Symmetric and Antisymmetric Wavefunction

[fog37]

Hello,

My understanding is that, for a multi-particle system, the overall wavefunction HAS to be either symmetric or antisymmetric. A wavefunction that is neither symmetric or antisymmetric must be converted into one that is one of the two types depending on the type of particles. For example, if the particles are fermions, the wavefunction MUST be antisymmetric. If the particles are bosons, the wavefunction must be symmetric.

• Is this true for both non interacting and interacting particles of the same type? What if the particles were interacting? Would the antisymmetric/symmetric argument still hold?
• What if some of the particles were bosons and some were fermions? What type of total wavefunction would be required?

The total wavefunction must include the spatial part and a spin part: $\Psi(r,t,s) = \Phi(r,t) \chi (s)$ and this is true also for multi-particle systems. If the overall wavefunction must be antisymmetric, it is possible for the spatial part to be symmetric while the spin part is antisymmetric.

• In the previous posts, when discussing entanglement for two particles, we mention that a joint two-particle state that is separable, i.e. the product of two functions, is not entangled. Is that separable wavefunction symmetric, antisymmetric or neither? Bt isn't a product solution not an acceptable solution since if we interchanged the position variables $x_1$ and $x_2$ the pdf that we would obtain would change when it should instead remain the same since quantum particles are indistinguishable?

Thanks for any comments.

 

[PeterDonis]

My understanding is that, for a multi-particle system, the overall wavefunction HAS to be either symmetric or antisymmetric.

This is true for multi-particle systems in three dimensions. However, it is not true for some systems that are under constraints so the effective number of dimensions is fewer. For example, there are systems in which the degrees of freedom are effectively restricted to two dimensions, where these degrees of freedom can be "anyons", which don't have to exhibit boson or fermion (i.e., symmetric or antisymmetric) statistics. See here for a quick overview:

https://en.wikipedia.org/wiki/Anyon

That said, for purposes of this discussion, we probably want to restrict attention to three-dimensional systems where only boson and fermion statistics are possible.

Is this true for both non interacting and interacting particles of the same type?

It's not a matter of particle "type". It's a matter of the total spin of the whole system. If the total spin is an integer, the system as a whole is a boson and its wave function is symmetric. If the total spin is a half integer, the system as a whole is a fermion and its wave function is antisymmetric.

What if some of the particles were bosons and some were fermions?

Same answer here; you look at the total spin of the whole system.

It might help to stop yourself from thinking of the constituents of the system as "particles" with particular statistics, and just think of them as degrees of freedom of the system as a whole.

If the overall wavefunction must be antisymmetric, it is possible for the spatial part to be symmetric while the spin part is antisymmetric.

Yes.

a joint two-particle state that is separable, i.e. the product of two functions, is not entangled. Is that separable wavefunction symmetric, antisymmetric or neither?

Everything I said above about states of a multi-particle system applies whether the state of the system as a whole is separable or entangled. But if the state is separable, it obviously becomes simpler to analyze because it is a single product of functions. The key question is: what would "particle exchange" correspond to if the state was separable? If you can answer that, you should be able to answer the question you ask above.

 

[vanhees71]

It's important to stress that what you discuss is a multi-particle system consisting of indistinguishable particles!

Another caveat concerns composite objects like an atom, which can be a boson if its total angular momentum (or spin for that matter) is an integer number (0, 1, 2,...). Then it behaves principly like a boson, but only if you don't look too close, i.e., if you don't investigate it with too high energy. If you do so you resolve the constituents which are fermions (protons, neutrons, and electrons), and this fermionic nature of the constituents leads to deviations from the behavior of an elementary bosonic particle/quantum (like photons and Higgs bosons according to the Standard Model).

 

[blue_leaf77]

It's not a matter of particle "type". It's a matter of the total spin of the whole system. If the total spin is an integer, the system as a whole is a boson and its wave function is symmetric. If the total spin is a half integer, the system as a whole is a fermion and its wave function is antisymmetric.

I think we need to be careful here especially in the use of the term "system". For instance if we take as an example of the "whole system" to be the electrons in an atom whose total spin is an integer, the wavefunction of the whole system i.e. the system of all electrons inside this atom will still be antisymmetric. This is because the constituents of the system, which are electrons, follow fermionic statistic. If however we put an ensemble of this atom into an isolated system so that now the whole system is a collection of identical atoms each of which has integer spin, then the wavefunction of this big system is approximately (in view of vanhees comment above) a bosonic system. Therefore, the sentence "It's a matter of the total spin of the whole system" would be better off amended to "It's a matter of the total spin of the constituent/sub system".

 

[fog37]

Hello again,

If the two identical particles composing the system are non-interacting, the composite wavefunction must be separable, i.e. it becomes the product of two wavefunctions (one for each particle): $$\Psi_{total} =\phi_1 \beta_2$$

where $\phi_1$ is the state of particle 1 and $\beta_2$ is the state of particle 2. Since the particles are identical, the exchange should not change the physics:

$$\Psi_{total} =\phi_2 \beta_2$$

but the products $\phi_1 \beta_2$ and $\phi_2 \beta_2$ are actually different and give rise to different probability distribution functions $|\Psi|^2$. That is why we need to create either a symmetric or an antisymmetric function to make sure that $|\Psi|^2$ does not change if we exchange the particles and put one in the state of the other.

Where does the word symmetric come from? In 1D, a simple function $f(x)$ is symmetric, or even, if it is reflected about some axis. In 2D or 3D, it becomes harder, I guess, to define a symmetric or antisymmetric function. What type of symmetry or antisymmetry are we talking about in this quantum mechanical context? Is there some axis about which the composite wavefunction is symmetric about?

Thanks!

 

[PeterDonis]

If the two identical particles composing the system are non-interacting, the composite wavefunction must be separable

Not if the particles are bosons or fermions. "Non-interacting" just means there is no interaction term in the Hamiltonian for the system; it does not require that the state be separable.

Since the particles are identical, the exchange should not change the physics:

What you wrote isn't an exchange; that would be $\phi_2 \beta_1$ or something like that, possibly with a coefficient. However, that's a minor point compared with the above.

 

[fog37]

Thanks PeterDonis,

Does it automatically means that the particle's state are entangled if there is an interaction term in the Hamiltonian?

 

[blue_leaf77]

Where does the word symmetric come from?

In this context, (anti)symmetry with respect to particle exchange.

 

[PeterDonis]

Does it automatically means that the particle's state are entangled if there is an interaction term in the Hamiltonian?

No.

 

[vanhees71]

Yes and no!

Yes, indistinguishable particles (I don't like the term "identical" in this context, which has no precise meaning) are always in entangled states, i.e., they are never in product states but in symmetrized (bosons) or antisymmetrized (fermions) superpositions of product states.

No, also non-interacting particles can be entangled. E.g. when a neutral pion decays into two photons their polarization states are entangled to get total angular momentum 0. To a very good approximation you can consider the two photons as non-interacting, but still they are entangled!

 

[fog37]

Ah! Thank you vanhees71.

I always thought that the presence of an interaction was a requirement for entanglement to take place. I guess not. So the electrons in the our body and in ordinary objects are in entangled states. If so, we don't need to go far to look for entanglement.

So, particles can be interacting, non interacting, dependent or independent. Would the case of two particles that are non-interacting but dependent imply entanglement?

 

[vanhees71]

It's a very important point that entanglement is not due to an interaction but due to the particular way how a state of a quantum system is described in the formalism of quantum theory. The entanglement describes a particular "quantum kind" of correlations, and it is utmost important to understand that correlations are not necessarily related by cause and effect as interactions. That's, why there is no "spooky action at a distance" involved in the possibility of the correlations described by entanglement. It's well worth to think very hard about this, because it's the most mind bogling point of QT and it's the very point where QT is radically different from classical deterministic physics.