diagram shown on page 410. 1. First pulse. We start with Iz and get quickly to +Iy = I1y I2y with the first (/2)x pulse. Since these terms contain both I+ and I-, we draw the diagram going to both the -1 and +1 levels. That is, we can change order in either direction. We have both - and + non-vanishing matrix elements of . 2. The first free-run. We know that each coherence just picks up phase, so the lines don't change. So ignoring these phases, but in general we end up with messy stuff as we saw in COSY when you run this full free propagator. After this free run, I no longer know the matrices without actually doing them. But, imagine that you DID know the matrices, and that you wrote them out in full detail. You could have terms of the following kind (all of which have order 1 or you might say 1). = aI1+ + bI1- + cI2+ + dI2- 3. The ()x pulse out in the middle. Our usual rules are Rx pulse are and we want = so exp(- i Jx) J exp(+ i Jx) = exp(- i Jx) [ Jx iJy] exp(+ i Jx) = Jx i exp(- i Jx) Jy exp(+ i Jx) =Jx i [ Jy cos + Jz sin] = Jx∓ i Jy = J∓ = a[I1+] + b[I1-] + c[I2+ ] + d[ I2-] a[I1-] + b[I1+] + c[I2- ] + d[ I2+] This says that we have a complete + interchange through this pulse, and now for the very first time I understand why he draws what he does on 410, the coherence flows swap! 4. The second free run. Coherences just phase, so diagram shows nothing. At the end of this time period we still have something of the form = aI1+ + bI1- + cI2+ + dI2- but these are not the same as before because more phases were picked up. To do this right, need to add the projectors, but I don't need to do that here. 5. The third pulse which is again (/2)x. Let's now run our coherence stream into the next (/2)x pulse, starting again with this -- but let's add some extra stuff = aI1+ + bI1- + cI2+ + dI2- + eIz + This time we have = /2 so get exp(- i Jx) J exp(+ i Jx) = Jx i [ Jy cos + Jz sin] = Jx iJz = (J+ + J-)/2 iJz How does this give only order 2 Well, the terms I show above seem to "leak through", giving 1 on the RHS. So I think one has to back up and look more carefully at the first free run. 6. First free run revisited. We come into this thing with = I1y I2y. I know from my COSY notes that the free run is going to result in this where = J and 1= 01: = I1yCC1 + 2I1xI2z SC1 + I1x C S1 + 2I1yI2z S S1 + (12) Not sure whether to use = J = /2. If we do, then we get this since C = 0 S = 1 = + 2I1xI2z C1 + 2I1yI2z S1 + (12) 7. The pulse revisited. Do this now with = , so exp(- i Jx) Jy exp(+ i Jx) = Jy cos + Jz sin = Jy exp(- i Jx) Jz exp(+ i Jx) = Jz cos Jy sin = Jz and we get = I1yCC1 + 2I1xI2z SC1 + I1x C S1 + 2I1yI2z S S1 + (12) + I1yCC1 2I1xI2z SC1 + I1x C S1 + 2I1yI2z S S1 + (12) and in our special case we get 2I1xI2z C1+ 2I1yI2z S1 + (12) Part III. Start over and use the SES rules. 1. First (/2)x pulse. After the first pulse we know we have = I1y I2y 2. The SES. The next two free runs and pulse form a SES. But top page 382 says we can replace this as shown where the free run has only J. The table middle of page 383 gives the total effect of the "tuned" SES, and it is easy to apply it to the above. We get = I1y I2y 2I1xI2z 2I1zI2x So this then is what enters the last /2 pulse shown on page 410. Our rules are (with = /2) exp(- i Jx) Jy exp(+ i Jx) = Jy cos + Jz sin = + Jz exp(- i Jx) Jz exp(+ i Jx) = Jz cos Jy sin = Jy So we get = 2I1xI2z 2I1zI2x + 2I1xI2y + 2I1yI2x This agrees with page 411 where I have circled ? mark. Am I missing something? Let's write this guys out: = + 2I1xI2y + 2I1yI2x = + 2 (I1+ + I1-)/2(I2+ I2-)/(2i) + (12) = (1/2i) { I1+I2+ +I1-I2+ I1+I2- I1-I2- + I1+I2+ +I2-I1+ I2+I1- I1-I2- } = (1/i) { I1+I2+ I1-I2- } so mystery resolved! Well, there it is now as 13.6, I missed it! So NOW I see that what emerges from the SES is only order = 2, just as he says. 3. The final /2 pulse. This is the one that gets cycled. Start with I1+I2+ I1-I2- without the -i. (/2)x: Start with these rules: exp(- i Jx) J exp(+ i Jx) = Jx i [ Jy cos + Jz sin] = Jx iJz I1+I2+ I1-I2- ( J1x + iJ1z)( J2x + iJ2z) ( J1x iJ1z)( J2x iJ2z) = (x+y)(a+b) (x-y)(a-b) = ( xa + ya + xb + yb) - ( xa - ya - xb + yb) = xa + ya + xb + yb - xa + ya + xb - yb = + ya + xb + + ya + xb = 2(ya + xb) I1+I2+ I1-I2- 2( iJ1z J2x + J1x iJ2z) = 2i (J1z J2x + J1x J2z ) = order 1 I think this was the hard way. Use instead = + 2I1xI2y + 2I1yI2x Then apply (/2)x just in your head to get = 2I1xI2y + 2I1yI2x 2I1xI2z + 2I1zI2x Then apply ()x just in your head to get = 2I1xI2y + 2I1yI2x - 2I1xI2y - 2I1yI2x Then apply (-/2)x just in your head to get = 2I1xI2y + 2I1yI2x - 2I1xI2z - 2I1zI2x Then apply (0)x to get = 2I1xI2y + 2I1yI2x 2I1xI2y + 2I1yI2x Now what happens if I just add all these second lines? 2I1xI2z + 2I1zI2x - 2I1xI2y - 2I1yI2x - 2I1xI2z - 2I1zI2x + 2I1xI2y + 2I1yI2x All terms cancel and the result is 0. BUT I did this all wrong. Start over = + 2I1xI2y + 2I1yI2x Apply (/2)y = 2I1xI2y + 2I1yI2x - 2I1zI2y - 2I1yI2z Then apply (/2)-x = 2I1xI2y + 2I1yI2x 2I1xI2z 2I1zI2x Then apply (/2)-y = 2I1xI2y + 2I1yI2x 2I1zI2y + 2I1yI2z Then apply (/2)x = 2I1xI2y + 2I1yI2x 2I1xI2z + 2I1zI2x Now add up all the results: - 2I1zI2y - 2I1yI2z 2I1xI2z 2I1zI2x + 2I1zI2y + 2I1yI2z + 2I1xI2z + 2I1zI2x And try to cancel terms. Again, they all cancel. So, it seems that phase cycling the last /2 pulse in this way blocks the specific combination coming in which was + 2I1xI2y + 2I1yI2x = (1/i) { I1+I2+ I1-I2- } So in this cycling experiment with an AX system, nothing gets through! BUT I forgot the effect of the digital phase cycling at the same time. So then we really get this: (- 2I1zI2y - 2I1yI2z)1 + ( 2I1xI2z 2I1zI2x ) (+i) + (2I1zI2y + 2I1yI2z )(-1) + (2I1xI2z + 2I1zI2x)(-i) = - 4I1zI2y - 4I1yI2z - i 4I1xI2z - i 4 I1zI2x = 4 ( - I1zI2y - i I1zI2x I1yI2z - i I1xI2z) = 4 { I1z (-I2y - i I2x) + I2z (-iI1x - I1y) } = (4i) { I1z (-iI2y + I2x) + I2z (I1x -i I1y) } = 2i { 2 I1z I2- + 2 I2z I1-} And now divide by 4 because we are averaging each cycle position, = (i/2) {2 I1z I2- + 2I2z I1-} So I would say this was the "effective" density matrix at the output with full cycling. This finally is what I want to be seeing. This means all the "energy" has been funneled into the - coherences. Now we can go to the actual coherences themselves: 2I1z I2- = 2 I2 I1z (I1 + I1) = I2 (I1 - I1) = I2 I1 I2 I1 = I1 I2 I1 I2 So our complete answer is this (before final FID) = (i/2) { I1 I2 I1 I2 + (12) } This is still in matrix form, but we can quickly get the matrix elements as follows (at FID start) - = i/2 a = 2i = -1 - = -i/2 a = -1 when phase considered, etc etc - = i/2 -= -i/2 No recall that the first and third are very close and separated by 2J so we get four peaks and they are in doublets and the picture shown is pretty good on page 410. Now look at the flow picture on page 410. Anything that comes in on the +2 or -2 lines is going to get funneled by the cycling into the -1 line! Finally that drawing is making sense.