### Re: [Kwant] Units of the wavefunction and velocity

```Hi Harshad,

> I was able to obtain the velocities from the modes object. I confirmed
> that modes.velocities is just the inverse of integral |휓|2 over the unit
> cell. If I did the math correctly, I got the units of modes.velocities to be
> eV (energy units of my Hamiltonian). How can I convert this to lets say m/s?

Sorry that I forgot to clarify that the unit of length in this case is
the side of the lead unit cell. Multiplying by that size should allow
to easily translate your velocity into m/s.

Anton

```

### Re: [Kwant] Units of the wavefunction and velocity

```Can I just do 1/2 m*v^2 = modes.velocities, assuming parabolic dispersion?

On Mon, Jan 23, 2017 at 3:13 PM, Harshad Sahasrabudhe
wrote:

> Hi,
>
> You could do that, but this is already done in Kwant, and you can read the
>> velocities off from the modes object [1].
>
>
> I was able to obtain the velocities from the modes object. I confirmed
> that modes.velocities is just the inverse of integral |휓|2 over the unit
> cell. If I did the math correctly, I got the units of modes.velocities to
> be eV (energy units of my Hamiltonian). How can I convert this to lets say
> m/s?
>
> Thanks,
> Harshad
>
> On Mon, Jan 23, 2017 at 10:18 AM, Harshad Sahasrabudhe <
> hsaha...@purdue.edu> wrote:
>
>> Great, thanks a lot!
>>
>> On Mon, Jan 23, 2017 at 10:15 AM, Anton Akhmerov <
>> anton.akhmerov...@gmail.com> wrote:
>>
>>> > Thank you. Then to calculate the velocity, should I just divide the
>>> > probability current by the integral of |휓|2 over the unit cell?
>>>
>>> You could do that, but this is already done in Kwant, and you can read
>>> the velocities off from the modes object [1].
>>>
>>> Best,
>>> Anton
>>>
>>> [1]: https://kwant-project.org/doc/1/reference/generated/kwant.ph
>>> ysics.PropagatingModes#kwant.physics.PropagatingModes
>>>
>>
>>
>

```

### Re: [Kwant] Units of the wavefunction and velocity

```Hi,

You could do that, but this is already done in Kwant, and you can read the
> velocities off from the modes object [1].

I was able to obtain the velocities from the modes object. I confirmed that
modes.velocities is just the inverse of integral |휓|2 over the unit cell.
If I did the math correctly, I got the units of modes.velocities to be eV
(energy units of my Hamiltonian). How can I convert this to lets say m/s?

Thanks,
Harshad

On Mon, Jan 23, 2017 at 10:18 AM, Harshad Sahasrabudhe
wrote:

> Great, thanks a lot!
>
> On Mon, Jan 23, 2017 at 10:15 AM, Anton Akhmerov <
> anton.akhmerov...@gmail.com> wrote:
>
>> > Thank you. Then to calculate the velocity, should I just divide the
>> > probability current by the integral of |휓|2 over the unit cell?
>>
>> You could do that, but this is already done in Kwant, and you can read
>> the velocities off from the modes object [1].
>>
>> Best,
>> Anton
>>
>> [1]: https://kwant-project.org/doc/1/reference/generated/kwant.ph
>> ysics.PropagatingModes#kwant.physics.PropagatingModes
>>
>
>

```

### Re: [Kwant] Units of the wavefunction and velocity

```Great, thanks a lot!

On Mon, Jan 23, 2017 at 10:15 AM, Anton Akhmerov <
anton.akhmerov...@gmail.com> wrote:

> > Thank you. Then to calculate the velocity, should I just divide the
> > probability current by the integral of |휓|2 over the unit cell?
>
> You could do that, but this is already done in Kwant, and you can read
> the velocities off from the modes object [1].
>
> Best,
> Anton
>
> [1]: https://kwant-project.org/doc/1/reference/generated/kwant.
> physics.PropagatingModes#kwant.physics.PropagatingModes
>

```

### Re: [Kwant] Units of the wavefunction and velocity

```>
> Yes, that is correct. The equation that you quoted is indeed the probability
> current

Thank you. Then to calculate the velocity, should I just divide the
probability current by the integral of |휓|2 over the unit cell?

On Mon, Jan 23, 2017 at 10:00 AM, Anton Akhmerov <
anton.akhmerov...@gmail.com> wrote:

> > Sorry I made a mistake in the units of 'I'. Isn't 'I' here the
> probability current and not the charge current? We would just have to
> multiply 'I' by e to get the charge current, right?
>
> Yes, that is correct. The equation that you quoted is indeed the
> probability current.
>

```

### Re: [Kwant] Units of the wavefunction and velocity

```> Sorry I made a mistake in the units of 'I'. Isn't 'I' here the probability
> current and not the charge current? We would just have to multiply 'I' by e
> to get the charge current, right?

Yes, that is correct. The equation that you quoted is indeed the
probability current.

```

### Re: [Kwant] Units of the wavefunction and velocity

```Hi Prof. Akhmerov,

Sorry I made a mistake in the units of 'I'. Isn't 'I' here the probability
current and not the charge current? We would just have to multiply 'I' by e
to get the charge current, right?

Thanks,
Harshad

On Mon, Jan 23, 2017 at 9:53 AM, Anton Akhmerov  wrote:

> Dear Hardshad,
>
> I might be missing some factors here, but the unit of current is
> elementary charge / unit of time. The unit of time is hbar / unit of energy
> that you used in defining your tight-binding model.
>
> Best,
> Anton
>
> On Mon, Jan 23, 2017 at 3:49 PM, Harshad Sahasrabudhe  > wrote:
>
>> Hi,
>>
>> I was wondering what units would the wavefunction obtained from Kwant
>> have?  I was thinking they would have the units 1/sqrt(nm.eV) (since my
>> energies are in eV and lengths are in nm) as the modes are normalized
>> according to
>>
>> [image: Inline image 1]
>>
>> How would I calculate the velocity of the mode, if the modes are
>> normalized to carry unit current?
>>
>> Thanks,
>> Harshad
>>
>
>

```

### Re: [Kwant] Units of the wavefunction and velocity

```Dear Hardshad,

I might be missing some factors here, but the unit of current is elementary
charge / unit of time. The unit of time is hbar / unit of energy that you
used in defining your tight-binding model.

Best,
Anton

On Mon, Jan 23, 2017 at 3:49 PM, Harshad Sahasrabudhe
wrote:

> Hi,
>
> I was wondering what units would the wavefunction obtained from Kwant
> have?  I was thinking they would have the units 1/sqrt(nm.eV) (since my
> energies are in eV and lengths are in nm) as the modes are normalized
> according to
>
> [image: Inline image 1]
>
> How would I calculate the velocity of the mode, if the modes are
> normalized to carry unit current?
>
> Thanks,
> Harshad
>

```