# I Eigenstates & Eigenvalues

1. Feb 20, 2017

### Jimmy87

Hi,

I have come across two definitions of eigenstates (and eigenvalues), both of which I understand but I don't understand how the two are related:

1) An eigenstate is one where you get the original function back, usually with some multiple, which is called the eigenvalue.

2) An eigenstate is a state of a particle with a definite momentum or position. For example, in the single slit experiment monochromatic light is used. This light has a single wavelength (and colour) thus it has a definite momentum. Therefore, the photons have a momentum eigenstate.

I don't get how 1 and 2 can be the same?

Thanks,

2. Feb 20, 2017

### Agni101

Both are correct statements since 1st one is definition and 2nd one is the physical significance of eigenstate in atomic and nucleur physics.

3. Feb 20, 2017

### Jimmy87

Thanks. Just to check I understand the connection. Let's say you have some simple wavefunction with a range of different possibilities for position like this one:

Let's say you detect the particle somewhere in the middle where it is most likely to be found. The wavefunction has now collapsed as you have identified where the particle is. Therefore you now have a position eigenstate. Are you saying that if you compare the original wavefunction (the graph above) to the function when it has collapsed, the collapsed function will be some multiple of the original wavefunction?

4. Feb 20, 2017

### PeroK

I admire your efforts, but this is why you need at least basic linear algebra to study QM.

5. Feb 20, 2017

### Jilang

Jimmy, your graph suggests that the particle is not in an eigenstate of position. It only becomes in an eigenstate once it is measured and localised.

6. Feb 20, 2017

### Jimmy87

Yeh I know - underneath the graph I discussed about measuring it and collapsing the wavefunction into a position eigenstate.

7. Feb 20, 2017

### Jilang

So why are you trying to relate the graph of what was before measurement to what is after? The wavefunctions will not be the same.

8. Feb 20, 2017

### Jimmy87

Will the collapsed function not have some multiple of the original function as per the first definition in my OP? What does it then mean about being some multiple of the original function? What does it mean by the original function?

9. Feb 20, 2017

### Staff: Mentor

This is not true in general. As an example, the energy eigenstates in atoms do not have a definite momentum or position.
It depends on the operator you use to find eigenstates.
Only if your state was an eigenstate before, but then nothing collapsed. A collapse (if you really want to use that concept - it is not necessary) always changes the wave function in weird ways.

10. Feb 20, 2017

### PeroK

Because of the normalisation requirement an eigenstate/function is unchanged by a further measurement of the observable in question.

The eigenvalue (multiple) is the measurement obtained.

11. Feb 20, 2017

### Jimmy87

Thanks. Oh I understand now. So if you have a wavefunction like the graph in the other post of mine then you measured and located the particle it would collapse and be in a position eigenstate. This new wavefunction can have various operations done on it. Any operation yielding some multiple of the original collapsed wavefunction eigenstate are also eigenstates? Is that right?

12. Feb 20, 2017

### Khashishi

If the state $\left|\psi_p\right>$ has a definite momentum $p$, then the observable $\hat{p}$ acts on the state to give
$\hat{p} \left|\psi_p\right> = p \left|\psi_p\right>$
In the matrix formulation, you can think of $\left|\psi_p\right>$ as an eigenvector and $\hat{p}$ as a matrix. So $p$ is an eigenvalue.
Now, let suppose you have a state $\left|\phi\right>$ which is a combination of two different states. $\left|\phi\right> = \frac{1}{\sqrt{2}} ( \left|\psi_{p_1}\right> + \left|\psi_{p_2}\right>)$
Then $\hat{p}\left|\phi\right> = \frac{1}{\sqrt{2}} ( p_1\left|\psi_{p_1}\right> + p_2\left|\psi_{p_2}\right>)$
If $p_1 \neq p_2$, you can't write the right side as a product $p \left|\phi\right>$

13. Feb 20, 2017

### Staff: Mentor

Any operator having position eigenstates as its eigenstates is (equivalent to) the position operator, but you can see what the wave function does if you apply it more than once, sure.

14. Feb 23, 2017

### stevendaryl

Staff Emeritus
You have to work through some examples to understand how this works.

An eigenstate is relative to a particular operator and a particular eigenvalue.

The function $[itex]\psi(x) = e^{i k x}$ satisfies the equation:

$(-i \hbar \frac{\partial}{\partial x}) \psi(x) = \hbar k \psi(x)$

So operating with $p \equiv (-i \hbar \frac{\partial}{\partial x})$ has the effect of multiplying by $\hbar k$. So we say that $\psi(x)$ is an eigenstate of $p$ with eigenvalue $\hbar k$.

There isn't really a function that is an eigenstate of position, but people loosely say that

$\psi(x) = \delta(x-x_0)$

is an eigenstate of position with eigenvalue $x_0$, because of the fact:

$x \delta(x-x_0) = x_0 \delta(x-x_0)$

15. Feb 23, 2017

### Karolus

I don't understand: in this case if you measured and obtained a defined momentum (eigenvalue momentum), don't you lose any information about the position? So the equation of the Dirac delta seems contradictory with the equation of eigenvalue of the momentum: In this way it seems that position and momentum are defined simultaneously, violating HUP (?)

16. Feb 23, 2017

### PeroK

Neither the position nor the momentum eigenstate given in post #14 is physically realisable. A particle, therefore, always has a range of position or momentum in any physically realisable state.

17. Feb 23, 2017

### Khashishi

Is the Dirac delta function an eigenstate of momentum? What is the Fourier transform of a Dirac delta function?

18. Feb 23, 2017

### vanhees71

The Dirac $\delta$ "function" is not a function but a distribution (in the sense of "generalized function"). It should be forbidden to all textbook writers to call it a function!

A pure state is described by a square integrable function, and neither the $\delta$ distribution nor the plane-wave $\exp(\mathrm{i} \vec{k} \cdot \vec{x})$ are square integrable. So these "generalized eigenfunctions" of position or momentum, respectively, do not represent pure states of a particle.

19. Feb 23, 2017

### stevendaryl

Staff Emeritus
I'm not sure what you mean. The function $\psi(x) = e^{ikx}$ describes a state with a definite value of momentum, namely $p = \hbar k$, but completely uncertain position. To see that the position is uncertain, take the square: $|\psi(x)|^2 = 1$, which has no position information at all. (It's the square of the wave function that tells the probability distribution.)

In contrast, the function $\psi(x) = \delta(x-x_0)$ describes a state with a definite value of position, namely $x=x_0$. That's because $\delta(x-x_0)$ is zero everywhere except at $x=x_0$; so it is certain that the particle will be at $x=x_0$.

The function $\psi(x) = \delta(x-x_0)$ corresponds to a state with a completely undetermined momentum, though. You can see that by writing the delta function in the following way:

$\delta(x-x_0) = \frac{1}{2\pi} \int e^{i k (x-x_0)} dk$

All momenta are equally likely.

20. Feb 23, 2017

### stevendaryl

Staff Emeritus
The Fourier transform of the delta function is a constant: $F(k) = \int dx \delta(x-x_0) e^{i kx} = 1$.