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4.8: Tensor Operators


Introduction: Cartesian Vectors and Tensors

Physics is full of vectors: x,L,S<math xmlns=""><semantics/></math> and so on.  Classically, a (three-dimensional) vector is defined by its properties under rotation: the three components corresponding to the Cartesian x,y,z<math xmlns=""><semantics/></math> axes transform as

ViRijVj<math xmlns=""><semantics/></math>,


with the usual rotation matrix, for example

Rz(θ)=cosθsinθ0sinθcosθ0001<math xmlns=""><semantics/></math>

for rotation about the z-<math xmlns=""><semantics/></math> axis.  (We’ll use (x,y,z) and (x1,x2,x3)<math xmlns=""><semantics/></math> interchangeably.)

tensor is a generalization of a such a vector to an object with more than one suffix, such as, for example, Tij or Tijk<math xmlns=""><semantics/></math> (having 9 and 27 components respectively in three dimensions) with the requirement that these components mix among themselves under rotation by each individual suffix following the vector rule, for example

TijkRilRjmRknTlmn<math xmlns=""><semantics/></math>

where R<math xmlns=""><semantics/></math> is the same rotation matrix that transforms a vector.  Tensors written in this way are called Cartesian tensors (since the suffixes refer to Cartesian axes).  The number of suffixes is the rank of the Cartesian tensor, a rank n<math xmlns=""><semantics/></math> tensor has of course 3n<math xmlns=""><semantics/></math> components.

Tensors are common in physics: they are essential in describing stress, distortion and flow in solids and liquids.  The inertial tensor is the basis for analyzing angular motion in classical mechanics.  Tensor forces are important in the dynamics of  the deuteron, and in fact tensors arise for any charge distribution more complicated than a dipole.  Going to four dimensions, and generalizing from rotations to Lorentz transformations, Maxwell’s equations are most naturally expressed in tensor form, and tensors are central to General Relativity.

To get back to non-relativistic physics, since the defining property of a tensor is its behavior under rotations, spherical polar coordinates are sometimes a more natural basis than Cartesian coordinates.  In fact, in that basis tensors (called spherical tensors) have rotational properties closely related to those of angular momentum eigenstates, as will become clear in the following sections. 

The Rotation Operator in Angular Momentum Eigenket Space

As a preliminary to discussing general tensors in quantum mechanics, we briefly review the rotation operator and quantum vector operators.  (A full treatment is given in my 751 lecture.)

Recall that the rotation operator turning a ket through an angle θ<math xmlns=""><semantics/></math> (the vector direction denotes the axis of rotation, its magnitude the angle turned through) is

U(R(θ))=eiθJ.<math xmlns=""><semantics/></math>

Since J<math xmlns=""><semantics/></math> commutes with the total angular momentum squared J2=j(j+1)2,<math xmlns=""><semantics/></math> we can restrict our attention to a given total angular momentum j,<math xmlns=""><semantics/></math> having as usual an orthonormal basis set |j,m<math xmlns=""><semantics/></math>, or |m<math xmlns=""><semantics/></math> for short, with 2j+1<math xmlns=""><semantics/></math> components, a general ket |α<math xmlns=""><semantics/></math> in this space is then:

|α=m=jjαm|m<math xmlns=""><semantics/></math>.

Rotating this ket,

|α|α=eiθJ|α<math xmlns=""><semantics/></math>

Putting in a complete set of states, and using the standard notation for matrix elements of the rotation operator,

|α=eiθJ|α=m,mαm|mm|eiθJ|m=m,mD(j)mm(R(θ))αm|m.<math xmlns=""><semantics/></math>

D(j)mm=m|eiθJ|m<math xmlns=""><semantics/></math> is standard notation (see the earlier lecture).

So the ket rotation transformation is

αm=mD(j)mmαm,or α=Dα.<math xmlns=""><semantics/></math>

with the usual matrix-multiplication rules.

Rotating a Basis Ket

Now suppose we apply the rotation operator to one of the basis kets |j,m<math xmlns=""><semantics/></math>, what is the result?

eiθJ|j,m=m|j,mj,m|eiθJ|j,m=m|j,mD(j)mm(R).<math xmlns=""><semantics/></math>

Note the reversal of mm′ compared with the operation on the set of component coefficients of the general ket.

(You may be thinking: wait a minute, |j,m<math xmlns=""><semantics/></math> is a ket in the space<math xmlns=""><semantics/></math>it can be written αmj,m<math xmlns=""><semantics/></math> with αm=δmm<math xmlns=""><semantics/></math>, so we could use the previous rule αm=mD(j)mmαm<math xmlns=""><semantics/></math> to get αm=D(j)mmαm=D(j)mmδmm=D(j)mm<math xmlns=""><semantics/></math>.  Reassuringly, this leads to the same result we just found.)

Rotating an Operator, Scalar Operators

Just as in the Schrödinger versus Heisenberg formulations, we can either apply the rotation operator to the kets and leave the operators alone, or we can leave the kets alone, and rotate the operators:

AeiθJAeiθJ=UAU<math xmlns=""><semantics/></math>


which will yield the same matrix elements, so the same physics.

scalar operator is an operator which is invariant under rotations, for example the Hamiltonian of a particle in a spherically symmetric potential. (There are many less trivial examples of scalar operators, such as the dot product of two vector operators, as in a spin-orbit coupling.) 

The transformation of an operator under an infinitesimal rotation is given by:


SU(R)SU(R) with  U(R)=1iεJ<math xmlns=""><semantics/></math>

from which

SS+[iεJ,S].<math xmlns=""><semantics/></math>

It follows that a scalar operator S,<math xmlns=""><semantics/></math> which does not change at all, must commute with all the components of the angular momentum operator, and hence must have a common set of eigenkets with, say, J2<math xmlns=""><semantics/></math> and Jz.<math xmlns=""><semantics/></math>    

Vector Operators:  Definition and Commutation Properties

quantum mechanical vector operator V<math xmlns=""><semantics/></math> is defined by requiring that the expectation values of its three components in any state transform like the components of a classical vector under rotation. 

It follows from this that the operator itself must transform vectorially,

Vi=U(R)ViU(R)=RijVj<math xmlns=""><semantics/></math>.

To see what this implies, it is easiest to look at a simple case.  For an infinitesimal rotation about the z-<math xmlns=""><semantics/></math> axis,

Rz(ε)=1ε0ε10001<math xmlns=""><semantics/></math>

the vector transforms

VxVyVz  1ε0ε10001VxVyVz  =VxεVyVy+εVxVz<math xmlns=""><semantics/></math>

The unitary Hilbert space operator U corresponding to this rotation U(Rz(ε))=1iεJz,<math xmlns=""><semantics/></math> so

UViU=(1+iεJz/)Vi(1iεJz/)=Vi+iε[Jz,Vi].<math xmlns=""><semantics/></math>

The requirement that the two transformations above, the infinitesimal classical rotation generated by Rz(ε)<math xmlns=""><semantics/></math> and the infinitesimal unitary transformation U(R)ViU(R)<math xmlns=""><semantics/></math>, are in fact the same thing yields the commutation relations of a vector operator with angular momentum:

i[Jz,Vx]=Vyi[Jz,Vy]=+Vx.<math xmlns=""><semantics/></math>

From this result and its cyclic equivalents, the components of any vector operator V<math xmlns=""><semantics/></math> must satisfy:

[Vi,Jj]=iεijkVk<math xmlns=""><semantics/></math>.

Exercise: verify that the components of x,L,S<math xmlns=""><semantics/></math> do in fact satisfy these commutation relations.


(Note:  Confusingly, there is a slightly different situation in which we need to rotate an operator, and it gives an opposite result.  Suppose an operator T acts on a ket |α<math xmlns=""><semantics/></math> to give the ket |α=T|α<math xmlns=""><semantics/></math>.  For kets |α<math xmlns=""><semantics/></math> and |α<math xmlns=""><semantics/></math> to go to U|α<math xmlns=""><semantics/></math> and U|α<math xmlns=""><semantics/></math> respectively under a rotation U,T<math xmlns=""><semantics/></math> itself must transform as TUTU<math xmlns=""><semantics/></math> (recall U=U1<math xmlns=""><semantics/></math> ).  The point is that this is a Schrödinger rather than a Heisenberg-type transformation: we’re rotating the kets, not the operators.)

Warning:  Does a vector operator transform like the components of a vector or like the basis kets of the space?  You’ll see it written both ways, so watch out!

We’ve already defined it as transforming like the components:




Vi=U(R)ViU(R)=RijVj<math display="block" xmlns=""><semantics/></math>


but if we now take the opposite rotation, the unitary matrix U(R)<math xmlns=""><semantics/></math> is replaced by its inverse U(R)<math xmlns=""><semantics/></math> and vice versa. Remember also that the ordinary spatial rotation matrix R<math xmlns=""><semantics/></math> is orthogonal, so its inverse is its transpose, and the above equation is equivalent to



Vi=U(R)ViU(R)=RjiVj<math display="block" xmlns=""><semantics/></math>



This definition of a vector operator is that its elements transform just as do the basis kets of the space<math xmlns=""><semantics/></math>so it’s crucial to look carefully at the equation to figure out which is the rotation matrix, and which is its inverse!

This second form of the equation is the one in common use.

Cartesian Tensor Operators

From the definition given earlier, under rotation the elements of a rank two Cartesian tensor transform as:

TijTij=RiiRjjTij.<math xmlns=""><semantics/></math>

where Rij<math xmlns=""><semantics/></math> is the rotation matrix for a vector.

It is illuminating to consider a particular example of a second-rank tensor, Tij=UiVj<math xmlns=""><semantics/></math>, where U<math xmlns=""><semantics/></math> and V<math xmlns=""><semantics/></math> are ordinary three-dimensional vectors.

The problem with this tensor is that it is reducible, using the word in the same sense as in our discussion of group representations is discussing addition of angular momenta.  That is to say, combinations of the elements can be arranged in sets such that rotations operate only within these sets.  This is made evident by writing:

UiVj=UV3δij+(UiVjUjVi)2+(UiVj+UjVi2UV3δij).<math xmlns=""><semantics/></math>

The first term, the dot product of the two vectors, is clearly a scalar under rotation, the second term, which is an antisymmetric tensor has three independent components which are the vectorcomponents of the vector product U×V<math xmlns=""><semantics/></math>, and the third term is a symmetric traceless tensor, which has five independent components.  Altogether, then, there are 1+3+5=9<math xmlns=""><semantics/></math> components, as required. 

Spherical Tensors

Notice the numbers of elements of these irreducible subgroups: 1,  3,5.<math xmlns=""><semantics/></math>  These are exactly the numbers of elements of angular momenta representations for = 0, 1, 2!  

This is of course no coincidence: as we shall make more explicit below, a three-dimensional vector is mathematically isomorphic to a quantum spin one, the tensor we have written is therefore a direct product of two spins one, so, exactly as we argues in discussing addition of angular momenta, it will be a reducible representation of the rotation group, and will be a sum of representations corresponding to the possible total angular momenta from adding two spins one, that is, j=0,1,2.<math xmlns=""><semantics/></math>   

As discussed earlier, the matrix elements of the rotation operator

U(R(θ))=eiθJ<math xmlns=""><semantics/></math>

within a definite j<math xmlns=""><semantics/></math> subspace are written

Djmm(R(θ))=j,m|eiθJ|j,m<math xmlns=""><semantics/></math>

so under rotation operator a basis state |j,m<math xmlns=""><semantics/></math> transforms as:

eiθJ|j,m=m|j,mj,m|eiθJ|j,m=m|j,mD(j)mm(R).<math xmlns=""><semantics/></math>

The essential point is that these irreducible subgroups into which Cartesian tensors decompose under rotation (generalizing from our one example) form a more natural basis set of tensors for problems with rotational symmetries.

Definition: We define a spherical tensor of rank k<math xmlns=""><semantics/></math> as a set of 2k+1<math xmlns=""><semantics/></math> operators  Tqk,q=k,k1,,k<math xmlns=""><semantics/></math> such that under rotation they transform among themselves with exactly the same matrix of coefficients as that for the 2j+1<math xmlns=""><semantics/></math> angular momentum eigenkets |m<math xmlns=""><semantics/></math> for k=j,<math xmlns=""><semantics/></math> that is,

U(R)TqkU(R)=qD(k)qqTqk<math xmlns=""><semantics/></math>.

To see the properties of these spherical tensors, it is useful to evaluate the above equation for infinitesimal rotations, for which D(k)qq(ε)=k,q|IiεJ/|k,q=δqqiεk,q|J/|k,q.<math xmlns=""><semantics/></math>

(The matrix element k,q|J/|k,q<math xmlns=""><semantics/></math> is just the familiar Clebsch Gordan coefficient in changed notation: the rank k<math xmlns=""><semantics/></math> corresponds to the usual j,<math xmlns=""><semantics/></math> and q<math xmlns=""><semantics/></math> to the “magnetic” quantum number m.<math xmlns=""><semantics/></math> )

Specifically, consider an infinitesimal rotation εJ=εJ+.<math xmlns=""><semantics/></math> (Strictly speaking, this is not a real rotation, but the formalism doesn’t care, and the result we derive can be confirmed by rotation about the x<math xmlns=""><semantics/></math> and y<math xmlns=""><semantics/></math> directions and adding appropriate terms.)

The equation is

(1iεJ+/)Tqk(1+iεJ+/)=q(δqqiεk,q|J+/|k,q)Tqk<math xmlns=""><semantics/></math>

and equating terms linear in ε,<math xmlns=""><semantics/></math>

[J±,Tqk]=±(kq)(k±q+1)Tq±1k[Jz,Tqk]=qTqk.<math xmlns=""><semantics/></math>

Sakurai observes that this set of commutation relations could be taken as the definition of the spherical tensors.

Notational note: we have followed Shankar here in having the rank k<math xmlns=""><semantics/></math> as a subscript, the “magnetic” quantum number q<math xmlns=""><semantics/></math> as a superscript, the same convention used for the spherical harmonics (but not for the D<math xmlns=""><semantics/></math> matrices!)  Sakurai, Baym and others have the rank above, usually in parentheses, and the magnetic number below.  Fortunately, all use k<math xmlns=""><semantics/></math> for rank and q<math xmlns=""><semantics/></math> for magnetic quantum number.

A Spherical Vector

The  j=1<math xmlns=""><semantics/></math> angular momentum eigenkets are just the familiar spherical harmonics

Y01=34πzr,Y±11=34πx±iy2r.<math xmlns=""><semantics/></math>

The rotation operator will transform (x,y,z)<math xmlns=""><semantics/></math> as an ordinary vector in three-space, and this is evidently equivalent to

|j=1,mm|j=1,mD(j)mm(R)<math xmlns=""><semantics/></math>

It follows that the spherical representation of a three vector (Vx,Vy,Vz)<math xmlns=""><semantics/></math> has the form:

T±11=Vx±iVy2=V±11,T01=Vz=V01.<math xmlns=""><semantics/></math>

In line with spherical tensor notation, the components (T11,T01,T11)<math xmlns=""><semantics/></math> are denoted Tq1.<math xmlns=""><semantics/></math>

Matrix Elements of Tensor Operators between Angular Momentum Eigenkets

By definition, an irreducible tensor operator Tqk<math xmlns=""><semantics/></math> transforms under rotation like an angular momentum eigenket |k,q<math xmlns=""><semantics/></math>.  Therefore, rotating the ket Tqk|j,m<math xmlns=""><semantics/></math>,


UTqk|j,m=UTqkU1U|j,m=qD(k)qqTqkmD(j)mm|j,m<math xmlns=""><semantics/></math>.

The product of the two D<math xmlns=""><semantics/></math> matrices appearing is precisely the set of coefficients to rotate the direct product of eigenkets |k,q|j,m<math xmlns=""><semantics/></math> where |k,q<math xmlns=""><semantics/></math> is the angular momentum eigenket having  j=k,m=q.<math xmlns=""><semantics/></math> 

We have met this direct product of two angular momentum eigenkets before: this is just a system having two angular momenta, such as orbital plus spin angular momenta.   So we see that Tqk<math xmlns=""><semantics/></math> acting on |j,m<math xmlns=""><semantics/></math> generates a state having total angular momentum the sum of (k,q)<math xmlns=""><semantics/></math> and (j,m).<math xmlns=""><semantics/></math>    

To link up (more or less) with Shankar’s notation: our direct product state |k,q|j,m<math xmlns=""><semantics/></math> is the same as |k,q;j,m<math xmlns=""><semantics/></math> in the notation |j1,m1;j2,m2<math xmlns=""><semantics/></math> for a product state of two angular momenta (possibly including spins). Such a state can be written as a sum over states of the form |jtot,mtot;j1,j2<math xmlns=""><semantics/></math> where this denotes a state of total angular momentum jtot,<math xmlns=""><semantics/></math>   z-<math xmlns=""><semantics/></math> direction component mtot,<math xmlns=""><semantics/></math> made up of two spins having total angular momentum j1,j2<math xmlns=""><semantics/></math> respectively.

This is the standard Clebsch-Gordan sum:


|j1,m1;j2,m2=jtot=|j1j2|j1+j2mtot=jtotjtot|jtot,mtot;j1,j2jtot,mtot;j1,j2|j1,m1;j2,m2.<math xmlns=""><semantics/></math>


The summed terms give a unit operator within this (2j1+1)(2j2+1)<math xmlns=""><semantics/></math> dimensional space, the term jtot,mtot;j1,j2|j1,m1;j2,m2<math xmlns=""><semantics/></math> is a Clebsch-Gordan coefficient.  The only nonzero coefficients have mtot=m1+m2,<math xmlns=""><semantics/></math> and jtot<math xmlns=""><semantics/></math> restricted as noted, so for given m1,m2<math xmlns=""><semantics/></math> we just set mtot=m1+m2,<math xmlns=""><semantics/></math> we don’t sum over mtot,<math xmlns=""><semantics/></math> and the sum over jtot<math xmlns=""><semantics/></math> begins at |mtot|.<math xmlns=""><semantics/></math>  


Translating into our |k,q|j,m<math xmlns=""><semantics/></math> notation, and cleaning up,

|k,q;j,m=jtot=|q+m|k+j|jtot,q+m;k,jjtot,q+m;k,j|k,q;j,m.<math xmlns=""><semantics/></math>

We are now able to evaluate the angular component of the matrix element of a spherical tensor operator between angular momentum eigenkets: we see that it will only be nonzero for  mtot=m1+m2,<math xmlns=""><semantics/></math> and  jtot<math xmlns=""><semantics/></math> at least |mtot|.<math xmlns=""><semantics/></math>  

The Wigner-Eckart Theorem

At this point, we must bear in mind that these tensor operators are not necessarily just functions of angle.  For example, the position operator is a spherical vector multiplied by the radial variable r,<math xmlns=""><semantics/></math> and kets specifying atomic eigenstates will include radial quantum numbers as well as angular momentum, so the matrix element of a tensor between two states will have the form

α2,j2,m2|Tqk|α1,j1,m1<math xmlns=""><semantics/></math>,

where the j<math xmlns=""><semantics/></math> ’s and m<math xmlns=""><semantics/></math> ’s denote the usual angular momentum eigenstates and the α<math xmlns=""><semantics/></math> ’s are nonangular quantum numbers, such as those for radial states.

The basic point of the Wigner<math xmlns=""><semantics/></math>Eckart theorem is that the angular dependence of these matrix elements can be factored out, and it is given by the Clebsch-Gordan coefficients

Having factored it out, the remaining dependence, which is only on the total angular momentum in each of the kets, not the relative orientation (and of course on the α<math xmlns=""><semantics/></math> ’s), is traditionally written as a bracket with double lines, that is,

α2,j2,m2|Tqk|α1,j1,m1=α2,j2Tkα1,j12j+1j2,m2|k,q;j1,m1.<math xmlns=""><semantics/></math>


The denominator is the conventional normalization of the double-bar matrix element. The proof is given in, for example, Sakurai (page 239) and is not that difficult.  The basic strategy is to put the defining identities

[J±,Tqk]=±(kq)(k±q+1)Tq±1k[Jz,Tqk]=qTqk<math xmlns=""><semantics/></math>

between |α,j,m<math xmlns=""><semantics/></math> bras and kets, then get rid of the J± and Jz<math xmlns=""><semantics/></math> by having them operate on the bra or ket.  This generates a series of linear equations for α2,j2,m2|Tqk|α1,j1,m1<math xmlns=""><semantics/></math> matrix elements with m<math xmlns=""><semantics/></math> variables differing by one, and in fact this set of linear equations is identical to the set that generates the Clebsch-Gordan coefficients, so we must conclude that these spherical tensor matrix elements, ranging over possible m<math xmlns=""><semantics/></math> and j<math xmlns=""><semantics/></math> values, are exactly proportional to the Clebsch-Gordan coefficients<math xmlns=""><semantics/></math>and that is the theorem.


A Few Hints for Shankar’s problem 15.3.3: that first matrix element comes from adding a spin j<math xmlns=""><semantics/></math> to a spin 1, writing the usual maximum m<math xmlns=""><semantics/></math> state, applying the lowering operator to both sides to get the total angular momentum j+1,m=j<math xmlns=""><semantics/></math> state, then finding the same m state orthogonal to that, which corresponds to total angular momentum j<math xmlns=""><semantics/></math> (instead of ). j+1<math xmlns=""><semantics/></math>  



For the operator J,<math xmlns=""><semantics/></math> the Wigner-Eckart matrix element simplifies because J<math xmlns=""><semantics/></math> cannot affect α,<math xmlns=""><semantics/></math> and also it commutes with J2,<math xmlns=""><semantics/></math> so cannot change the total angular momentum.

So, in the Wigner-Eckart equation, replace Tqk<math xmlns=""><semantics/></math> on the left-hand side by J01<math xmlns=""><semantics/></math>, which is just Jz.<math xmlns=""><semantics/></math>   The result of (1) should follow.

(2) First note that a scalar operator cannot change m.<math xmlns=""><semantics/></math> Since c<math xmlns=""><semantics/></math> is independent of A<math xmlns=""><semantics/></math> we can take

A=J<math xmlns=""><semantics/></math> to find c.<math xmlns=""><semantics/></math>