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Bosonic condensation in 1 spatial dimension

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JK423
#1
Dec8-12, 09:31 AM
P: 381
Hello guys,

I'm trying to understand bosonic condensation and i would really need your help.
The actual question, before getting into the details is:
"Why Bose-Einstein Condensation doesn't take place in a one-dimensional (1D) space?"

In what follows i'll give you my own (mis)understanding so that you will be able to see for any mistakes in my line of reasoning. I'll try to show, from what i have understood, that a BEC in 1D is possible.
Consider that i have a bosonic gas in a 1D infinite square well, hence with a non-zero ground state. The conservation of the total mean particle number is given by the condition
[itex]N = \sum\limits_i {\frac{1}{{{e^{\left( {{E_i} - \mu (T)} \right)/{k_B}T}}}-1}} \ \ \ (1) [/itex]
from which the chemical potential is determined. For simplicity choose e.g. the mass of the particles suitably so that i can write (1) as
[itex] N = \sum\limits_{n = 1}^\infty {\frac{1}{{{e^{\frac{1}{T}\left( {{n^2} - a(T)} \right)}}}-1}} \ \ \ \ (2)[/itex]
with [itex] - \infty < a(T) = \frac{\mu (T) }{{{k_B}}} \le {E_{ground\_state}}=1 \ \ \ (3)[/itex].
To my understanding, condensation occurs if there is a finite temperature below which the sum of the (mean) number of particles in all the excited states cannot reach [itex]N[/itex], i.e.
[itex] \sum\limits_{n > 1}^\infty {\frac{1}{{{e^{\frac{1}{T}\left( {{n^2} - a\left( T \right)} \right)}}}-1} \le N\,\,{\rm{for}}\,\,\,T \le {T_c}} \ \ \ \ \ (4)[/itex],
and start occupying the ground state.
Well, there is such a critical temperature even for a one-dimensional system! In particular, i did the sum (4) numerically for [itex]N = 10^3 [/itex], and found that for temperatures [itex] {T \le {T_c} = 1397}[/itex] which correspond to [itex]a\left( {{T_c}} \right) \le 1 [/itex] the sum (4) -that excludes the ground state- cannot reach [itex]N[/itex] since [itex]a[/itex] cannot be larger than 1.

Concluding, i gave you an example of 1D bosonic condensation without the need to go to the continuum and work with integrals etc as is usually done. So i have made no reference in the density of states etc, i just used the sum to find the critical temperature numerically.
Now, since i know that 1D condensation is not possible for reasons that i cannot understand, can you please tell at which point the reasoning above is mistaken?

Thank you in advance for your help!
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Cthugha
#2
Dec8-12, 11:58 AM
Sci Advisor
P: 1,624
There is no BEC in fewer than three dimensions for homogeneous systems (as assumed in the "historic" derivations when taking the thermodynamic limit) at finite temperature. As there will be phonon-like Goldstone modes with arbitrarily low energy for spontaneously broken continuous symmetries and the creation of these is favored, you end up with long-range fluctuations killing your condensate.

The situation is obviously different in a nonhomogeneous trapped system of lower dimensionality as the trap means you have introduced a minimal energy scale for excitations. Depending on what school of theory you follow this allows a BEC or a similar-to-BEC phase transition to happen. There are of course differences. There is obviously no spontaneous breaking of a continuous, but a discrete symmetry for example.
JK423
#3
Dec8-12, 12:17 PM
P: 381
Cthugha thank you for your answer!

Are you sure that the ground-state being non-zero makes the difference? Forget about the trap setting, even if i allow for zero ground energy my conclusions remain the same since
[itex] {N_{excited}} = \sum\limits_{n > 0}^\infty {\frac{1}{{{e^{\frac{1}{T}\left( {{n^2} - a\left( T \right)} \right)}}}-1} \le N = {{10}^3}\,\,\,\,{\rm{for}}\,\,\,\,\,T \le {T_c} \approx 66} [/itex]
since [itex] a\left( T \right) [/itex] cannot be larger than zero. The only thing that changed in this case is the critical temperature getting smaller.

Any ideas?

Edit:
I just wanted to note that the ordinary arguments against 1D condensation involve the energy density of states, since the number of excited particles get to infinity when the chemical potential approaches zero (i assume zero ground energy). And this infinity is the reason why they say no condensation happens. But If you look closer, the infinity is coming from the ground state (E=0).. I think that i have really misunderstood something quite badly..

Cthugha
#4
Dec8-12, 12:25 PM
Sci Advisor
P: 1,624
Bosonic condensation in 1 spatial dimension

Quote Quote by JK423 View Post
Are you sure that the ground-state being non-zero makes the difference?
No, it is not about the ground-state being zero or not, but the question whether the energy difference to the first excited state can be made arbitrarily small. In your example it cannot (if I get your terminology right). In homogeneous systems it can.
Mute
#5
Dec8-12, 12:45 PM
HW Helper
P: 1,391
Quote Quote by JK423 View Post
Hello guys,

I'm trying to understand bosonic condensation and i would really need your help.
The actual question, before getting into the details is:
"Why Bose-Einstein Condensation doesn't take place in a one-dimensional (1D) space?"

In what follows i'll give you my own (mis)understanding so that you will be able to see for any mistakes in my line of reasoning. I'll try to show, from what i have understood, that a BEC in 1D is possible.
Consider that i have a bosonic gas in a 1D infinite square well, hence with a non-zero ground state. The conservation of the total mean particle number is given by the condition
[itex]N = \sum\limits_i {\frac{1}{{{e^{\left( {{E_i} - \mu (T)} \right)/{k_B}T}}}}} \ \ \ (1) [/itex]
Why is your denominator ##\exp(\beta(E_i-\mu))## (where ##\beta = (kT)^{-1}##) rather than ##\exp(\beta(E_i-\mu))-1##, as it should be for the Bose-Einstein distribution?
JK423
#6
Dec8-12, 09:18 PM
P: 381
@Cthugha
Thank you, i will think about what you said.

@Mute
Yes you are right, i mistyped it. I edited it and now it's correct.


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