# Momentum in an Infinite Square Well

## Main Question or Discussion Point

Does the momentum space of a particle in an infinite square well have spikes or is it continuous?

I've heard so many differing opinions on this.
I go online and heaps of websites say that you just do a fourier transform on position to get the momentum, and if you go through with this you'll get a continuous momentum space. However, other sources say you can't do a Fourier transform and instead you must look at the superposition of free particle wavefunctions, leading to spikes in the momentum space.

So could I please get help in reaching a definite answer to this conundrum?

Thanks

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Meir Achuz
Homework Helper
Gold Member
If the wave function is that of a bound state, there will be delta functions in the FT of the sin or cos functions.

DrDu
The momentum operator is the generator of translations. As the wavefunction has to vanish outside the box and a translation of an allowed wavefunction will lead to a non-allowed wavefunction generically, the momentum operator doesn't exist for the infinite square well problem.

Hans de Vries
Gold Member
Does the momentum space of a particle in an infinite square well have spikes or is it continuous?

I've heard so many differing opinions on this.
I go online and heaps of websites say that you just do a fourier transform on position to get the momentum, and if you go through with this you'll get a continuous momentum space. However, other sources say you can't do a Fourier transform and instead you must look at the superposition of free particle wavefunctions, leading to spikes in the momentum space.

So could I please get help in reaching a definite answer to this conundrum?

Thanks

The Fourier transform of a wave function multiplied by a rectangular function, is equal to the convolution of the Fourier transform of the unbounded function
(which consists of deltas) with the sync function: http://en.wikipedia.org/wiki/Fourier_transform

So. You'll get sync functions as the Fourier transform.

------------

If you look at the instantaneous momentum, or the current density,
which is proportional to,

$$i\left(~\frac{\partial \psi^*}{\partial x} \psi ~-~ \psi^* \frac{\partial \psi}{\partial x}~\right)$$

then you'll get zero everywhere because there are two currents
with opposite directions which cancel. The instantaneous (local)
momentum of each individual current at any point within the
rectangular well is the one corresponding with the quantized
wavelength.

Regards, Hans

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Thanks, everyone :)

Hi.
So. You'll get sync functions as the Fourier transform.
No, sinc function is fourier transform of rectangular function. We should calculate Fourier transform of the wave function. i.e.
Asin ax for 0< x < pai/a,
0 for others

In Fig 3.2 of http://www.ecse.rpi.edu/~schubert/C...um mechanics/Ch03 Position&momentum space.pdf , you find calculated probability amplitude in momentum space.
The formula of probability amplitude in momentum space are (3.22) and (3.23).
Regards.

Last edited by a moderator:
Hans de Vries
Gold Member
Hi.

No, sinc function is fourier transform of rectangular function. We should calculate Fourier transform of the wave function. i.e.
Asin ax for 0< x < pai/a,
0 for others

In Fig 3.3 of http://www.ecse.rpi.edu/~schubert/C...um mechanics/Ch03 Position&momentum space.pdf , you find calculated probability amplitude in momentum space.
The formula of probability amplitude in momentum space are (3.22) and (3.23).
Regards.

The formula (3.22) and (3.23) are both build from two Sync functions resulting
from the the convolution of two Dirac pulses with the Sync function.

Check and you'll see that the momentum p is in the denominator shifted away
from zero by the same distance as the Dirac pulses are shifted away from zero.

The momentum space image in fig. 3.2. could me "more accurate". What looks
like Gausians in the image should really be Sync functions.

Also, the width of the pulses should not depend on the frequency, (not on n)
but solely on the the width of the infinite well (L)

Regards, Hans

Last edited by a moderator:
Hi. Thanks Hans.
I am glad to know we are sharing the same view. Now I know that there are at least three views on this problem. Please take a look at another recent thread: eigenvalue of momentum for particle in a box
Regards.