Curiosity about the wavefunction

[gentsagree]

Going back to the basics, I recall the wave function of quantum mechanics being dependent on space and time coordinates, such as [itex]\Psi (\overline{x},t)[/itex], however one says that quantum mechanics is a 0+1 dimensional (0 space, 1 time) QFT. So there is NO SPACE.

Now, I know there's the caveat that in QM position and momentum are operators on the Hilbert space of states, so in a way we could just have time as a physical coordinate, but how does this relate to the space dependence of the wave-function?

Sorry for the possibly trivial question, but this is bugging me!

Thanks

 

[atyy]

It's probably easiest to see in the path integral formulation. Let's talk about bosons, so that the things in the path integral have classical meanings.

In quantum mechanics, it is classical particle trajectories that enter the path integral. Classical particle trajectories are lines like x(t), so they are "0 dimensional".

In quantum field theory, it is classical field configurations that enter the path integral. Classical field configurations are things like A(x,t), so they are "1 dimensional".

 

[bhobba]

Going back to the basics, I recall the wave function of quantum mechanics being dependent on space and time coordinates, such as [itex]\Psi (\overline{x},t)[/itex], however one says that quantum mechanics is a 0+1 dimensional (0 space, 1 time) QFT. So there is NO SPACE.

Now, I know there's the caveat that in QM position and momentum are operators on the Hilbert space of states, so in a way we could just have time as a physical coordinate, but how does this relate to the space dependence of the wave-function?

Its expanded in eigenfunctions of the position operator which is how 'space' comes into it. There is no 'space' in normal QM as you correctly point out - that expansion is entirely arbitrary - you could expand it in eigenfunctions of momentum if you like. However Schroedinger's equation uses that form so it usually done that way.

Thanks
Bill

 

[stevendaryl]

Going back to the basics, I recall the wave function of quantum mechanics being dependent on space and time coordinates, such as [itex]\Psi (\overline{x},t)[/itex], however one says that quantum mechanics is a 0+1 dimensional (0 space, 1 time) QFT. So there is NO SPACE.

Now, I know there's the caveat that in QM position and momentum are operators on the Hilbert space of states, so in a way we could just have time as a physical coordinate, but how does this relate to the space dependence of the wave-function?

Sorry for the possibly trivial question, but this is bugging me!

Thanks

I'm not sure whether this has already been answered, but the connection between quantum mechanics and 0+1 dimensional field theory is a little subtle.

For a massive free spin-zero field, you have the field equations:

[itex]\hbar^2 (\frac{\partial^2 }{(\partial t)^2} - \frac{1}{c^2} (\frac{\partial^2 }{(\partial x^1)^2} + \frac{\partial^2 }{(\partial x^2)^2} + ...)) \phi = -m^2 c^4 \phi [/itex]

If there are no spatial dimensions, then this is just

[itex]\hbar^2 \frac{d^2 }{d t^2}\phi = - m^2 c^4 \phi [/itex]

That's the same equation as the 1-D harmonic oscillator equation

[itex]\frac{d^2}{dt^2} X = - \omega^2 X[/itex]

if we identify [itex]\omega[/itex] with [itex]\frac{mc^2}{\hbar}[/itex]

Quantizing the 0+1 dimensional field [itex]\phi[/itex] means treating [itex]\phi[/itex] and [itex]\frac{d}{dt} \phi[/itex] as operators, just like quantizing the classical equations of motion for the harmonic oscillator means treating [itex]X[/itex] and [itex]\frac{d}{dt} X[/itex] as operators.

But notice that even though the two theories are mathematically the same, they have different interpretations. In the case of the harmonic oscillator, [itex]X[/itex] is a location in space, while in the case of 0+1 dimensional field theory, [itex]\phi[/itex] is a field strength, and is not a location.