Dear all, I have two simple questions concerning electronic transport with spins with no interaction: If I take the program spin_orbit.py from the tutorial and call the system sys=make_system ( alpha=0, ez=0, W=2, L=2) # no spin orbit interaction print matrix.round(kwant.smatrix(sys,energy=1.7).data , 2 ) I choose the energy E=1.7 because one mode is closed. the result is: [[0.000.j 0.000.j 0.41+0.43j 0.740.33j] [ 0.000.j 0.000.j 0.80+0.08j 0.52+0.28j] [ 0.41+0.43j 0.740.33j 0.000.j 0.000.j ] [0.80+0.08j 0.52+0.28j 0.000.j 0.000.j ]] What surprises me is that there is scattering between spins knowing that the Hamiltonians for the two spins are separated without interaction ! This seems to appear only when one mode is closed. To understand better and check this result, I rewrote the program using two lattices (as for electrons and holes). the program is attached with the Email. The result for the scattering matrix is [[ 0.000.j 0.16+0.99j 0.00+0.j 0.00+0.j ] [0.16+0.99j 0.00+0.j 0.00+0.j 0.00+0.j ] [ 0.00+0.j 0.00+0.j 0.00+0.j 0.16+0.99j] [ 0.00+0.j 0.00+0.j 0.16+0.99j 0.00+0.j ]] The result is now better because there is no scattering between spins. My Question is :what am I missing? The second question concerns the position of the coefficients in the scattering matrix. We have the Hamiltonian [image: H=h \otimes 1_{2\times2}] where [image: \otimes] is the Kronecker product and [image: h] the spinlessHamiltonian. The scattering matrix is given by : [image: S=1+2\pi i W^\dagger \frac{1}{EH\Sigma} W] [image: W] is the coupling matrix and [image: \Sigma] the self energy If we look at the simple system [image: ( L=2,W=2)] all the matrices are square and we can easily proove, just by the properties of the Kronecker poduct that the scattering matrix can be written as: [image: S=s\otimes 1_{2\times2}] Where [image: s] is the scattering matrix without spin. [image: s= \left(\begin{array} {c c} r & t \\ t^\prime & r^\prime \end{array}\right)] [image: t] is the transmission matrix. So I am expecting a scattering matrix with the followin form: [image: S= \left(\begin{array} {c c c c} r &0 & t & 0 \\ 0 &r & 0 & t \\ t^\prime &0 & r^\prime & 0 \\ 0& t^\prime & 0 & r^\prime \end{array}\right)] This is consistent with the result with two lattices but still the position of the coefficients are different from the one obtained from the Hamiltonian base. So I suggest that you are using different base. The question I ask is: in order to know the position of coefficients corresponding to different modes or different spins, which formula do you use to obtain the scattering matrix. is it a matrix inversion or some recursive method ? Thank you in advance. Regards, A. Abbout
Hi Abbout, The coefficients of the scattering matrix are just the amplitudes of modes in the lead. In presence of full spin rotation, any linear superposition of lead modes has the same momentum, and so Kwant chooses just some arbitrary basis that may be different for different leads. In this basis there is no guarantee that a certain mode will correspond to a certain spin expectation. Take a look at a closely related question http://thread.gmane.org/gmane.comp.science.kwant.user/77 Best regards, Anton Akhmerov On Sat, Mar 8, 2014 at 8:09 AM, Abbout Adel <abbout.adel@gmail.com> wrote:
Dear all, I have two simple questions concerning electronic transport with spins with no interaction:
If I take the program spin_orbit.py from the tutorial and call the system
sys=make_system ( alpha=0, ez=0, W=2, L=2) # no spin orbit interaction print matrix.round(kwant.smatrix(sys,energy=1.7).data , 2 )
I choose the energy E=1.7 because one mode is closed. the result is: [[0.000.j 0.000.j 0.41+0.43j 0.740.33j] [ 0.000.j 0.000.j 0.80+0.08j 0.52+0.28j] [ 0.41+0.43j 0.740.33j 0.000.j 0.000.j ] [0.80+0.08j 0.52+0.28j 0.000.j 0.000.j ]]
What surprises me is that there is scattering between spins knowing that the Hamiltonians for the two spins are separated without interaction ! This seems to appear only when one mode is closed.
To understand better and check this result, I rewrote the program using two lattices (as for electrons and holes). the program is attached with the Email. The result for the scattering matrix is
[[ 0.000.j 0.16+0.99j 0.00+0.j 0.00+0.j ] [0.16+0.99j 0.00+0.j 0.00+0.j 0.00+0.j ] [ 0.00+0.j 0.00+0.j 0.00+0.j 0.16+0.99j] [ 0.00+0.j 0.00+0.j 0.16+0.99j 0.00+0.j ]]
The result is now better because there is no scattering between spins.
My Question is :what am I missing?
The second question concerns the position of the coefficients in the scattering matrix.
We have the Hamiltonian [image: H=h \otimes 1_{2\times2}] where [image: \otimes] is the Kronecker product and [image: h] the spinlessHamiltonian. The scattering matrix is given by : [image: S=1+2\pi i W^\dagger \frac{1}{EH\Sigma} W]
[image: W] is the coupling matrix and [image: \Sigma] the self energy
If we look at the simple system [image: ( L=2,W=2)] all the matrices are square and we can easily proove, just by the properties of the Kronecker poduct that the scattering matrix can be written as: [image: S=s\otimes 1_{2\times2}]
Where [image: s] is the scattering matrix without spin.
[image: s= \left(\begin{array} {c c} r & t \\ t^\prime & r^\prime \end{array}\right)]
[image: t] is the transmission matrix.
So I am expecting a scattering matrix with the followin form:
[image: S= \left(\begin{array} {c c c c} r &0 & t & 0 \\ 0 &r & 0 & t \\ t^\prime &0 & r^\prime & 0 \\ 0& t^\prime & 0 & r^\prime \end{array}\right)]
This is consistent with the result with two lattices but still the position of the coefficients are different from the one obtained from the Hamiltonian base. So I suggest that you are using different base. The question I ask is: in order to know the position of coefficients corresponding to different modes or different spins, which formula do you use to obtain the scattering matrix. is it a matrix inversion or some recursive method ?
Thank you in advance.
Regards, A. Abbout
Dear Anton, Thank you for your answer. Actually my main question was about the fact that the transmission matrix contains 4 non vanishing elements knowing that there is no interaction in the system ( alpha=0 and ez=0) and one mode is closed (at E=1.7). In this case we expect the scattering from spin up to spin down ( and down to up) to vanish and therefore we should obtain only two non vanishing elements in the transmission matrix. The model with two lattices confirms this expectation and moreover, the scattering coefficients are different (beside their position). The model in the two programs is the same, it is just described differently. Best regards Adel Abbout On Sun, Mar 16, 2014 at 1:58 PM, Anton Akhmerov <anton.akhmerov@gmail.com>wrote:
Hi Abbout,
The coefficients of the scattering matrix are just the amplitudes of modes in the lead. In presence of full spin rotation, any linear superposition of lead modes has the same momentum, and so Kwant chooses just some arbitrary basis that may be different for different leads. In this basis there is no guarantee that a certain mode will correspond to a certain spin expectation. Take a look at a closely related question http://thread.gmane.org/gmane.comp.science.kwant.user/77
Best regards, Anton Akhmerov
On Sat, Mar 8, 2014 at 8:09 AM, Abbout Adel <abbout.adel@gmail.com> wrote:
Dear all, I have two simple questions concerning electronic transport with spins with no interaction:
If I take the program spin_orbit.py from the tutorial and call the system
sys=make_system ( alpha=0, ez=0, W=2, L=2) # no spin orbit interaction print matrix.round(kwant.smatrix(sys,energy=1.7).data , 2 )
I choose the energy E=1.7 because one mode is closed. the result is: [[0.000.j 0.000.j 0.41+0.43j 0.740.33j] [ 0.000.j 0.000.j 0.80+0.08j 0.52+0.28j] [ 0.41+0.43j 0.740.33j 0.000.j 0.000.j ] [0.80+0.08j 0.52+0.28j 0.000.j 0.000.j ]]
What surprises me is that there is scattering between spins knowing that the Hamiltonians for the two spins are separated without interaction ! This seems to appear only when one mode is closed.
To understand better and check this result, I rewrote the program using two lattices (as for electrons and holes). the program is attached with the Email. The result for the scattering matrix is
[[ 0.000.j 0.16+0.99j 0.00+0.j 0.00+0.j ] [0.16+0.99j 0.00+0.j 0.00+0.j 0.00+0.j ] [ 0.00+0.j 0.00+0.j 0.00+0.j 0.16+0.99j] [ 0.00+0.j 0.00+0.j 0.16+0.99j 0.00+0.j ]]
The result is now better because there is no scattering between spins.
My Question is :what am I missing?
The second question concerns the position of the coefficients in the scattering matrix.
We have the Hamiltonian [image: H=h \otimes 1_{2\times2}] where [image: \otimes] is the Kronecker product and [image: h] the spinlessHamiltonian. The scattering matrix is given by : [image: S=1+2\pi i W^\dagger \frac{1}{EH\Sigma} W]
[image: W] is the coupling matrix and [image: \Sigma] the self energy
If we look at the simple system [image: ( L=2,W=2)] all the matrices are square and we can easily proove, just by the properties of the Kronecker poduct that the scattering matrix can be written as: [image: S=s\otimes 1_{2\times2}]
Where [image: s] is the scattering matrix without spin.
[image: s= \left(\begin{array} {c c} r & t \\ t^\prime & r^\prime \end{array}\right)]
[image: t] is the transmission matrix.
So I am expecting a scattering matrix with the followin form:
[image: S= \left(\begin{array} {c c c c} r &0 & t & 0 \\ 0 &r & 0 & t \\ t^\prime &0 & r^\prime & 0 \\ 0& t^\prime & 0 & r^\prime \end{array}\right)]
This is consistent with the result with two lattices but still the position of the coefficients are different from the one obtained from the Hamiltonian base. So I suggest that you are using different base. The question I ask is: in order to know the position of coefficients corresponding to different modes or different spins, which formula do you use to obtain the scattering matrix. is it a matrix inversion or some recursive method ?
Thank you in advance.
Regards, A. Abbout
 Abbout Adel
Dear Abbout, The scattering coefficients are only different by a phase if you introduce two sublattices. Their absolute values must necessarily be identical because of spin rotation symmetry. The reason why the scattering matrix in the case with only one sublattice has nonvanishing elements is exactly what I wrote before: the modes calculated by Kwant are arbitrary superpositions of two spins, that are in addition different in different leads. You can directly verify that this is so by examining the lead_info attribute of the resulting scattering matrix. It contains the wave functions of the modes in the lead, and you will find that these wave functions are superpositions of two spins. Best regards, Anton Akhmerov On Mon, Mar 17, 2014 at 2:49 AM, Abbout Adel <abbout.adel@gmail.com> wrote:
Dear Anton,
Thank you for your answer. Actually my main question was about the fact that the transmission matrix contains 4 non vanishing elements knowing that there is no interaction in the system ( alpha=0 and ez=0) and one mode is closed (at E=1.7). In this case we expect the scattering from spin up to spin down ( and down to up) to vanish and therefore we should obtain only two non vanishing elements in the transmission matrix. The model with two lattices confirms this expectation and moreover, the scattering coefficients are different (beside their position). The model in the two programs is the same, it is just described differently.
Best regards Adel Abbout
On Sun, Mar 16, 2014 at 1:58 PM, Anton Akhmerov <anton.akhmerov@gmail.com> wrote:
Hi Abbout,
The coefficients of the scattering matrix are just the amplitudes of modes in the lead. In presence of full spin rotation, any linear superposition of lead modes has the same momentum, and so Kwant chooses just some arbitrary basis that may be different for different leads. In this basis there is no guarantee that a certain mode will correspond to a certain spin expectation. Take a look at a closely related question http://thread.gmane.org/gmane.comp.science.kwant.user/77
Best regards, Anton Akhmerov
On Sat, Mar 8, 2014 at 8:09 AM, Abbout Adel <abbout.adel@gmail.com> wrote:
Dear all, I have two simple questions concerning electronic transport with spins with no interaction:
If I take the program spin_orbit.py from the tutorial and call the system
sys=make_system ( alpha=0, ez=0, W=2, L=2) # no spin orbit interaction print matrix.round(kwant.smatrix(sys,energy=1.7).data , 2 )
I choose the energy E=1.7 because one mode is closed. the result is: [[0.000.j 0.000.j 0.41+0.43j 0.740.33j] [ 0.000.j 0.000.j 0.80+0.08j 0.52+0.28j] [ 0.41+0.43j 0.740.33j 0.000.j 0.000.j ] [0.80+0.08j 0.52+0.28j 0.000.j 0.000.j ]]
What surprises me is that there is scattering between spins knowing that the Hamiltonians for the two spins are separated without interaction ! This seems to appear only when one mode is closed.
To understand better and check this result, I rewrote the program using two lattices (as for electrons and holes). the program is attached with the Email. The result for the scattering matrix is
[[ 0.000.j 0.16+0.99j 0.00+0.j 0.00+0.j ] [0.16+0.99j 0.00+0.j 0.00+0.j 0.00+0.j ] [ 0.00+0.j 0.00+0.j 0.00+0.j 0.16+0.99j] [ 0.00+0.j 0.00+0.j 0.16+0.99j 0.00+0.j ]]
The result is now better because there is no scattering between spins.
My Question is :what am I missing?
The second question concerns the position of the coefficients in the scattering matrix.
We have the Hamiltonian where is the Kronecker product and the spinless Hamiltonian. The scattering matrix is given by :
is the coupling matrix and the self energy
If we look at the simple system all the matrices are square and we can easily proove, just by the properties of the Kronecker poduct that the scattering matrix can be written as:
Where is the scattering matrix without spin.
is the transmission matrix.
So I am expecting a scattering matrix with the followin form:
This is consistent with the result with two lattices but still the position of the coefficients are different from the one obtained from the Hamiltonian base. So I suggest that you are using different base. The question I ask is: in order to know the position of coefficients corresponding to different modes or different spins, which formula do you use to obtain the scattering matrix. is it a matrix inversion or some recursive method ?
Thank you in advance.
Regards, A. Abbout
 Abbout Adel
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Abbout Adel

Anton Akhmerov