# Wavefunction for n-particles

• A
Gold Member
Hi, in Bohm's "Quantum Theory" David Bohm writes:

for n-particles the wavefunction is:

G (_{N}) = Ae^{ip \eta /\hbar} + B e^{-ip \eta /\hbar}

But this is the same as a wavefunction in one dimension (x) given in Atkins and Friedman "Molecular Quantum Mechanics", just with a different variable:

\psi (x) = Ae^{ip x /\hbar} + B e^{-ip x /\hbar}

Unless I Have typed something wrong here, how does this come about? ##\eta## particles equal dimension x?

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hilbert2
Gold Member
The only explanation I can think of is that the ##\eta## is a vector containing the degrees of freedom of the ##n## particles as its components, and that ##p\eta## is actually a dot product of an ##n##-particle momentum vector with ##\eta##.

That kind of a wavefunction describes an energy eigenstate of a system with free non-interacting particles. In the general case, the wavefunction can be just about anything, but for non-energy-eigenstates the probability distribution doesn't remain constant in time.

SemM
Gold Member
So for a system with 4 non-interacting particles, the Bohm-variant would yield:

G (_{4}) = Ae^{i4p /\hbar} + B e^{-i4p /\hbar}

which simply has a more narrow oscillation than the 1-state (accounting for p be preserved)?

I take no probability density can be derived from the G(N) model of Bohm?

Does this in any case represent a 4-particle wavefunction in Hilbert space?

hilbert2
Gold Member
The ordinary letter n (meaning the number of particles) and the Greek letter ##\eta## (eta) are not the same thing. For 3 particles moving in three dimensions, the vectors ##\eta## and ##p## would need to have 9 components (x, y and z for each particle).

A problem with interpreting ##\eta## as a vector is that with several degrees of freedom there exist more "directions" than only the positive and negative direction that appear as the plus and minus signs in ##e^{\pm ip\eta /\hbar}## terms of the given wavefunction, so the solution here would not be the most general one. For example, if you make a combination of all plane waves with wavevector of absolute value ##k## and any direction, you get something like a Bessel function with ##r## (distance from origin) in its argument.

I don't have the Bohm's book available so I can't check what this is about.

SemM
Gold Member
The ordinary letter n (meaning the number of particles) and the Greek letter ##\eta## (eta) are not the same thing. For 3 particles moving in three dimensions, the vectors ##\eta## and ##p## would need to have 9 components (x, y and z for each particle).

A problem with interpreting ##\eta## as a vector is that with several degrees of freedom there exist more "directions" than only the positive and negative direction that appear as the plus and minus signs in ##e^{\pm ip\eta /\hbar}## terms of the given wavefunction, so the solution here would not be the most general one. For example, if you make a combination of all plane waves with wavevector of absolute value ##k## and any direction, you get something like a Bessel function with ##r## (distance from origin) in its argument.

I don't have the Bohm's book available so I can't check what this is about.

Can you by chance give an example on how this function would look like for 3 particles?

( I know that in the p term, Bohm uses the momentum formula with the mass of the particles (he writes ##m_1+m_2## in the mass part within p)

hilbert2
Gold Member
If you have a vector that contains the positions of three 2-dimensional particles, it would be like ##\eta = (x_1 , y_1 , x_2 , y_2 , x_3 , y_3)## and the momentum vector would be ##p = (p_{x1} , p_{y1} ,p_{x2} ,p_{y2},p_{x3} ,p_{y3})##. The inner product ##x\cdot p## would have value

##\eta \cdot p = x_1 p_{x1} + y_1 p_{y1} + x_2 p_{x2} + y_2 p_{y2}+ x_3 p_{x3} + y_3 p_{y3}##.

However, I'm not at all certain whether the quatities in the function are actually this kind of vectors, as there's no ##"\cdot "## sign denoting inner product. Bohm had some quite unconventional theories about the nature of quantum wavefunctions, and it's possible that the one here is describing something that doesn't even exist in mainstream QM.

Gold Member
If you have a vector that contains the positions of three 2-dimensional particles, it would be like ##\eta = (x_1 , y_1 , x_2 , y_2 , x_3 , y_3)## and the momentum vector would be ##p = (p_{x1} , p_{y1} ,p_{x2} ,p_{y2},p_{x3} ,p_{y3})##. The inner product ##x\cdot p## would have value

##\eta \cdot p = x_1 p_{x1} + y_1 p_{y1} + x_2 p_{x2} + y_2 p_{y2}+ x_3 p_{x3} + y_3 p_{y3}##.

However, I'm not at all certain whether the quatities in the function are actually this kind of vectors, as there's no ##"\cdot "## sign denoting inner product. Bohm had some quite unconventional theories about the nature of quantum wavefunctions, and it's possible that the one here is describing something that doesn't even exist in mainstream QM.

Let me quote his book directly a little later. I will get back to you on this on this post.

Gold Member
However, I'm not at all certain whether the quatities in the function are actually this kind of vectors, as there's no ##"\cdot "## sign denoting inner product. Bohm had some quite unconventional theories about the nature of quantum wavefunctions, and it's possible that the one here is describing something that doesn't even exist in mainstream QM.
Okey, here is what he writes (p 337 - Quantum Theory):

quote:
"

\frac{-\hbar^2}{2(m_1+m_2)} \frac{\nabla^2G(n)}{G(n)} = E_0 = constant

The above equation, however is exactly the same as the Schrödinger equation for a free particle of mass ##m_1 + m_2##. The wave function for the center of mass, therefore, behaves exactly as if the system were a single particle with kinetic energy ##E_0## and mass equal to the total mass of the system. (...) The quantum result can also be generalized to an arbitrary number of particles. The function G(n) is given by

G(n) = Ae^{i \dot{p} \eta/\hbar} + BA^{-i \dot{p} \eta/\hbar}

where A and B are arbitrary constant and ##|p| = \sqrt{2(m_1+m_2}E_0.##
:quote end

hilbert2