Understanding Spherical Tensor Operators

in this video we will explain spiracle tensor operators they are defined like this a spherical tensor operator tkq with rank K is a collection of two K plus 1 operators that are numbered by the index Q which transform under rotations in the same way as spherical harmonics do now let's try to understand this definition since this will be a rather long video we will explain spherical tensor operators in three steps first we need to understand tensor operators in particular Cartesian tensor operators a tensor always has to do with rotations and operators are important in quantum mechanics so we will first discuss rotations in quantum mechanics then we can go to the spherical basis which allows us to rigorously defined spherical tensor operators and after that we will investigate how to calculate components of spherical tensor operators from Cartesian components this video should have everything you need to know about this topic so without further ado here we go Part 1 tensor operators what characterizes a tensor is the way it transforms under rotations since we want to investigate tensor operators let's look at how rotations work in quantum mechanics where they are operators that act on States we distinguish between two different types of rotation active and passive ones in active rotations the state vector sy gets rotated using a unitary operator D whereas for passive rotations the operators get rotated in a certain way we can now make the following observation in one experiment we rotate the state vector and in another experiment we rotate our lab equipment that is the operator but in the opposite direction as you can see this has the same result so let's write this a little bit more mathematically the expectation value of this operator when the state gets rotated will be the same as the expectation value for when we rotate the operator in the opposite direction from this we can see that the operator which was rotated in the opposite way is given by D dagger o D and if we look for the opposite rotation we get D oh D dagger since the unitary operator D for a rotation and the other way is just the dagger a classical vector VI transforms under rotation as R IJ VJ in quantum mechanics the same relations hold for the expectation value of a vector operator however we also know from our previous calculation that this is equivalent to rotating the vector operator itself but in the opposite way therefore D dagger VI D is equal to RI J VJ note here that the unitary operator D acts in the Hilbert space whereas RI J acts in coordinate space now that we know how accurate agent vector operator transforms under rotations we can generalize this to a Cartesian tensor operator which just gets more copies of the rotation matrix R but again all of this was in the Cartesian basis let us now change to the spherical basis part 2 spherical basis the spherical basis is defined such that the transformation behavior under rotations is the same as how spherical harmonics transform under rotations before we consider general tensors let's see how the basis vectors in the spherical basis look like the spherical harmonics Y 1 1 Y 1 0 and y 1 -1 are proportional to x y&z therefore we make the following anzats that the spherical basis vectors are given in terms of the Cartesian basis vectors like this in order to determine the proportionality factors alpha beta and gamma we use the condition that the basis vectors should be orthonormal this is a rather straightforward calculation however we have to keep in mind that since we are now dealing with a complex basis we have to perform a complex conjugation on one of the vectors in an inner product the absolute value of alpha beta and gamma can be obtained rather quickly and concerning their face we choose it the same way as in the spherical harmonics this means a minus sign for alpha and a plus sign for beta and gamma using these three basis vectors we can now construct vectors in the spherical basis let's take the vector a and write it in terms of its components strictly speaking those are contravariant components since their index is upstairs the distinction between contravariant and covariant components will become very important later so let's do this carefully if we write our spherical basis in terms of the Cartesian basis we can read off the Cartesian components a X a Y and a set by inverting these relations we can define the contravariant spherical components of a vector like this now what about the covariant components the covariant vector components have their index downstairs and the basis vectors now have an upper index this is because they are elements of the dual basis which is defined like this but we already know a relation of basis vectors which yields a Kronecker Delta using the complex conjugate basis vectors therefore the dual basis vectors are just the complex conjugated versions of the original basis vectors using this information we can also calculate the covariant components of a spherical vector finally we get the following relation between covariant and contravariant components of a spherical vector before we continue here are some additional useful relations concerning the spherical basis first we have the following relations for the cross-product of basis vectors second let's talk about the inner product of two vectors in the spherical basis if a is a contravariant vector and B is a covariant vector we get the following relation since we know how to switch between covariant and contravariant components we can also write it like this and using a sum over Q like this but enough about vectors what about general tensors a spherical tensor and transforms in the same way as spherical harmonics do so we first need to know how spherical harmonics transform under rotations a spherical harmonic is an eigenstate of the angular momentum operator projected onto the coordinate basis note that this is slightly different from ylm of theta and Phi which are defined as LM projected onto a unit length basis and the difference is just a factor of R to the power of L which ylm we use does not matter for us since both transform in the same way so we just used the first one after performing a rotation we get yl m of r prime which is defined like this our prime and r are related by a unitary transformation d so we get d dagger here next we put in a completeness relation with L Prime and M Prime the first term here is why L prime M prime and the second one is a matrix element of the unitary transformation operator d note that since we have a dagger here we can write this as the undergird version of d for a rotation r inverse since this unitary rotation will not stretch or compress anything L will be the same as L prime so we get Delta L L Prime the remaining matrix element is called the vignati matrix this D stands for - tell on the German word for representation the indices of d R M prime m since we have M Prime on the left and M on the right and since we have the matrix element of D with our inverse we denote this and brackets let's see what happens for a rotation and the opposite way on the left hand side we get d ylm T dagger and on the right hand side we get the via D matrix for our there is one more way in which we can write down this transformation behavior starting with the definition of the vegan er D matrix we can first investigate what happens for an inverse rotation but also what happens if we take the hermitian conjugate of the whole equation since the left-hand side is just a number we just get a complex conjugate but the right-hand side gets conjugated properly if we now compare these two terms we can make a connection between the complex conjugated vigener d matrix and the Vigna d matrix for an inverse rotation so finally the transformation behavior of spherical harmonics under rotations can be written in these two equivalent ways now comes the highlight of this video we define a spherical tensor operator tkq as a set of 2 k plus 1 operators that transform under rotations exactly like the spherical harmonics did compare this to the definition of a cartesian tensor operator where the number of rotation matrices is determined by the rank of the tensor whereas for a spherical tensor operator we only have one big nerdy matrix using the definition of the unitary rotation operator we can write down an infinitesimal version of the first relation which leads us to the following commutator relations the detailed calculations how to arrive at these commutators are rather long and since this video is already quite long we refer here to one of our references in the description we can further simplify the first two commutator relations using the definition of ladder operators which results in another commutator relation note that both the definition using the vignette D matrices as well as the definition using commentators are equivalent both completely define spherical tensor operators last but not least let us talk about hermitian conjugate of spherical tensor operators if we take the hermitian adjoint of the commutator relations we get the following two new commutator relations in terms of T dagger you can easily verify that if we define the taggert version of t KQ like this we get back our original commutator relations the value of a is arbitrary and sometimes set to 0 also whether we use plus or minus q in the exponent is up to us as long as we are consistent in our choice we now have a good definition of spherical tensor operators but how do we actually calculate them say if we have a rank 2 cartesian tensor tij what are the components of TK Q what is K and what is Q this will be answered in the final part of this video part 3 examples to conclude this video will show you how to switch from Cartesian tensor components to spherical tensor components first a-rank zero tensor that is a scalar is the same in both coordinate systems next a Rank 1 tensor which is a vector we already talked about spherical vector operators but one question remains which is the real spherical tensor component t11 the contravariant or the covariant component to solve this we can use the commutation relations in particular that the commutator of J plus with t11 should be 0 since T 1 1 already is the highest state in its multiplet the calculation is not too difficult and the result is that only the covariant vector components are spherical tensor operators as indicated by their index this makes sense since the covariant vector components are those that transform similar to the basis vectors and those were constructed from the spherical harmonics but don't be confused a plus is still a spherical tensor operator however it is equal to minus a minus so a plus is the negative of 81 minus 1 spiracle tensor operator even though the index S Plus next onto rank two tensors here it's not obvious how to calculate tkq from t IJ we know that T IJ has 3 times 3 so 9 components if they form an irreducible representation we should represent those nine components as the nine components of T for Q where Q can take on nine values from minus four to four however a rank two tensor is reducible so this is wrong instead we can use a trick to deal with rank two tensors and in fact all higher rank tensors this trick is to write T IJ as AI BJ so as the direct product of to rank want answers since both a and B are tensors T IJ is guaranteed to transform as a tensor as well so how does this help us since spherical tensor operators are basically spherical harmonics we can couple them like angular momenta using clebsch-gordon coefficients we already covered clebsch-gordon coefficients in a different video so if you'd like you can watch our video on that topic still here is a very quick run-through of clubs gordan coefficients starting with a state of definite angular momentum L M we can insert a completeness relation to expand this state in the product state L 1 m 1 l 2 m 2 the bracket element here is precisely a cleped gordon coefficient we can write the same relation in the coordinates based spaces which tells us how to couple spherical harmonics and in the very same way we can couple to spherical tensor operators now back to our example of a rank two tensor since a and B are Rank 1 K 1 and K 2 are equal to 1 this means we need the following clebsch-gordan table the notation is such that the coupled state is on top and the product states are on the left as you can see from this table if we couple to Rank 1 spherical tensors we can get a rank 2 a rank 1 and even a rank 0 tensor as a result the components agree we have five components from a rank two tensor three from a rank one tensor and one from a rank zero tensor which all add up to nine components as an example let's calculate the t10 component of the Cartesian tij tensor according to the table this is equal to a11 b1 minus 1 times the square root of 1/2 plus a1 minus 1 B 1 1 times minus the square root of 1/2 it is important to use the covariant components of a and B here as we discussed earlier the result is given by I over the square root of 2 times a X B Y minus a YB X as a final step we change back to T IJ which leads to this result and there you have it this is how you start with a Cartesian tensor operator tij and change the basis to a spherical tensor operator tkq which K and Q are possible is completely decided by the clebsch-gordan table the other components look like this and can be calculated in exactly the same way note that the rank 0 tensor is proportional to the trace of T IJ the rank 1 tensor is proportional to anti-symmetric components and the rank two tensor is proportional to symmetric components this is due to the fact that we can split up a general cartesian rank two tensor into a diagonal part proportional to its trace a part that is anti-symmetric and a part that is symmetric and traceless finally higher-ranked answers can be calculated in exactly the same way for a rank 3 tensor T ijk you could write it as a IJ BK and use the results from the rank 2 case in order to couple a and B and that's pretty much it for this video thanks for watching

in this video we will explain spiracle tensor operators they are defined like this a spherical tensor operator tkq with rank K is a collection of two K plus 1 operators that are numbered by the index Q which transform under rotations in the same way as spherical harmonics do now let&#39;s try to understand this definition since this will be a rather long video we will explain spherical tensor operators in three steps first we need to understand tensor operators in particular Cartesian tensor operators a tensor always has to do with rotations and operators are important in quantum mechanics so we will first discuss rotations in quantum mechanics then we can go to the spherical basis which allows us to rigorously defined spherical tensor operators and after that we will investigate how to calculate components of spherical tensor operators from Cartesian components this video should have everything you need to know about this topic so without further ado here we go Part 1 tensor operators what characterizes a tensor is the way it transforms under rotations since we want to investigate tensor operators let&#39;s look at how rotations work in quantum mechanics where they are operators that act on States we distinguish between two different types of rotation active and passive ones in active rotations the state vector sy gets rotated using a unitary operator D whereas for passive rotations the operators get rotated in a certain way we can now make the following observation in one experiment we rotate the state vector and in another experiment we rotate our lab equipment that is the operator but in the opposite direction as you can see this has the same result so let&#39;s write this a little bit more mathematically the expectation value of this operator when the state gets rotated will be the same as the expectation value for when we rotate the operator in the opposite direction from this we can see that the operator which was rotated in the opposite way is given by D dagger o D and if we look for the opposite rotation we get D oh D dagger since the unitary operator D for a rotation and the other way is just the dagger a classical vector VI transforms under rotation as R IJ VJ in quantum mechanics the same relations hold for the expectation value of a vector operator however we also know from our previous calculation that this is equivalent to rotating the vector operator itself but in the opposite way therefore D dagger VI D is equal to RI J VJ note here that the unitary operator D acts in the Hilbert space whereas RI J acts in coordinate space now that we know how accurate agent vector operator transforms under rotations we can generalize this to a Cartesian tensor operator which just gets more copies of the rotation matrix R but again all of this was in the Cartesian basis let us now change to the spherical basis part 2 spherical basis the spherical basis is defined such that the transformation behavior under rotations is the same as how spherical harmonics transform under rotations before we consider general tensors let&#39;s see how the basis vectors in the spherical basis look like the spherical harmonics Y 1 1 Y 1 0 and y 1 -1 are proportional to x y&amp;z therefore we make the following anzats that the spherical basis vectors are given in terms of the Cartesian basis vectors like this in order to determine the proportionality factors alpha beta and gamma we use the condition that the basis vectors should be orthonormal this is a rather straightforward calculation however we have to keep in mind that since we are now dealing with a complex basis we have to perform a complex conjugation on one of the vectors in an inner product the absolute value of alpha beta and gamma can be obtained rather quickly and concerning their face we choose it the same way as in the spherical harmonics this means a minus sign for alpha and a plus sign for beta and gamma using these three basis vectors we can now construct vectors in the spherical basis let&#39;s take the vector a and write it in terms of its components strictly speaking those are contravariant components since their index is upstairs the distinction between contravariant and covariant components will become very important later so let&#39;s do this carefully if we write our spherical basis in terms of the Cartesian basis we can read off the Cartesian components a X a Y and a set by inverting these relations we can define the contravariant spherical components of a vector like this now what about the covariant components the covariant vector components have their index downstairs and the basis vectors now have an upper index this is because they are elements of the dual basis which is defined like this but we already know a relation of basis vectors which yields a Kronecker Delta using the complex conjugate basis vectors therefore the dual basis vectors are just the complex conjugated versions of the original basis vectors using this information we can also calculate the covariant components of a spherical vector finally we get the following relation between covariant and contravariant components of a spherical vector before we continue here are some additional useful relations concerning the spherical basis first we have the following relations for the cross-product of basis vectors second let&#39;s talk about the inner product of two vectors in the spherical basis if a is a contravariant vector and B is a covariant vector we get the following relation since we know how to switch between covariant and contravariant components we can also write it like this and using a sum over Q like this but enough about vectors what about general tensors a spherical tensor and transforms in the same way as spherical harmonics do so we first need to know how spherical harmonics transform under rotations a spherical harmonic is an eigenstate of the angular momentum operator projected onto the coordinate basis note that this is slightly different from ylm of theta and Phi which are defined as LM projected onto a unit length basis and the difference is just a factor of R to the power of L which ylm we use does not matter for us since both transform in the same way so we just used the first one after performing a rotation we get yl m of r prime which is defined like this our prime and r are related by a unitary transformation d so we get d dagger here next we put in a completeness relation with L Prime and M Prime the first term here is why L prime M prime and the second one is a matrix element of the unitary transformation operator d note that since we have a dagger here we can write this as the undergird version of d for a rotation r inverse since this unitary rotation will not stretch or compress anything L will be the same as L prime so we get Delta L L Prime the remaining matrix element is called the vignati matrix this D stands for - tell on the German word for representation the indices of d R M prime m since we have M Prime on the left and M on the right and since we have the matrix element of D with our inverse we denote this and brackets let&#39;s see what happens for a rotation and the opposite way on the left hand side we get d ylm T dagger and on the right hand side we get the via D matrix for our there is one more way in which we can write down this transformation behavior starting with the definition of the vegan er D matrix we can first investigate what happens for an inverse rotation but also what happens if we take the hermitian conjugate of the whole equation since the left-hand side is just a number we just get a complex conjugate but the right-hand side gets conjugated properly if we now compare these two terms we can make a connection between the complex conjugated vigener d matrix and the Vigna d matrix for an inverse rotation so finally the transformation behavior of spherical harmonics under rotations can be written in these two equivalent ways now comes the highlight of this video we define a spherical tensor operator tkq as a set of 2 k plus 1 operators that transform under rotations exactly like the spherical harmonics did compare this to the definition of a cartesian tensor operator where the number of rotation matrices is determined by the rank of the tensor whereas for a spherical tensor operator we only have one big nerdy matrix using the definition of the unitary rotation operator we can write down an infinitesimal version of the first relation which leads us to the following commutator relations the detailed calculations how to arrive at these commutators are rather long and since this video is already quite long we refer here to one of our references in the description we can further simplify the first two commutator relations using the definition of ladder operators which results in another commutator relation note that both the definition using the vignette D matrices as well as the definition using commentators are equivalent both completely define spherical tensor operators last but not least let us talk about hermitian conjugate of spherical tensor operators if we take the hermitian adjoint of the commutator relations we get the following two new commutator relations in terms of T dagger you can easily verify that if we define the taggert version of t KQ like this we get back our original commutator relations the value of a is arbitrary and sometimes set to 0 also whether we use plus or minus q in the exponent is up to us as long as we are consistent in our choice we now have a good definition of spherical tensor operators but how do we actually calculate them say if we have a rank 2 cartesian tensor tij what are the components of TK Q what is K and what is Q this will be answered in the final part of this video part 3 examples to conclude this video will show you how to switch from Cartesian tensor components to spherical tensor components first a-rank zero tensor that is a scalar is the same in both coordinate systems next a Rank 1 tensor which is a vector we already talked about spherical vector operators but one question remains which is the real spherical tensor component t11 the contravariant or the covariant component to solve this we can use the commutation relations in particular that the commutator of J plus with t11 should be 0 since T 1 1 already is the highest state in its multiplet the calculation is not too difficult and the result is that only the covariant vector components are spherical tensor operators as indicated by their index this makes sense since the covariant vector components are those that transform similar to the basis vectors and those were constructed from the spherical harmonics but don&#39;t be confused a plus is still a spherical tensor operator however it is equal to minus a minus so a plus is the negative of 81 minus 1 spiracle tensor operator even though the index S Plus next onto rank two tensors here it&#39;s not obvious how to calculate tkq from t IJ we know that T IJ has 3 times 3 so 9 components if they form an irreducible representation we should represent those nine components as the nine components of T for Q where Q can take on nine values from minus four to four however a rank two tensor is reducible so this is wrong instead we can use a trick to deal with rank two tensors and in fact all higher rank tensors this trick is to write T IJ as AI BJ so as the direct product of to rank want answers since both a and B are tensors T IJ is guaranteed to transform as a tensor as well so how does this help us since spherical tensor operators are basically spherical harmonics we can couple them like angular momenta using clebsch-gordon coefficients we already covered clebsch-gordon coefficients in a different video so if you&#39;d like you can watch our video on that topic still here is a very quick run-through of clubs gordan coefficients starting with a state of definite angular momentum L M we can insert a completeness relation to expand this state in the product state L 1 m 1 l 2 m 2 the bracket element here is precisely a cleped gordon coefficient we can write the same relation in the coordinates based spaces which tells us how to couple spherical harmonics and in the very same way we can couple to spherical tensor operators now back to our example of a rank two tensor since a and B are Rank 1 K 1 and K 2 are equal to 1 this means we need the following clebsch-gordan table the notation is such that the coupled state is on top and the product states are on the left as you can see from this table if we couple to Rank 1 spherical tensors we can get a rank 2 a rank 1 and even a rank 0 tensor as a result the components agree we have five components from a rank two tensor three from a rank one tensor and one from a rank zero tensor which all add up to nine components as an example let&#39;s calculate the t10 component of the Cartesian tij tensor according to the table this is equal to a11 b1 minus 1 times the square root of 1/2 plus a1 minus 1 B 1 1 times minus the square root of 1/2 it is important to use the covariant components of a and B here as we discussed earlier the result is given by I over the square root of 2 times a X B Y minus a YB X as a final step we change back to T IJ which leads to this result and there you have it this is how you start with a Cartesian tensor operator tij and change the basis to a spherical tensor operator tkq which K and Q are possible is completely decided by the clebsch-gordan table the other components look like this and can be calculated in exactly the same way note that the rank 0 tensor is proportional to the trace of T IJ the rank 1 tensor is proportional to anti-symmetric components and the rank two tensor is proportional to symmetric components this is due to the fact that we can split up a general cartesian rank two tensor into a diagonal part proportional to its trace a part that is anti-symmetric and a part that is symmetric and traceless finally higher-ranked answers can be calculated in exactly the same way for a rank 3 tensor T ijk you could write it as a IJ BK and use the results from the rank 2 case in order to couple a and B and that&#39;s pretty much it for this video thanks for watching

Transcript for:Understanding Spherical Tensor Operators

Transcript for:
Understanding Spherical Tensor Operators