Speciality of position momentum representation

[neelakash]

Hello Everyone.I want to know if there is anything special about the position or the momentum representation in quantum mechanics.Every book deals with them.Why do not they work with Energy representation or time representation?Do not they exist?Basically,I feel it hard to imagine that a wavefunction \psi(\ t) is a Fourier transform of \phi(\ E). Is it possible?Basically time-energy or azimuthal angle-angular momentum all statisfy the uncertainty relation as done by position and mometum.Still we generally do not use such representations (if possible)?

Can you people suggest why?

 

[kexue]

People use energy representation, for example for the harmonic oscillator. Time is not a operator and there is no time representation in qm.

 

[tiny-tim]

I want to know if there is anything special about the position or the momentum representation in quantum mechanics.Every book deals with them.Why do not they work with Energy representation or time representation?

Hello neelakash!

the position or the momentum representation in QM is used to integrate over all possible states (eigenfunctions) …

but the same energy, for example, can correspond to different momenta, and therefore to different states …

so an energy representation doesn't cover every possiblity

 

[neelakash]

I suppose there is no degeneracy for mometum operator;For all examples that I can remember now,has their momentum eigenstates non-degenerate.

 

[DeShark]

Hello Everyone.I want to know if there is anything special about the position or the momentum representation in quantum mechanics.Every book deals with them.Why do not they work with Energy representation or time representation?Do not they exist?Basically,I feel it hard to imagine that a wavefunction \psi(\ t) is a Fourier transform of \phi(\ E). Is it possible?Basically time-energy or azimuthal angle-angular momentum all statisfy the uncertainty relation as done by position and mometum.Still we generally do not use such representations (if possible)?

Can you people suggest why?

I think it all really depends on what it is that you're interested in. Essentially, the representation, as I think you know, is which base states to use to describe the system. Since the base states of classical mechanics are position and momentum (i.e. if you know the position and momentum of a particle at t=0, you can model its evolution). So that's what we seek most of the time.

However, when working with the hydrogen atom for example, we can use the energy, angular momentum and z-component of angular momentum to describe the system (ignoring spin). From this we can extract different values easily, but we can also cast the representation to find out what the picture looks like in a position (or Schroedinger) basis.

So we do use an energy basis (or representation) too, but energy is not always enough to describe a state (since states can be degenerate and this is why we must use the other quantum numbers). In a SHO situation, we can use a position basis, because this makes it easy to describe the potential, but we can then cast these bases into a momentum basis, because this basis is discrete.

The essential point is that whatever basis we choose, it has to be complete. I.e. Every possible state has to be able to be written as a sum of the bases.

As for a time representation, since time is not an observable, it doesn't make much sense to speak about the system being described by a sum of time states.

 

[DeShark]

I suppose there is no degeneracy for mometum operator;For all examples that I can remember now,has their momentum eigenstates non-degenerate.

The total linear momentum operator can be degenerate! There are multiple values of p_x, p_x and p_z which all correspond to the same total momentum. Thus the momentum is not enough to define a state entirely.

Also, two states can have the exact same components of p_x, p_y and p_z and yet describe two completely different states. You might have a spin 1/2 particle and a spin 1 particle of the same mass, but different spin. These are two different states, which will be evident if you apply a magnetic field to the particle(s), destroying the degeneracy.

 

[Demystifier]

Time is not a operator and there is no time representation in qm.

Don't be so sure:
http://xxx.lanl.gov/abs/0811.1905 [accepted for publication in Int. J. Quantum Inf.]

 

[neelakash]

Let me tell you the basic source of confusion.Depending on in which representation my wave functions are,I have to integrate over space co-ordinates or momentum co-ordinates when I am trying to calculate the expectation value of some operator.However,we never integrate the wave functions \psi(\ E) or \psi( t) over energy or over time respectively.Truly speaking I have not seen anythong like that.

One of my friends suggest we may use Energy and time as well.He argues that Energy-time are canonically conjugate variables like position and momentum.Same is the case for Angular momentum and azimuthal angle.So,he says the formulae may be well written using energy/time/angular momentum or the angle as the variables to integrate over.Now I am sceptical as I cannot remember if I have seen \psi(\ E) or \psi( t)---wave functions only function of time or enegy.How can you check thenormalization condition for \psi(\ E) or for \psi(\ t)?What will be the limits to use? Can you get my point?

It seems that there is something convenient in the position or momentum representation. Because just imagine you are writing the Schrodinger's equation in energy/time representation.Also think that your wave function \psi(\ E) is Fourier transform of \psi(\ t)---and what that means I do not know.You cannot have representation of operators in terms of canonically conjugate variables---whereas in position-momentum case you have reciprocal representation of operators...thant's the thing.It looks weird to me in some sense...

 

[DeShark]

Don't be so sure:
http://xxx.lanl.gov/abs/0811.1905 [accepted for publication in Int. J. Quantum Inf.]

Interesting bit of info. But it still specifies that eigenvalues of this time operator are not physical observables. Which leaves me in doubt as to whether these eigenstates could be used as a base. The current discussion is about why position and momentum are used so much. But thanks for the link.

Depending on in which representation my wave functions are,I have to integrate over space co-ordinates or momentum co-ordinates when I am trying to calculate the expectation value of some operator.

Well, you don't *have* to find expectation values in the position or momentum co-ordinates. To find the expectation (as far as I've been told/read), you decompose the state into its (orthonormal) bases and sum over the bases contained in the state. i.e.
\sum_{n} <n|\hat{A}|n>

If your state is represented in the position representation, then it is composed of base states corresponding to delta functions over the entirety of space. So when you sum the product of the operator with respect to this base, it corresponds to taking the integral. Similarly for the momentum representation, since it is continuous.

However,we never integrate the wave functions \psi(\ E) or \psi( t) over energy or over time respectively.Truly speaking I have not seen anythong like that.

Well, that's because the wave function only makes sense in the position (read shroedinger) representation. In other representations, the state is represented in another way (as are all the operators).

It seems that there is something convenient in the position or momentum representation. Because just imagine you are writing the Schrodinger's equation in energy/time representation.

I'm not sure what the energy-time representation would look like to be honest. A state could be described in terms of energy base states, but only if there is no degeneracy. You can do this with a single particle in the harmonic oscillator or particle in an infinite square well, because there is no other quantum number needed. The "energy basis" is known as the {N}-Representation.

Schroedinger's equation as it is usually put is in terms of wave functions, but in its general form (independent of base), it can be written

i\hbar \frac{\partial |\phi>}{\partial t} = \hat{H} |\phi>

So in the position basis, one obtains the usual Schroedinger equation.