Tensor operators

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Okay, I've read and re-read the section on tensor operators and the Wigner-Eckart theorem in Sakurais book, but I'm still confused. Could anyone explain to me how to think about vector and tensor operators and the significance of the Wigner-Eckart theorem? :confused:

Thanks.
 

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  • #2
vanesch
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broegger said:
Okay, I've read and re-read the section on tensor operators and the Wigner-Eckart theorem in Sakurais book, but I'm still confused. Could anyone explain to me how to think about vector and tensor operators and the significance of the Wigner-Eckart theorem? :confused:

Thanks.
I wrote the following about this part of Sakurai's book a few years ago:

10. Tensor Operators

Vector Operator

In classical physics, a vector is somehow a quantity that has "magnitude and direction". Analytically, vectors are represented by 3 coordinates in Euclidean space, but that doesn't mean that 3 numbers form a vector. The essential property of a vector is that it transforms under a rotation just as one expects: the same magnitude, and the direction is "rotated" as required. Mathematically, this means that the 3 coordinates of a vector before and after rotation have to undergo a transformation, which is nothing else but the matrix multiplication with the 3-dimensional rotation matrix that corresponds to the rotation under consideration.
In quantum mechanics, things are the same. Only, now a vector is not something that is described with 3 numbers, giving "magnitude and direction", it consists of 3 operators on hilbert space. But not just any three operators form a vector. In order for them to form a vector, their transformation properties must be such, that the expectation value of the 3 operators under a rotated state must transform as a classical 3-dim vector from the expectation value of the 3 operators under the unrotated state. The funny thing here is that the 3 operators stay the same, and it are the state kets that are rotated. Working the condition out for arbitrary states, one arrives at equation (3.10.3) for a vector operator. Using the infinitesimal representation of the rotation operator as a function of angular momentum, this condition is equivalent to (3.10.8).

Cartesian Tensors versus Irreducible Tensors

In classical physics, cartesian tensors are nothing else but "multiple index vectors", and nothing stops us from defining the equivalent in quantum mechanics. However, looking at it from another point of view, we can see that a cartesian tensor of rank r consists then of 3^r operators, and then these operators should form a 3^r dimensional representation of the rotation group if we apply the generalisation of equation (3.10.3). But we already know all irreducible representations of the rotation group ! It turns out that cartesian tensors (their components) can be written as linear combinations of things that correspond to irreducible representations. This is actually no surprise: all representations of a group are irreducible representations, or can be written as a linear combination of irreducible representations. This is illustrated in equations (3.10.12) and (3.10.13), for a specific case of a rank 2 cartesian tensor. We can write it as the sum of a scalar (spin 0), a vector (spin 1) and a thing that will turn out to be a spin-2 representation with 5 components. So we see that what the group representation properties is concerned, these cartesian tensors are quite messy objects. We therefore define spherical tensors to be things that are irreducible (fixed l) representations of the rotation group.
First some illustrations are used (such as putting a vector as argument of the set of spherical harmonics of given l) to suggest the definition of a spherical tensor, which is a set of operators. That definition is formulated in equation (3.10.22). So, in a similar way as for a cartesian tensor, a spherical tensor is defined by its properties under rotation ; it turns out that the components of a spherical tensor under rotation transform in a very similar way as do transform the eigenstates (j,m) (look at equation 3.5.49). In fact, they transform in the opposite way (inverse rotation). The infinitesimal condition for a spherical tensor is then given in equation (3.10.25).

Product of Tensors

The rather complicated-sounding theorem in fact says that we can make "products of spherical tensors" by using a similar linear superposition as those that occur when we combine two sets of spin kets into the "sum of spins".

Matrix Elements of Tensor Operators: The Wigner-Eckart Theorem

Finally, a famous theorem: The Wigner-Eckart theorem. It states that the matrix representation of a spherical tensor in a set of basis kets of angular momentum takes on a specific form: each block between a set of j' and j (and alpha' and alpha, standing for other operators defining a complete set of commuting operators together with angular momentum) consists of elements that are the "sum of spin" elements as if we added together the spin j with the spin of the tensor to obtain the spin j', and a scale factor which depends on the particularity of the tensor, but which is the same in the entire block. This theorem is not very surprising, but it is practical when having to calculate matrix representations of tensor operators.

You can find it also here:
http://perso.wanadoo.fr/patrick.vanesch/nrqmJJ/Summary_Chapter_3_JJ.html
 
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Wow, this is great, thanks!
 
  • #4
reilly
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For my money, the best reference on spherical tensors is in Edmonds, Angular Momentum in Quantum Mechanics. he discusses the group theoretic aspects, and gives numerous examples -- quadrapole moments, emission of radiation and so forth. For a long time Edmonds was the bible for angular momentum, it's a great book with a good balance between theory and application.
Regards,
Reilly Atkinson
 
  • #5
Tom Mattson
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reilly said:
For my money, the best reference on spherical tensors is in Edmonds, Angular Momentum in Quantum Mechanics.
What do you think of Elementary Theory of Angular Momentum by Rose?

I've recently purchased it.
 
  • #6
Haelfix
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Yea, although I hate visualizing things like that. I much prefer to think of vectors in the *usual* way but in an abstract space, and thinking about projecting operators onto vector components. Eg here we think of states as rays in Hilbert space.
 
  • #7
reilly
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Tom -- I always preferred Edmonds, but Rose is good, and preferred by some. It's a matter of taste. As I remember, Edmonds is a bit briefer than Rose. So, enjoy.

Haelfix -- To some degree the whole matter of spherical tensors is strongly connected to the multipole expansions for E and B in classical E&M. which typically avoids Hilbert Spaces and such, and is strongly embedded in spherical coordinates. (Of course, the Ylm's form complete sets, but since they are finite polynomials, they clearly span their appropriate spaces,)
Regards,
Reilly Atkinson
 

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