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Symmetric wave functions

by aaaa202
Tags: functions, symmetric, wave
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aaaa202
#1
Feb10-13, 08:13 PM
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For multibosonic systems, as I understand, the wave function must always be symmetric (antisymmetric for fermionic, which this question easily generalizes to).
But as far as I can see for N>2 you can easily construct alot of other wave functions which are symmetric rather than the one my book finds (which essentially is the slater determinant one with + instead of -),if you allow for the fact that each product only contains product of two wave functions.

Say for instance you have 3 particles with wavefunction a1, b2, c3

Then we could choose:

ψ = a1b2 + a2b1 + a1b3 + a3b1 + (combination of a and c, combination of b and c in the same way)

The wave function above is invariant under any switch between the number 1,2 and 3 (which will represent the three coordinate sets for our bosons).
Why is it then, that a wave function of this kind is not acceptable?
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fzero
#2
Feb10-13, 09:22 PM
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You are constructing a wavefunction composed of states that have only 2 of the 3 particles present at a time. Your textbook is describing wavefunctions corresponding to having all 3 particles present at the same time.
aaaa202
#3
Feb11-13, 04:48 AM
P: 1,005
why is my ones necessarily only describing 2 particles present? Yes, each product only contains two wavefunctions, but all three particles' wavefunctions are included in the product as a whole.

Bill_K
#4
Feb11-13, 06:15 AM
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Symmetric wave functions

You're writing ψ as a coherent superposition of states, with a probability amplitude for being in each state. In the first two states, particle 3 does not exist. In the third and fourth state, particle 2 does not exist, and so on. This is a clear contradiction, each state must contain the same particles.
aaaa202
#5
Feb12-13, 02:17 AM
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why exactly is that a contradiction? For a three bosonic system the wave function must be symmetric. I don't see any loss of generality in dropping out on one of the particles in one as long as its compensated for in another of the terms.
Maybe you can elaborate on that superposition thing, as far as I understood the symmetric formula with the slater determinant (switch minuses to plus) is not a superposition, its a requirement that the wave function for an n-bosonic system satisfies it.
DrClaude
#6
Feb12-13, 02:36 AM
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Try calculating [itex]\langle \psi | \psi \rangle[/itex] for your wave function. I think you will see why it is not acceptable.
cgk
#7
Feb12-13, 04:25 AM
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OP, a single permanent wave function (i.e., the symmetrized orbital product) is *not* the general bosonic wave function, just as a single Slater determinant is not a general fermionic wave function. However, general bosonic wave functions can be decomposed into linear combinations of permanents. If |phi_i> is a basis of the one-particle space, then the set of all permanents that can be formed from the |phi_i> is a basis of the Fock space. That is, all bosonic wave functions can be decomposed into sums of permanents, but in general not into a single one.

If you are familiar with electronic structure: Choosing a single permanent is analogous to the Hartree-Fock approximation, decomposing a wave function into sums of permanents is analogous to the configuration interaction (CI) method.
aaaa202
#8
Feb13-13, 08:18 PM
P: 1,005
Can you elaborate please? Isn't it true no matter what that for an n-bosonic or fermionic system the wave function describing it must NO MATTER WHAT be symmetric or antisymmetric and that can only be done (apparantly) with the formulas provided by the slater determinant.
DrClaude
#9
Feb14-13, 03:51 AM
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Indeed, the wave function for an ensemble of identical bosons (fermions) must be symmetric (anti-symmetric) with respect to the interchange of two particles.

The Slater determinant is a simple method to construct a wave function for n fermions that ensures that the properties of the fermionic wave functions are satisfied. But there are many Slater derterminants that you can write for the same nfermion system. An keep in mind that a Slater determinant only produces a wave function that is a linear combination of products of one-fermion wave functions, which in itself is an approximation.
aaaa202
#10
Feb15-13, 06:54 PM
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What you mean it can produce more than one antisymmetric combination? And what do you mean by the approximation?
DrClaude
#11
Feb16-13, 06:59 AM
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When you write a multi-electron wave function as a Slater determinant, you are by construction writing an n-electron wave function as a product of n one-electron wave functions, and that is an approximation. The real n-electron wave function will be a complicated function of the coordinates of all electrons.

Also, you can write multiple Slater determinants for n electrons by choosing different one-electron wave functions to put in the derterminant.

Take for example He. You can construct the Slater determinant ψA using 1sα and 1sβ. You can also construct it using 1sα and 2sβ (ψB), or 1sα and 2pβ (ψC), etc. You can then take as the actual wave function a linear combination of these Slater derterminants, ψ = cA ψA + cB ψB + cC ψC. This is in essence what is called configuration interaction.


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