# Different wave function for different observers?

1. Feb 12, 2016

### friend

Can a singe particle have a different wave function for different observers?

Suppose someone at rest prepares some electrons with an (close to exact) momentum. In his rest frame the position of the electron is not known. But what kind of wave function will a person see traveling in the same direction and speed of the electron? Would he see that the electron has a (close to exact) position and not know its momentum? Is all that given by one wave function? Or does the wave function change for different observers?

2. Feb 12, 2016

### andrewkirk

People don't observe wave functions. Wave functions are a mathematical construct used to predict observations. The wave function is not defined relative to any observer.
In very loose terms, the answer to the question is No.

When relativistic effects come into play - eg because of very high relative velocities - one has to move from QM to Quantum Field Theory. In that case, differences between coordinate frames may be relevant. But I get the sense that you are not asking about Quantum Field Theory.

3. Feb 13, 2016

### Jilang

The wavefunction would be different at the expectation value of the momentum would be zero. It would be the standard Gaussian of a stationary electron.

4. Feb 13, 2016

### Bruno81

It has been done before in various double slit set ups. If the path is known(for single electrons), you will not get an interference pattern. It doesn't matter who has the ability to tell where the particle landed but that in principle it is knowable.

5. Feb 13, 2016

### bhobba

No.

Thanks
Bill

6. Feb 13, 2016

### Jilang

No? Doesn't the moving spreading Gaussian become just a stationary spreading Gaussian?

7. Feb 13, 2016

### bhobba

Moving Gaussian? Wave-packets spread - but is the same for all observers.

Thanks
Bill

8. Feb 13, 2016

### Jilang

The spread is the same agreed, but the spreading Gaussian travels if the particle travels. If you move to a frame where you are travelling at the same velocity as the particle the Spreading Gaussian does not travel.

9. Feb 13, 2016

### bhobba

Ever heard of coordinate transformation? Transforming to different coordinates changes nothing - its just a different representation.

Thanks
Bill

10. Feb 13, 2016

### Jilang

Agreed. It won't change the physics, just the representation.

11. Feb 13, 2016

### vanhees71

Of course, the wave function is dependent on the reference frame used to describe the usual kinematics. Since you talk about wave functions, it's implied that you talk about non-relativistic quantum theory, i.e., the Schrödinger wave function transforms under the Galilei transformation with the well-known unitary ray representation, i.e., if an observer in the reference frame $\Sigma$ describes it as $\psi(t,\vec{x})$, an observer moving with velocity $\vec{v}$ relative to him, he'll describe the same state by the wave function
$$\psi'(t',\vec{x}')=\exp \left (\mathrm{i} \vec{v}' \cdot \vec{x}'-\frac{\mathrm{i}m}{2} \vec{v}^2 t' \right) \psi(t',\vec{x}'-\vec{v} t'),$$
where $t'=t$, $\vec{x}'=\vec{x}-\vec{v} t.$

12. Feb 13, 2016

### friend

In the still frame the particle with known momentum has a wave function that is represented by a plane wave that extends throughout all of space. In the frame of reference travelling along side the particle the wave function is represented by a localized gaussian. How can a single particle have two wave functions. Or is this just a matter of expressing the one wave function in different bases? Or is there some transformation that changes a plane wave to a gaussian?

13. Feb 13, 2016

### vanhees71

No, the wave function of a particle is never represented by a plane wave, because the plane wave is not square integrable. I've given the transformation of the wave function from one to another reference frame, connected by a Galileo boost. A Gaussian wave in one frame goes over to a Gaussian wave in the other.

14. Feb 13, 2016

### friend

You might be right. Let me ask a few questions. The leading exponential terms seems to only be a phase factor. How does that account for the going from what may look to be a damped plane wave to a gaussian? It seems we need to go from the oscillations of a plane wave (perhaps damped to form a square integrable) to a gaussian that has no oscillations. Does the phase factor introduce oscillations in amplitude? I know it does if you take just the real or complex part of the phase factor - those oscillate.

Let me see. On further thought, as I recall, the plane waves of QM are described by the complex exponential, right? So, yes, I think I can see how your transformation works.

Last edited: Feb 13, 2016
15. Feb 13, 2016

### andrewkirk

I would rather say that the coordinate representation of the representation of the wave function in a particular basis is dependent on the the reference frame, because I think of the 'wave function' as being the vector in the Hilbert space, which is coordinate-independent.

I have not convinced myself whether there should be two 'representation' steps there or only one, but there is at least one. The reason I think there might be two is that, to get a complex-valued function, first we have to choose a basis for the Hilbert space (eg position, momentum or energy), and then we need to choose a coordinate basis to express the chosen basis elements as complex-valued functions. What I'm not sure of is whether that coordinate system is always implied by the choice of Hilbert space basis.

'Wave function' is not a well-defined term, so perhaps sometimes it is used to refer to the representation in a coordinate system, rather than to the Hilbert space element. With that interpretation, the wavefunction is coordinate-dependent. But I have not seen that interpretation used.

16. Feb 14, 2016

### vanhees71

The wave function always refers to a basis of eigenvectors of a minimal complete set of compatible observables and are thus dependent on the reference frame used to measure these observables, $\psi(t,o_1,\ldots,o_j)=\langle o_1,\ldots,o_j|\psi,t \rangle$. The state vector, $|\psi,t \rangle$ is of course independent of the coordinates (everything written in the Schrödinger picture of time evolution).

17. Feb 14, 2016

### friend

I can see how the exponential adds oscillations. But how does it widen towards a plane wave. Does the $\vec{x}'-\vec{v} t'$ in $\psi(t',\vec{x}'-\vec{v} t')$ widen out the $\psi$ function so that the exponential oscillations extend a bit throughout space so that it starts to look like a plane wave?