# Eigenstates, Eigenblahs, and Eigenpoofs? HELP O.O

## Main Question or Discussion Point

Okay, so here's my knowledge of quantum mechanics: *crickets chirping in the background*. Here's my mathematical knowledge: basic integral calculus, some multivariable calculus, decent amount of linear algebra (knowledge of subspaces, projections, eigenvalues of matrices, eigenvectors, etc.), a decent amount of diff. eqs (not PDEs, though), all of high school math (extremely well... I was like 3 points of the usamo this year).

So anyways, I started learning quantum mechanics like a week ago, and pretty much have learned:

the construction of the Schrodinger equation (time dependant and independant)
the overall concept of a Hermitian operator
the infinite well situation

I have been running into the words: eigenvalues, eigenfunctions, and eigenstates many times. I am pretty sure I understand what an eigenvalue and eigenfunction is, but I'm a bit shaky on what an eigenstate is. Just to clarify, in the equation

$$\hat{A}\psi = a \psi$$

for some $$a$$, we have that $$\psi$$ is an eigenfunction (which could be vector-valued?) of $$\hat{A}$$ and $$a$$ is an eigenvalue of $$\hat{A}$$. Is this correct?

If relevant or helpful at all, I am currently a senior in high school who is going to enter college as a freshman this coming fall.

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Fredrik
Staff Emeritus
Gold Member
Just to clarify, in the equation

$$\hat{A}\psi = a \psi$$

for some $$a$$, we have that $$\psi$$ is an eigenfunction (which could be vector-valued?) of $$\hat{A}$$ and $$a$$ is an eigenvalue of $$\hat{A}$$. Is this correct?
Yes, it is. The "which could be vector-valued" comment might be an indication that you have misunderstood one thing though. If you define the sum of two functions by (f+g)(x)=f(x)+g(x) and the product of a number and a function by (af)(x)=af(x), then you have turned the set of all functions into a vector space. My point is that it's not the values that the functions take that are vectors. The functions are the vectors, according to the definition of a vector space. The functions you encounter in quantum mechanics are going to be complex-valued functions of 1-3 real variables.

Yes, it is. The "which could be vector-valued" comment might be an indication that you have misunderstood one thing though. If you define the sum of two functions by (f+g)(x)=f(x)+g(x) and the product of a number and a function by (af)(x)=af(x), then you have turned the set of all functions into a vector space. My point is that it's not the values that the functions take that are vectors. The functions are the vectors, according to the definition of a vector space. The functions you encounter in quantum mechanics are going to be complex-valued functions of 1-3 real variables.
Yes, by this I meant that $$\psi$$ takes in a complex valued vector and outputs another complex valued vector... each component of the input is fed into the respective component of $$\psi$$.

Fredrik
Staff Emeritus
Gold Member
The "input" is real-valued, and the "output" is just one complex number. The input represents a position in space and time. So I shouldn't have said 1-3 variables. It's 4 variables unless we ignore some of the spatial dimensions and/or the time dependence. $\psi:\mathbb{R}^4\rightarrow\mathbb{C}$

When your book starts talking about spin, you will probably see expressions like $\psi(\vec x)\chi$, where $\psi$ is the same as before and $\chi$ is a $n\times 1$ matrix (n, the number of rows, depends on the spin of the particle).

vanesch
Staff Emeritus
Gold Member
If I may add something, trying to de-confuse the OP.

In quantum mechanics, the state of a system (be it a particle, a molecule, or anything else we want to handle with quantum mechanics) is described by, well, a quantum state. For single particles, that quantum state corresponds to "a wavefunction", that is, a complex-valued function over space: psi(x,y,z). Historically, Schroedinger thought it was going to be something similar to the electric potential, V(x,y,z): a kind of field in space. But later, people realized that what was important, was the fact that in the set of quantum states, the superposition principle holds: if A is a state, and B is a state, then a A + b B must also be a state. Turns out that that works pretty well for "wavefunctions": if psi1(x,y,z) is a quantum state, and psi2(x,y,z) is a quantum state, then psi3(x,y,z) = a psi1(x,y,z) + b psi2(x,y,z) is also a quantum state, a wavefunction. But it will turn out that quantum states are not always just "functions over x,y and z" ; it is just in the single particle case that this is so. What is important, is that superposition principle holds, which means that any quantum system must have a set of quantum states which forms a vector space.

It is from that realization that we call the set of quantum states, the "state space" and a quantum state also a "state vector". In the case of a single particle, the state vector is nothing else but a wavefunction of the form psi(x,y,z).

The more "sophisticated" quantum mechanics one does, the less one thinks of "wave functions", and the more one thinks of "vectors in a vectorspace".

Note that the vector space of wavefunctions psi(x,y,z) is infinite-dimensional.

In fact, the vector space of quantum states will turn out to have some other mathematical properties (apart from just being a vector space), and mathematically, a set obeying these properties is called a Hilbert space.

It is from that realization that we call the set of quantum states, the "state space" and a quantum state also a "state vector". In the case of a single particle, the state vector is nothing else but a wavefunction of the form psi(x,y,z).
Wouldn't it be fair to add that for $N$ particle systems, we have a wave function of the form
$$\psi:\mathbb{R}^{3N+1}\rightarrow\mathbb{C}$$​
with one triple of coordinates for every particle and a time coordinate (usually) added? Correct me if I am wrong.

Harald.

vanesch
Staff Emeritus
Wouldn't it be fair to add that for $N$ particle systems, we have a wave function of the form
$$\psi:\mathbb{R}^{3N+1}\rightarrow\mathbb{C}$$​