Orthonormal Basis of Wavefunctions in Hilbert Space

  • #1
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0
Hello,

I've a fundamental question that seems to keep myself confused about the mathematics of quantum mechanics. For simplicity sake I'll approach this in the discrete fashion. Consider the countable set of functions of Hilbert space, labeled by [itex] i\in \mathbb{N} [/itex]. This set [itex] \left \{u_{i}(\overrightarrow{r}) \right \} [/itex] is orthonormal and canonical (can it be?). So my question is regarding the position vector of the basis: What is the explicit expression that accounts for the position in this case? The wavefunction has a value that depends on the position that can be expressed by the basis multiplied by a complex value, so the information of the variation of this constant by position must be in the basis function. What am I failing to see here?

Your help will be much appreciated!
 

Answers and Replies

  • #2
3
0
Hello,

I've a fundamental question that seems to keep myself confused about the mathematics of quantum mechanics. For simplicity sake I'll approach this in the discrete fashion. Consider the countable set of functions of Hilbert space, labeled by [itex] i\in \mathbb{N} [/itex]. This set [itex] \left \{u_{i}(\overrightarrow{r}) \right \} [/itex] is orthonormal and canonical (can it be?). So my question is regarding the position vector of the basis: What is the explicit expression that accounts for the position in this case? The wavefunction has a value that depends on the position that can be expressed by the basis multiplied by a complex value, so the information of the variation of this constant by position must be in the basis function. What am I failing to see here?

Your help will be much appreciated!

I think I see now what I failed to see, the discrete approach does not simplify, it is the source of the problem. I still didn't get this 100%, why is there a closure relation? What is the need to express a basis that is orthonormal for every continuous value?
 
  • #3
fresh_42
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The position vector is irrelevant as far as it concerns the basis. It is merely a variable of the functions involved. However, depending on how the inner product is defined, it plays a role for that, e.g. ##\langle u_i,u_j \rangle = \int_M u_i(\vec{r})u_j^\dagger(\vec{r}) \,d\vec{r}##. But as soon as you have a orthonormal basis, the variable has nothing to do with it.

As to whether such a basis can be chosen canonically, well, you can apply the Gram-Schmidt algorithm in case of a separable space.
 

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