# I Position representation of the state of the system

1. Jan 23, 2017

### fog37

Hello Forum,
My understanding is that the state of the system is $|\Psi>$. We can take the inner product between the state $|\Psi>$ and the eigenstates of the position operator $\hat{x}$:
$$<x|\Psi>=\Psi(x)$$
The function $\Psi(x)$ is the wave function we are initially introduced to in beginner quantum mechanics. We call this the position representation of the state $|\Psi>$. In position space, the eigenfunctions of the position operator $\hat{x}$ are delta functions $\delta(x-x_0)$ located at $x_0$.
We could also take the inner product between the state $|\Psi>$ and the eigenstates of the momentum operator $\hat{p_x}$ to get a new function:
$$<p|\Psi>=\Psi(p)$$
It turns out that $\Psi(p)$ is the Fourier transform of $\Psi(x)$. The eigenfunctions of the momentum operator in momentum space are delta functions $\delta(p-p_0)$ located at $p_0$, correct?
What function would be get by taking inner product between the state $|\Psi>$ and the eigenstates of the energy operator $\hat{H}$ also called Hamiltonian:
$$<H|\Psi>=\Psi(E)?$$
Would we get a function like $\Psi(E)$?
In which circumstance could we get a function like $\Psi(x, p_y, E)$? What type of inner product would we need to calculated between the the operators $\hat{x}$, $\hat{p_y}$, $\hat{H}$ and $\Psi(x)$? Would it be possible if the operators were pairwise commuting with each others?

2. Jan 23, 2017

### Staff: Mentor

Yes. The eigenstates of any operator can be used in this way to generate a wave function.

Any circumstance in which all of those operators commute. Note that if there is a nonzero potential energy, the momentum operators will in general not commute with the Hamiltonian.

You've forgotten your own description of how this works. You don't compute inner products between operators; you compute them between states. So if we have a set of mutually commuting operators, we first have to find a complete set of states that are eigenstates of all of them. Then we label each of the states by their eigenvalues for each of the operators, and generate the wave function by taking the inner product of each state with $\Psi$, and calling the result $\Psi(a, b, c, ...)$, where $a, b, c, ...$ are the eigenvalues labeling the state.

3. Jan 23, 2017

### fog37

Cool. Thanks PeterDonis.
Ok, the Hamiltonian does not commute with the momentum operators $\hat{p_x}$ , $\hat{p_y}$, $\hat{p_z}$. So let's stick with the energy operator and the position operator only. I agree that the inner product is between states, like the state $|\Psi>$ and the eigenstates of a certain operators.

We know that the position operators and the energy operators commute. What type of inner product can we perform that is equal to something like $\Psi(x, E)$ ? I am used to think of the inner product as happening between two states. In the case of $\Psi(x, E)$ we have the state $|\Psi>$, the position eigenstates and the energy eigenstates. How do we perform such inner product with apparently three states to output $\Psi(x, E)$?

Assuming we do that, I have never seen the wavefunction $\Psi(x,y,z,E)$ in textbooks. Why is it uncommon to have a wave function that depends on spatial coordinates and the energy E?

I guess the intrinsic spin operator does not commute with any operator. Is it possible to find a set of common eigenstates when the intrinsic spin operator is involved?

Some important commutation relations are:
• Position operators all mutually commute.
• All three position operators commute with the potential energy (position dependent only).
• The linear momentum operators all mutually commute. Position operators and linear momentum operators in the same direction do not commute.
• Orbital angular momentum operators do not mutually commute.
• The angular momentum components do all commute with the square angular momentum operator.
• Position and orbital angular momentum operators in the same direction commute but those in different directions do not.
This means that we can only know with perfect precision the values of a limited number of observables of a system that is in a certain state at time t. The other observables will be uncertain.

4. Jan 23, 2017

### Staff: Mentor

Actually, on thinking it over, I don't think they do. If two operators commute, then it must be possible to find a set of states which are simultaneous eigenstates of both of them. But position eigenstates are not energy eigenstates, and vice versa.

Let's try an example where we know the operators do commute: for example, energy and angular momentum for an electron in a hydrogen atom. Then there will be a set of states which are each eigenstates of both energy and angular momentum. We can label each of these eigenstates with labels $E, L$, which are its eigenvalues for energy and momentum, respectively; so we can write each eigenstate as a ket $\vert E, L \rangle$. Then we generate the wave function for an arbitrary state $\Psi$ by taking its inner product with each eigenstate, i.e., $\langle E, L \vert \Psi \rangle = \Psi( E, L )$. (In the literature these eigenstates are usually labeled with an energy quantum number $n$, which gives the number of the energy level, and an angular momentum quantum number $l$, which gives the index of the spherical harmonic that appears in the wave function. This is just a difference in labeling, the physics is the same.)

5. Jan 23, 2017

### fog37

Great!! Very clear. I am making great strides with your help). Thanks!

So, based on the mentioned notation, a ket like $|E,L>$ is a simultaneous eigenstate of both the energy and angular momentum operators.
There are eigenstates expressed like $|l,m>$ which are the eigenstates of the squared magnitude angular momentum operator $L^2$, which in the spherical position representation are equal to $<l,m|\Psi>=Y_{lm}(\theta,\phi)$. These are the spherical harmonics. To what operators do the symbols $l$ and $m$ correspond to? From now on, everytime I see a ket with two indices it should indicate a simultaneous eigenstate of two operators (or multiple operators if there are multiple symbols).

The eigenstates of a certain operator are all Delta function in their own observable representation, correct? For instance, in the linear momentum representation, the eigenstates have the form $\delta(p-p_0)$

Can we perform the inner product of the state $\Psi>$ with the two eigenstates of the spin operator?
$$<s|\Psi>=\Psi(s)$$
? Or is there something special about the spin operator? I know spin can only have two eigenstates which have two different eigenvalues.

6. Jan 23, 2017

### Staff: Mentor

Yes.

Actually, it should be $\vert n, l, m \rangle$, where $n$ is the energy level.

$l$ corresponds to the (squared) total angular momentum operator $L^2$; I used $L$ for this eigenvalue in my previous post.

$m$ corresponds to the angular momentum operator in a particular direction, conventionally the $z$ direction. An eigenstate of $L^2$ is also an eigenstate of the angular momentum operator in a particular direction, but only one direction; once we've picked a direction, such as $z$, eigenstates of $L^2$ and $L_z$ (the angular momentum operator in the $z$ direction) are not eigenstates of the angular momentum operator in any other direction (because measurements of angular momentum in different directions don't commute).

(Note: by "angular momentum in a particular direction", I mean angular momentum measured about an axis oriented in that direction.)

No. This is true only of the position and linear momentum operators.

Yes. You can do it with the eigenstates of any operator. You just have to remember that in cases like spin, the index $s$ is discrete, not continuous.

Note that the "two values" part is only true for certain particles. There are particles that have more than two spin eigenstates. (There are also particles that have only one, i.e., they have no spin.)

7. Jan 23, 2017

### fog37

All good.

Only doubt (and I am done for tonite, thanks a lot for your help. I will be back for more), is that only the eigenstates of a position and momentum operators are Delta function in their own base representation. An eigenstate has a unique eigenvalue. For the position operator, the values x along the x-axis are the eigenvalues of the various position eigenstates. Why can't reason the same way for any other operator? For instance, the angular momentum operator eigenstates could be $\delta(L-L_0)$ where $L_0$ is the eigenvalue of a particular angular momentum eigenstate....

8. Jan 23, 2017

### Staff: Mentor

Because the delta function doesn't just mean "the observable has this value". There is a lot of mathematical complexity lurking beneath the delta function that is not applicable to observables other than position and momentum. For one thing, the "delta function" isn't even a function, strictly speaking.

9. Jan 24, 2017

### fog37

Ok.
I am too curious to know how an eigenfunction of, say, the angular momentum operator $L_x$ looks like in the angular momentum representation basis. Wouldn't this be equivalent to calculate the inner product of the eigenfunction with itself, which means projecting the eigenfunction of itself? That should produce a wavefunction representation that depends on $L_x$...

10. Jan 24, 2017

### Staff: Mentor

It looks like a member of a finite set of discrete states. For example, suppose our quantum system is a single spin-1/2 particle, like an electron. Then there are two eigenstates of $L_x$, which correspond to the two possible results you can get when you measure the spin of the electron in the $x$ direction; so the "angular momentum representation basis" is just those two states. (Note that this basis should really be called the "angular momentum about the x axis" basis; there are an infinite number of possible sets of two basis states for this system, each corresponding to angular momentum about one of the possible directions the axis can point.) There isn't really any useful sense in which such a state is an "eigenfunction in the angular momentum representation basis", since the spectrum of the operator is discrete rather than continuous.

The argument of a wave function, even in the continuous case where there is one in a useful sense, is not an operator. It's a variable--a label that gets put on eigenstates to designate their eigenvalues.

11. Jan 24, 2017

### fog37

Great. I see how it does not make sense to talk about a wavefunction when the spectrum of the operator is discrete. The wavefunction, being a function, presumes a continuity of values of the eigenvalues....

I was just looking at my introductory physics book and its section about the hydrogen atom. For $n=3$, there are various (nine if we ignore the duplicity generated by the 4th spin quantum number) allowed states are represented (in position space) by wavefunctions $\Psi_{nlm_l}$. All these states have the same energy, dictated by the quantum number $n=3$. These 9 states are eigenfunctions of the energy operator (Hamiltonian), correct? The other two quantum numbers specify the amount of angular momentum ($l$) and the orientation of the angular momentum vector ($m_l$) associated to these states.

These wavefunction have the form $\Psi_{nlm_l}= R_n(r) Y(\theta, \phi)_{lm_l}$. Are these entire functions $\Psi_{nlm_l}$ also eigenfunctions of the orbital angular momentum operator since the functions $Y(\theta, \phi)_{lm_l}$ are the spherical harmonics and are eigenfunctions of such operator?

12. Jan 24, 2017

### Staff: Mentor

Yes and yes.

Note that this should be $\Psi_{nlm_l}= R_{nl}(r) Y(\theta, \phi)_{lm_l}$: the radial part also depends on the $l$ quantum number.

13. Jan 24, 2017

### Orodruin

Staff Emeritus
A function can have a discrete domain ...

14. Jan 24, 2017

### fog37

Thank you.

@Orodruin, so the eigenfunctions of operators other than the position and linear momentum don't really have a representation in their own basis that is described as a wavefunction. In the own basis, they appear as values in column vector, I guess...

15. Jan 24, 2017

### fog37

Also, in general, is it really true that can can build all Hermitian operators starting from the position operator and the momentum operator or are the exceptions? Maybe the spin is an exception...

16. Jan 24, 2017

### Orodruin

Staff Emeritus
You cannot say Hermitian operator so generally. It depends what Hilbert space you are considering. Spin is not described by the same Hilbert space as position and momentum.

What do you mean by "representation in their own basis"? If you have a discrete basis of eigenfunctions of some operator, the eugenfunctions Could be represented by a column vector with one one entry and the rest zeros.

17. Jan 24, 2017

### fog37

Hi Orodruin,

What I mean is that the state $|\Psi>$ has the by position representation $\Psi(x)$. The same state $|\Psi>$ has the by momentum representation $\Psi(p)$.
The position eigenstates have the position representation $\delta(x-x_0)$. The momentum eigenstates have the momentum representation $\delta(p-p_0)$ or a position representation $e^{ikx}$.

My initial dilemma,which I initially discussed with PeterDonis, was: what is the angular momentum representation of the angular momentum eigenstates?

18. Jan 24, 2017

### Staff: Mentor

As Orodruin said (and as you guessed), they are column vectors (each with one element 1 and the rest 0).

Also, pay careful attention to what Orodruin said about different Hilbert spaces. The state of a particle with spin includes both a position/momentum part and a spin part; the spin part does not have a "position representation" or a "momentum representation".

19. Jan 24, 2017

### fog37

Yes, I am processing.

Orodruin also mention about different Hilbert spaces that we can be considering. I have asked before this same questions and I am still struggling with the answers that were kindly provided. The state $|\Psi>$, conceptually a vector, lives in a single linear Hilbert vector space. However, each observable (operator) has its own basis of eigenvectors which seem to construct a specific vector space. So for each operator and its basis there seems to be a vector space. But the actual state the physical system lives in is one and only one vector space. How do we reconcile all these vector spaces (one per operator) with the single vector space the state $|\Psi>$ lives in?

20. Jan 24, 2017

### Staff: Mentor

Yes. But this space might have subspaces; for example, for a particle with spin, the full Hilbert space contains a position/momentum subspace and a spin subspace. The full Hilbert space is the tensor product of these two subspaces.

Which will be either the full Hilbert space of the system or a subspace of it. For example, for a particle with spin, the position and momentum operators each have eigenvectors that form a basis of the position/momentum subspace, and the spin operators each have eigenvectors that form a basis of the spin subspace. A basis for the full Hilbert space can then be constructed by multiplying eigenvectors of each subspace.

21. Jan 24, 2017

### Truecrimson

Just a technical nitpicking to be clear to the OP, the vector spaces for the spatial part and the spin part are not technically subspaces, but rather subsystems of the tensor product space. :)

22. Jan 24, 2017

### fog37

Ok, so I should look at these vector spaces (one per operator) as "subspaces" of the full Hilbert vector space. The tensor product of these subspaces then gives the full Hilbert vector space. But these vector spaces are not really subspaces in the elementary sense, i.e. subsets of a host vector space where a particular vector can be either a direct member of a subset or have its projection be in that subset....

The tensor product is more sophisticated but would it be conceptually correct to see the full state $|\Psi>$ of the system, which is a single vector in the full vector spaces, as projecting on each one of these subspaces since the basis of each operator for that subspace can represent the state of the system as a linear superposition?

$|\Psi>$

23. Jan 24, 2017

### fog37

Thanks Truecrimson.

It is an important detail I don't need to miss.

24. Jan 24, 2017

### fog37

But, as far as the tensor product between spaces and between the eigenvectors from different bases goes, I read that in the case of entanglement the wavefunction is not factorizable, i.e. the state is not a weighted superposition of the product eigenstates of this new tensor product space. Does it mean that the tensor product is not a viable path to represent the full state $|\Psi>$ in the case of entanglement ?

25. Jan 24, 2017

### Truecrimson

The product eigenstates form a basis for the tensor product space, so any state, including entangled states, can be written as a superposition of these product states. This does not imply, however, that entangled states are factorizable.