scratch4

en = |en> = = "vector" enT = <en| = ( (en)1, (en)2, (en)3) = "transpose vector" **************** = RniRnj = (RTR)ij = δij Here is how orthogonality (1.1.1) and completeness (1.1.9) appear in Dirac notation <en|em> = δnm orthogonality 1 = |ei><ei| completeness The second notation may seem odd, but consider Notice how the dot product (scalar product, inner product) works in the following examples, where on the left we use expressions from (1.1.8), δn,m = (em)n = en em = enTem = ( (en)1, (en)2, (en)3) = <en|em> δn,m = (e'm)'n = e'n e'm = e'nTe'm = ( (e'n)1, (e'n)2, (e'n)3) = <e'n|e'm> Rmn = (e'm)n = en e'm = enTe'm = ( (en)1, (en)2, (en)3) = <en|e'm> Since our space H is real (not complex), we know that We now imagine that T is some operator in H . We can associate various matrices with T as follows Tnm = <en|T|em> = <en| Tem> |Tem> is a certain vector in H T'nm = <e'n|T|e'm> = <e'n| Te'm> |Te'm> is some other vector in H T"nm = <e"n|T|e"m> = <e"n| Te"m> |Te"m> is still some other vector in H In this Dirac language, there is only one operator T but it is associated with three different matrices, one for each basis used to span the space H. Formally T is a rank-2 tensor, which is represented by three different matrices in the three different bases. One can expand T on the en basis in this way T = Tij |ei><ej| = Tij eiejT <en|T|em> = Tij <en|ei><ej|em> = Tijδniδjm = Tnm Notice that, whereas eiTej = δij is a number, eiejT = |ei><ej| = is a 3x3 matrix which is an operator in the space H. One can write the above line in the other bases, such as T = T'ij |e'i><e'j|. One operator in H of special interest of special interest is the unity operator 1 such that 1 |a> = | 1a> = |a> for any vector |a> in H. In the Dirac notation one writes 1 = |ei><ei| = |e'i><e'i| = |e"i><e"i| Each of these is just a statement of completeness as in (1.1.9). For example, <en| 1 |em> = <en| e'i><e'i| em> = (e'i)n (em)'i Then for example <en| 1 |em> = <en| e'i><e'i| em> = (e'i)n (em)'i = RinRmi = (RTR)mn = δmn ***************** We have shown that the vector [Te'n] = T |e'n> is associated with three different matrices in the three bases en, e'n and e"n : Tmn = <em|T |en> = <em| Ten> T'mn = <e'm|T |e'n> = <e'm| Te'n> T"mn = <e"m|T |e"n> = <e"m| Te"n> > We have three matrices R, R' and R" associated with this operator R, Rmn = <em|R |en> = <em| Ren> R'mn = <e'm|R |e'n> = <e'm| Re'n> R"mn = <e"m|R |e"n> = <e"m| Re"n> So we have hopefully clarified the idea that one can have multiple matrices R, R' and R" associated with the same vector |Ren> . ************************ Consider now the following object Re'n This object is a certain vector whose name is Re'n. Suppose we had Frames S, S' and S" each having their own bases en, e'n and e"n. We would express the components of the vector Re'n in each of these three coordinate systems as follows [Re'n]i = Rij(e'n)j Frame S components [Re'n]'i = R'ij(e'n)'j Frame S' components [Re'n]"i = R"ij(e'n)"j Frame S" components On the right sides we see three matrices which we would write as R, R' and R". All three of these matrices (which in general are not all the same) are associated with the symbol R in vector Re'n. This symbol R in the name Re'n is not itself a matrix, but in each basis it is represented by a cetain matrix. One way to clarify this seeming ambiguity in the meaning of symbol R is to use the Dirac Notation which makes explicit the fact that there is an underlying "Hilbert Space" H which is spanned by any of the three sets of basis vectors shown above, and which is inhabited by "operators". We feel that this is the only "bulletproof" way to understand the meaning of the casual notation Re'n in an environment where there are multiple coordinate systems of interest (such as in discussing rotating frames of reference). WE shall now digress momentarily to discuss this Dirac notation. ***************************************** for some constants Anm. Close with < Then |e'i><e'i| R|e'n> = |en> |e'i> Rin We now compute R'nm = <e'n | R |e'm> = <e'n | 1R 1 |e'm> = <e'n |ei><ei|R |ej><ej| e'm> = Rni <ei|R |ej> Rmj = Rni Rij Rmj = Rni Rij(R-1)jm = Rni(RR-1)im = Rniδim = Rnm Thus we find that the two matrices R and R' associated with the operator R are the same. This is consistent with ** which says T' = RTR-1 R' = RRR-1 = R Assume now ** which states |e'n> = Rnm |em> Apply R to get R|e'n> = Rnm R|em> Since I am trying to prove that R|e'n> = |en> is true, I have to show that Rnm R|em> = |en> How do I show this??? Rnm R|em> = Rnm |ei><ei|R|em> = RnmRim |ei> = δni Now consider R |e'n> = |Re'n> so then |Re'n> = R |e'n> = |e'm><e'm| R |e'n> = |e'm> R'mn = |e'm> Rmn Then R |e'n> = |Re'n> so <em | R |en> = <e'm |Re'n> Using 1.1.8 and ** we can write this as want to show that |Re'm> = |em> <e'n| em> = <en|Rem>  want to show that R |e'm> = |em> We can write this as |e'n> = |em><em | R |en> We now finally return to our discussion of the meaning of Re'n back in **. It is this Re'n = |Ren> = R|en> where R is the rotation operator in the Hilbert Space H . So far, we have used R or R to represent an arbitrary rotation and en, e'n and e"n were three arbitrary bases. We now consider R and R to define the specific relationship between en and e'n such that en = Re'n or |en> = R |e'n> In this situation In this situation the matrices R and R' are the same, since according to **, R' = RRR-1 = R so <em| R|en> = <e'm| R|e'n> We now have a simple thoerem of interest: The Basis Rule: This simple rule states that (proof below), en = Re'n e'n = Rnm em (1.1.10) The right equation involves a linear combination of basis vectors. The left equation is a statement that the vector en is the same as the vector Re'n. In Dirac notation the Basis Rule is made clearer, |en> = |Re'n> = R|e'n> |e'n> = Rnm |em> Proof |en> = R|e'n> = 1R|e'n> = |e'm><e'm |R|e'n> |e'n> = 1 |e'n> = |em><em |e'n> = |em>Rnm = Rnm |em> STOP. Easy to show that all three matrices are the same, so what was the point of all this effort?? You have to have a situation where |e'n> = R |en> and |e"n> = S |e'n> T'mn ≡ <e'm|T |e'n> so then [Te'n]'m = T'mn T"mn ≡ <e"m|T |e"n> so then [Te"n]"m = T"mn Although the matrices change, the symbol T says put inside [Ten] because this means Trying to get to this: Re'n Here the matrix Tmn represents the operator T in the en basis. We now imagine that T is some operator in H . We can associate various matrices with T as follows Tnm = <en | T | em> = <en | Tem> |Tem> is a certain vector in H T'nm = <e'n | T | e'm> = <e'n | Te'm> |Te'm> is some other vector in H T"nm = <e"n | T | e"m> = <e"n | Te"m> |Te"m> is still some other vector in H In the Dirac language, there is only one operator T but it is associated with three different matrices, one for each basis used to span the space H. Formally T is a rank-2 tensor, which is represented by three different matrices in the three different bases. One can expand T on the en basis in this way, T = Tij |ei><ej| = Tij eiejT <en|T|em> = Tij <en|ei><ej|em> = Tijδniδjm = Tnm For the e'n basis one would write T = T'ij |e'i><e'j| . One operator in H of special interest is the unity operator 1 such that 1 |a> = | 1a> = |a> for any vector |a> in H. In the Dirac notation one can write, for example in the e'n basis, 1 = |e'n><e'n| // implied sum on n !! This is in fact the statement of completeness in the e'n basis. If we "close" with <ei| on the left and |ej> on the right, we find δij = <ei|ej> = <ei| 1 |ej> = <ei| e'n><e'n| ej> = (e'n)i(e'n)j and this replicates the completeness statement (1.1.9). Consider now the object <a | T | b > = <a | Tb> where T is an operator in H . The meaning of | Tb> is that it is the vector in H obtained by applying the operator T to the vector We can return now to our discussion of the meaning We are now going to prove a crucial "rule" which seems to have no official name: The Basis Rule: This simple rule states that (proof below), en = Re'n e'n = Rnm em (1.1.10) The right equation involves a linear combination of basis vectors and its meaning us unambiguous. The left equation is a statement that the vector en is the same as the vector [Re'n]. In this notation [Re'n] the object R is not a matrix, it is part of the name of the vector Re'n . However, to find out We can learn about this vector by taking its components in Frame S or in Frame S': [Re'n]i = Rij(e'n)j Frame S components [Re'n]'i = R'ij(e'n)'j Frame S' components In the Frame S world, the object R in Re'n is represented by a matrix One can take components of en = Re'n to get, (en)i = Rij (e'n)j Frame S components (en)'i = R'ij (e'n)'j Frame S' components (1.1.11) The matrix Rij is a matrix in the Frame S basis, while R'ij is in the Frame S' basis. It turns out that these two matrices are the same R' = R (see below), but if R were not a rotation, they likely would not be the same, see below. We wish to stress the fact that in e'n = Rnm em the implied sum on m creates a linear combination of basis vectors, whereas the implied sums on j in (1.1.11) are sums over the component indices of basis vectors. One must not accidentally confuse these two kinds of equations (it is easy to do). Once the Basis Rule is proven, both equations in (1.1.10) can be inverted to give e'n = R-1e'n en = (R-1)nm e'm . (1.1.12) Each e'n is thus a "back-rotated" version of en by R-1. Later we shall discuss "normal vectors" which have the property a' = Ra where a is rotated into a' by R. Since e'n = R-1en is an exception to this normal transformation, we say it is back-rotated. More on this later. Proof : Show that e'n = Rnm em en = Re'n We prove that en = Re'n by showing that both sides have the same components in Frame S: en = Re'n ? (en)i = (Re'n)i ? δn,i = Rij (e'n)j ? δn,i = Rij Rnj ? // from (1.1.8) which is based on e'n = Rnm em δn,i = Rij RTjn ? δn,i = (RRT)in yes ! // from (1.1.2) that 1 = RRT Thus our proof here consists of just reversing the above steps. QED Proof : Show that en = Re'n e'n = Rnm em Each e'n must be some linear combination of the en since a basis is by definition complete. So we write e'n = Anm em . We know that e'n = R-1en so then R-1en = Anm em . Taking components in Frame S (R-1en)i = Anm (em)i (R-1)ij(en)j = Anm (em)i Rji δn,j = Anm δm,i // R-1 = RT Rni = Ani . Therefore Anm = Rnm and so e'n = Rnm em. QED vvv ************* Rank-2 tensors, Dirac notation and Completeness Whereas a normal vector transforms under rotation R as a' = Ra , a rank-2 tensor transforms as T' = RTR-1 . (1.1.16) See e.g. Lucht Tensor ***. If we stick T = R into this formula, we get R' = RRR-1 = R. So for the tensor R which defines the rotation of interest, we get this simplification. For other tensors this will generally not be the case, unless the tensor T happens to commute with R, [T,R] = TR-RT = 0. We find the Dirac Notation helpful in understanding equations like (1.1.16). In this notation, basis vectors like en are written |en> and enT = <en|. Matrices like T' and T above are regarded as matrices of the same abstract operator T (the tensor) evaluated in the two different bases e'n and en. In Dirac notation statements above appear as follows: en em = δn,m (= enTem) e'n e'm = δn,m . // orthonormality (1.1.1) <en | em> = δn,m <e'n | e'm> = δn,m (1.1.17) (en)i(en)j = δij (e'n)i(e'n)j = δij // completeness (1.1.9) |en><en| = 1 |e'n><e'n| = 1 (1.1.18) The last line equations do seem unusual, but the "1" is the unity operator in the space in which T lives. To verify |e'n><e'n| = 1 one can "close" each side as follows <ei |e'n><e'n| ej> = <ei| 1 | ej> = <ei| ej> or (e'n)i(e'n)j = δij . Finally one can see then that <en | e'm> = en e'm = (e'm)n = Rmn = e'm en = <e'm | en> so then <en | e'm> = <e'm | en> = Rmn . (1.1.19) One says that the matrix Rmn is the "basis change matrix" between Frame S and Frame S' coordinates. We now interpret the matrices T and T' in the following manner, Tij = <ei | T | ej > T'ij = <e'i | T | e'j > . (1.1.20) Matrix T is the Dirac sandwich of the rank-2 tensor T in the Frame S basis. Matrix T' is the Dirac sandwich of the rank-2 tensor T in the Frame S' basis. These sandwich objects are referred to as "matrix elements of the operator T" in quantum mechanics. Consider now, (RTR-1)in = Rij Tjk (R-1)kn = Rij <ej | T | ek > Rnk = <e'i | ej><ej | T | ek > <ek | e'n> = <e'i | ej><ej | T | ek > <ek | e'n> = <e'i | 1 T 1 | e'n> = <e'i | T | e'n> = T'in , Thus we provide a derivation of sorts for the claim (1.1.16) that T' = RTR-1. In this same language we may write (1.1.15) (1.1.12) Rij = <ei | R | ej > = <ei | Rej > = ei [Rej] = [R-1ei] ej = e'i ej = <e'i | ej> R'ij = <e'i | R | e'j > = <e'i | Re'j > = e'i [Re'j] = e'i ej = <e'i | ej> (1.1.21) (1.1.10) and we recover the fact that R = R' alluded to earlier.