Dear Victor,
Thank you again, but, there is an interesting thing, in your replay, you
mentioned that"The direction of the spin is given in spherical
coordinates,
i. e. theta is the angle between the spin and the z axis "you know that x,
y,and z axis is identical for the systerm. How can I determine a referened
axis to build the spherical coordinates? I guess that the referenced axis
is the line butween atom1 and atom2 ? Does it right? But for atom2 and
atom3, how can I do?
I can not understand it, so, I have to trouble you again beacuse I want
to get a fully understand of this question and I think over it for longer
time. Thank you.
Your's Sincerely: Xianlong
> Dear Xianlong,
>
> The direction of the spin is given in spherical coordinates, i. e. theta
> is the angle between the spin and the z axis (it varies between 0 and
> 180 degrees) and phi is the angle between the projection of the spin on
> the xy plane and the x axis (it varies between 0 and 360 degrees)
>
> For example, for spins along z, x and y (1, 2 and 3):
>
> 1 1.0 0.0 0.0
> 2 1.0 90.0 0.0
> 3 1.0 90.0 90.0
>
> Regards,
>
> Victor
>
>
> -------------------------------------------------------------
> Dr. Victor Manuel Garcia Suarez
> Research Assistant | Tlf: 0044 - (0)1524 593 995
> Physics Department | Fax: 0044 - (0)1524 844 037
> Lancaster University | e-m: [EMAIL PROTECTED]
> | [EMAIL PROTECTED]
> Lancaster LA1 4YB |
> United Kingdom |
> www: http://condmat.uniovi.es/victor
> -------------------------------------------------------------
>
>
>
> -----Original Message-----
> From: Siesta, Self-Consistent DFT LCAO program, http://www.uam.es/siesta
> on behalf of John B. Baba Sent: Wed 12/20/2006 11:25
> To: SIESTA-L@listserv.uam.es
> Subject: Re: [SIESTA-L] NonCollinearSpin
>
> Dear Victor,
> First, thank you for your reply. Just as you say, I have test a atom
> line
> just now and find that the results are not infulenced by the direction
> of the atom line.
> But if I want to do a NonCollinearSpin calculations about a atom line,
> how can I determine the theta and phi? In other words, how can I use the
> theta and phi to determine the relative spin between theose atoms. Can
> you give me a modle about a system contained two atoms.
> In my mind, if there is not a fixed direction, it is difficult to use
> the theta and phi beacuse I do not know what they are meaning.
> Please help me, I am pullzed by this question and interesting about
> it.
> %block DM.InitSpin
> 1 + theta phi
> 2 + theta phi
> %endblock DM.InitSpin
> Yours Sincerely Xianlong
>> Dear Xianlong,
>>
>> If there is no spin orbit the direction of the spin is independent of
>> the spatial orientation. What only matters is the relative orientation
>> between different spins. I think that the possibility of including
>> spin orbtit will be included in the official release of Siesta soon
>> but in the meanwhile you may talk to Jaime Ferrer
>> ([EMAIL PROTECTED]) about it.
>>
>> Victor
>>
>> -------------------------------------------------------------
>> Dr. Victor Manuel Garcia Suarez
>> Research Assistant | Tlf: 0044 - (0)1524 593 995
>> Physics Department | Fax: 0044 - (0)1524 844 037
>> Lancaster University | e-m: [EMAIL PROTECTED]
>> | [EMAIL PROTECTED]
>> Lancaster LA1 4YB |
>> United Kingdom |
>> www: http://condmat.uniovi.es/victor
>> -------------------------------------------------------------
>>
>>
>>
>> -----Original Message-----
>> From: Siesta, Self-Consistent DFT LCAO program,
>> http://www.uam.es/siesta on behalf of John B. Baba Sent: Wed
>> 12/20/2006 07:02
>> To: SIESTA-L@listserv.uam.es
>> Subject: [SIESTA-L] NonCollinearSpin
>>
>> Dear All:
>> I am a new SIESTA user. I am puzzled by the question about the
>> SIESTA
>> funtion of NonCollinearSpin.
>> My question: When I set one atom with plus spin in the input file,
>> but
>> I don't know which direction is the tolerent direction in SIESTA. If I
>> put atoms in line along the Z-axis and set them with a plus spin, then
>> which direction is the spins point at? Do they all point at z-dirction
>> or other directions?
>> Hope for your help. Thank you very much.
>>
>> Sincerely: Xianlong Wang
