# Quantum mechanics: wave function (momentum space vs real space)

1. May 6, 2014

### Nikitin

Hi. I never understood why the momentum wave function $\phi (p)$ is the fourier transform/integral of the real space wave function $\psi (x)$. Basically, the second pair of formulas here http://quantummechanics.ucsd.edu/ph130a/130_notes/node82.html, and (and the rest of the text, for that matter) makes little sense to me. Is there any intuitive reason behind the formulas?

Could you guys also help me understand the dirac delta normation?

Thanks for any help!

Last edited: May 6, 2014
2. May 6, 2014

### Nikitin

a few more questions, if I may:

Often in QM texts they speak of THE almighty wave function that completely describes a system.

But...

(1) Why then are there sets of eigenfunctions for each operator? Is "THE" wave function the set of eigenfunctions that solve the schroedinger equation?

(2) Also, in the case of larger atoms with multiple electrons: What kind of wave functions do they have? Is there one and only one wave function for the entire system, describing how all the electrons behave? Shouldn't there be multiple wavefunctions; one for each electron?

thanks

3. May 6, 2014

### strangerep

All of these questions involve very lengthy answers (i.e., several chapters in a textbook).

What level of calculus, linear algebra and group theory are you comfortable with. (I could, of course, simply recommend Ballentine's textbook "QM -- A Modern Development", but that depends on your current level of math knowledge.)

To pass from a classical to a quantum description, one essentially asks that the observables in the classical system (i.e., functions on phase space) be represented as self-adjoint operators on a Hilbert space such that the Poisson brackets of the classical system correspond to commutators between operators on the Hilbert space. (There's a lot more to it, of course, but that's the basic idea.)

Sticking with the basics, the elements of the Hilbert space are state vectors, and can be superposed as linear combinations, just as in any vector space. There are theorems involving self-adjoint operators on Hilbert spaces that make this framework useful for physics. E.g., the eigenvalues of such operators are real, hence can correspond to measurement results. Also, the eigenfunctions of a self-adjoint operator span the Hilbert space (i.e., any state vector can be expressed as a linear combination of those eigenfunctions).

If one has multiple such operators (that don't mutually commute), then there are distinct sets of eigenfunctions. That's essentially what's going on with the position and momentum operators. (There is a technical subtlety that these operators are better suited to a generalized framework, known as "Rigged Hilbert Space", but I'll skip that for now.) In performing a Fourier transform, you're really just doing a change of basis from position eigenstates to momentum eigenstates.

For the multiparticle case, one forms larger state vectors by a technique called "tensor product" (also explained in Ballentine).

Anyway, if you can get hold of Ballentine, and if you can handle the math in ch1, that's probably the best advice...

[Edit: this thread could probably go in the QM forum.]

4. May 7, 2014

### Nikitin

I'm only in my 2nd year doing an intro class called 'Quantum Physics', so my math level is probably far too low, and I haven't taken Hamiltonian mechanics yet either. Though I think we will start with the formal stuff next semester in the 'Quantum Mechanics I' class. So I guess there's no need to explain everything in detail; I'm just looking for some intuition on this.

Anyway, despite struggling with the details, I *think* I understood the gist of what you said; That both the momentum and real space wave functions are just different forms of the "THE wave function", and you use fourier transforms to switch between them (don't know why fourier transforms work, though).

Another thing:
What do you mean exactly? Isn't the real space wave function $\psi(\vec{r})$ an eigenfunction to both the momentum operator $\hat{p}=\frac{\hbar}{i} \nabla$ and the position operator $\vec{r}=[x,y,z]$? And in momentum space you similarly got both position and momentum operators (like momentum operator $\vec{p}=[p_x,p_y,p_z]$) that work on $\phi(\vec{p})$ and return the relevant eigenvalues.

Oh and thanks for the answer!

Last edited: May 7, 2014
5. May 7, 2014

### Fredrik

Staff Emeritus
"Eigenfunctions" of position are of the form $\vec x\mapsto \delta^{(3)}(\vec x-\vec y)$. Eigenfunctions of momentum are plane waves, i.e. functions of the form $\vec x\mapsto e^{i\vec p\cdot\vec x/\hbar}$. So a position "eigenfunction" is infinitely sharply peaked, and a momentum eigenfunction is evenly spread out over all of space. (The absolute value is the same everywhere). Wavefunctions are typically somewhere in between these two extremes.

Note that the equality $\psi(x)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{+\infty} \phi(p)e^{-ipx/\hbar}dp$ is saying that $\psi$ is a kind of generalized linear combination of momentum eigenfunctions (with all possible eigenvalues).

Last edited: May 7, 2014
6. May 7, 2014

### Nikitin

thanks for the reply! :) I really it!

What does this notation mean? Are you saying the position eigenfunction a dirac-delta function?
So you are saying an eigenfunction must always produce a sharp eigenvalue?

Well, even if the wave function $\psi(x) = e^{i\vec p\cdot\vec x/\hbar}$, isn't it still an eigenfunction to the position operator $x$? I mean, $\hat{x} \psi(x) = x \cdot \psi(x)= x\psi(x)$ which makes $\psi(x)$ an eigenfunction to $x$ even if the probability to find the particle is the same everywhere in space.

Indeed. And I think I will give up trying to understand why.

7. May 7, 2014

### MisterX

No, for it to be an eigenvector of the position operator $x$, we must have $\hat{x} \psi(x) = \lambda \psi(x)$, where $\lambda$ is constant (does not depend on x).

8. May 7, 2014

### Matterwave

Are you familiar with the Dirac notation? If so, the derivation is not so difficult.

A wave function in the position representation is defined as: $\psi(x,t)\equiv \left<x|\Psi(t)\right>$

A wave function in the momentum representation is defined as: $\phi(p,t)\equiv \left<p|\Psi(t)\right>$

I can insert a resolution of the identity:

$$1=\int_{-\infty}^\infty dx\left|x\right>\left<x\right|$$

Into the second equation to get:

$$\phi(p,t)= \left<p\right|\int_{-\infty}^{\infty} dx \left|x\right>\left<x|\Psi(t)\right>$$

Therefore:

$$\phi(p,t)= \int_{-\infty}^{\infty} dx\left<p|x\right>\psi(x,t)$$

All we have to know then is $\left<p|x\right>$.

To find this we look back at how these vectors are defined. These vectors are defined as the eigenvectors of the momentum and position operators respectively:

$$\hat{p}\left|p'\right>=p'\left|p'\right>$$
$$\hat{x}\left|x'\right>=x'\left|x'\right>$$

In particular then, we have:

$$-i\hbar\partial_x\left<x|p\right>=\left<x|\hat{p}|p\right>=p\left<x|p \right>$$

This equation has solution (check it by taking the derivative):

$$\left<x|p\right>=Ce^{ipx/\hbar}$$

The constant of normalization $C$ can be found by the requirement:

$$\left<p'|p\right>=\delta(p'-p)$$

Again adding in a resolution of the identity we get:

$$\left<p'\right|\int_{-\infty}^{\infty} dx\left|x\right>\left<x|p\right>=|C|^2\int_{-\infty}^{\infty} e^{i(p-p')x/\hbar}dx$$

We know this integral from Fourier analysis (essentially, from Fourier's inversion theorem we know this must be true):

$$|C|^2\int_{-\infty}^{\infty} e^{i(p-p')x/\hbar}dx=2\pi\hbar|C|^2\delta(p-p')$$

Matching this condition with the normalization requirement gives:

$$C=\frac{1}{\sqrt{2\pi\hbar}}$$

Therefore, we finally have our final formula:

$$\phi(p,t)= \frac{1}{\sqrt{2\pi\hbar}}\int_{-\infty}^{\infty} e^{-ipx/\hbar}\psi(x,t)dx$$

9. May 7, 2014

### Fredrik

Staff Emeritus
$\vec x\mapsto \delta^{(3)}(\vec x-\vec y)$ is the "function" that takes an arbitrary $\vec x$ to $\delta^{(3)}(\vec x-\vec y)$.

Not sure if it's appropriate to say "yes" to that, because I'm sure that someone would say that there's only one delta function on $\mathbb R^3$, and it has its peak at $\vec 0$. The "function" identified by the notation above has its peak at $\vec y$ (and there's one such "function" for each $\vec y$). I'm inclined to call it a delta function, but I'm not sure everyone else would.

Not sure what you mean by "produce", but every eigenfunction has an eigenvalue. The specific case of the position operator is tricky though. It doesn't actually have an eigenfunction. The delta "function" isn't even a function. But it can be thought of as the wave"function" of a particle whose position has just been determined with perfect accuracy to be exactly at 0.

$\hat x\psi(x)=x\psi(x)$ for all $x\in\mathbb R$ does not imply that there's a number $\lambda$ such that $\hat x\psi=\lambda\psi$. Define $I:\mathbb R\to\mathbb R$ by $I(x)=x$ for all $x\in\mathbb R$. Let $\psi$ be an arbitrary function in the domain of $\hat x$. Let $x\in\mathbb R$ be arbitrary. We have
$$(\hat x\psi)(x)=x\psi(x)=I(x)\psi(x)=(I\psi)(x).$$ Since x is arbitrary, this implies that $\hat x\psi =I\psi$. But $I$ is a function, and it's not a constant function.

The following calculation is actually quite interesting. It sort of explains why the delta function is thought of as an eigenfunction of the position operator. Suppose that $\hat x$ has an eigenfunction. Denote it by f. Denote the eigenvalue by $\lambda$. This means that $\hat x f=\lambda f$. Let $x\in\mathbb R$ be arbitrary. We have
$$\lambda f(x)=(\lambda f)(x)=(\hat x f)(x)=xf(x).$$ This result tells us that if $f(x)\neq 0$, then $\lambda =x$. This is logically equivalent to if $x\neq\lambda$, then $f(x)=0$. So f takes every number to 0, except maybe $\lambda$.

It's essentially just that an integral can be approximated by a sum. For each p, define $u_p:\mathbb R\to\mathbb C$ by $u_p(x)=e^{ipx}$ for all $x\in\mathbb R$. (I'm using units such that $\hbar=1$). A linear combination of several of the $u_p$ functions would look like this: $\sum_k a_k u_{p_k}$.

$$\psi(x) =\frac{1}{\sqrt{2\pi}}\int \phi(p)e^{ipx}\mathrm dp =\frac{1}{\sqrt{2\pi}}\int\phi(p)u_{p}(x)\mathrm dp \approx \frac{1}{\sqrt{2\pi}}\sum_k \phi(p_k) u_{p_k}(x)\Delta p_k =\left(\sum_k \frac{\phi(p_k)\Delta p_k}{\sqrt{2\pi}}u_{p_k}\right)(x).$$

Last edited: May 7, 2014
10. May 7, 2014

### strangerep

No. To have common eigenfunctions, the 2 operators would have to commute. But position and momentum don't commute.

11. May 8, 2014

### Nikitin

Thanks for the replies! I especially appreciate that the eigenvalue thing was cleared up - things make allot more sense now. I actually didn't know that an eigenvalue had to be a constant, embarrassingly enough.

Well, I'm not sure about the controversies, but I understand why mathematically in one dimensional space it would have be a delta-function of the form $\delta(\lambda -x)$. It's because the delta function is zero everywhere except $x=\lambda$.
OK. With "produce" I meant use the eigenfunction on its operator so it creates(/produces) an eigenfunction. Probably bad way to say it though.

I'm well aware that an integral is an infinite sum, and I understand that the wave function $\psi(x)$ is a linear combination of momentum-eigenfunctions multiplied by $\phi(p)$ (the wave function in momentum space). What I don't understand is why the latter is true. But as matterwave has been so kind as to post a proof, I'll go learn bra-ket notation and return to that

On another note, I think I understand dirac-delta normalization now. I read a bit on the function, and found that the integral of a complex exponential function is the sinc function, which will converge to become the delta function if the integral of the complex exponential is taken from minus infinity to plus infinity.

Hence this trick can be used to normalize momentum eigenfunctions,,, even though it makes no sense from a physical standpoint. And all this stuff seems to go against the interpretation of QM where $|\psi \psi^*|$ equals the probability density. They made an exception for non-normalizable eigenfunctions like the momentum eigenfunction?

12. May 8, 2014

### Matterwave

There's no "they" in science who can make exceptions for a physical process! Science is dictated by reality, not by expert scientists. There can be no physical exception to be made for a momentum or position eigenfunction's non-normalizability. This tells you that particles can't REALLY be in a momentum or position eigenstate. They are approximations (think very large momentum spread = very sharply peaked position wave function), but the real world can never actually reach these idealizations. When you make a measurement of position in the real world, you never get one EXACT number x. You get a range x+dx depending on the accuracy of your measurement. You can never be perfectly exact.

Mathematically speaking, the momentum and position eigenfunctions are not within the Hilbert space. Therefore, if we are to be strictly dealing with Hilbert spaces mathematically, the equation $\hat{x}\left|x\right>=x\left|x\right>$ has no solution (i.e. $\left|x\right>$ does not exist).

Physicists are not usually so concerned with mathematical rigor as long as the correct physical results (which i practice are always approximations since experiments are never perfectly precise) are obtained. But mathematically, there are ways to make the physicist's statements rigorous. These have to do with the spectral theorem, or rigged Hilbert spaces. But these are rather formal subjects which, at this point in your training, you probably do not have to worry too much about. In practice, we just use $\left|x\right>$ as if it exists, but always remembering to keep Dirac delta functions inside integral signs.