Why do you need infinite size matrix which commute...

  • #1
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...to give a number?

https://ocw.mit.edu/courses/physics...g-2016/lecture-notes/MIT8_04S16_LecNotes5.pdf

On page 6, it says,
"
Matrix mechanics, was worked out in 1925 by Werner Heisenberg and clarified by Max Born and Pascual Jordan. Note that, if we were to write xˆ and pˆ operators in matrix form,they would require infinite dimensional matrices. One can show that there are no finite size matrices that commute to give a number times the identity matrix, as is required from (2.13). This shouldn’t surprise us: on the real line there are an infinite number of linearly independent wavefunctions, and in view of the correspondences in (2.14) it would suggest an infinite number of basis vectors. The relevant matrices must therefore be infinite dimensional.
"

Equation 2.13: [x, p] = ihbar

2.14 Correspondences:
operators ↔ matrices
wavefunctions ↔ vectors
eigenstates ↔ eigenvectors

The part that I don't get is, "on the real line, there are an infnite number of linear independent wavefunctions." Why is it so? If I think of the real line and think of wavefunctions as vectors, every single vector will be linearly *dependent* not independent, right? What am I missing here?
 

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  • #2
andrewkirk
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If I think of the real line and think of wavefunctions as vectors, every single vector will be linearly *dependent* not independent, right?
I think what you may be missing is the notion of what it means for two functions to be equal. Two functions are equal if and only if they have the same domain and, for every element of that domain, they give the same result. So the sine and cosine functions are unequal, even though they intersect infinitely many times on the real numbers (at ##x=(k+0.25)\pi## for any integer ##k##), because for all other inputs they give different results.

So for a set of functions to be dependent, there needs to be some linear combination of them that is the function that gives zero for every input - called the zero function.

The infinite set of functions:
$$\{x\mapsto \sin kx\ :\ k\in\mathbb Z\}$$
is linearly independent because there is no linear combination of them that gives the zero function.
 
  • #3
atyy
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Consider the usual 3d directions: x, y and z. These are linearly independent because there is no way to add a vector pointing in the x and a vector pointing in the y direction to get a vector pointing in the z direction.

On the real line, one can consider "wave functions" that are infinitely narrow and peaked or located at only a single position - let's call these position eigenfunctions. There is no way to add a position eigenfunction located at x1 and a position eigenfunction located at x2 to get a position eigenfunction located at x3. Thus the position eigenfunctions that are located at different positions are linearly independent.
 
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