Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

https://docs.google.com/open?id=0BxrBcN1-BZWUOXNxR1l4S0l2MjQ

http://www.2shared.com/complete/Qjy1_uzp/Quantum_Mechanics_in_Simple_Ma.html (I uploaded a pdf file that contains the parts of the textbook that I am having some difficulty understanding. Seven pages.) (two links point to the same pdf.)

A matrix $N$ has all entries except diagnoal ones zero, and first row, first column also zero, other diagonal entries depending on its column, row place; for example, second row, second column has one as its entry, third row, third column has two as its entry and so on.

It is known that $H = \hbar\omega (N + \frac{1}{2}) $.

The question is, my textbook says that if there are a continuous range of possible values, then for some states, $\langle (N - \langle N \rangle )^2 \rangle \leq [\frac{1}{2} (n - \langle N \rangle)]^2$ , where $n = 0,1,2,3,...,$ then progresses to say that since $\langle (N - \langle N \rangle )^2 \rangle \geq (n - \langle N \rangle)^2 (\langle I_0 \rangle + \langle I_1\rangle + ...)$ then $\langle (N - \langle N \rangle )^2 \rangle \geq (n - \langle N \rangle)^2$.

Then it says that this proves that there cannot be a continuous range of possible values, as $(0-\langle N \rangle)^2$, $(1 - \langle N \rangle)^2$ and so on cannot be smaller than $(n - \langle N \rangle)^2$.

I am not sure how one gets $\langle (N - \langle N \rangle )^2 \rangle \geq (n - \langle N \rangle)^2 (\langle I_0 \rangle + \langle I_1\rangle + ...)$. Can anyone show me the proof?

Also, how does one get the part - $(0-\langle N \rangle)^2$, $(1 - \langle N \rangle)^2$ and so on cannot be smaller than $(n - \langle N \rangle)^2$?

share|improve this question
    
What is the precise reference to the textbook? –  Qmechanic Sep 29 '12 at 23:03
    
@Qmechanic Quantum mechanics in simple matrix form old edition page 145. According to my search, it's available in 4shared, though..... it's not legal, I guess.... –  War Sep 29 '12 at 23:05
    
Author is Thomas F. Jordan? –  Qmechanic Sep 29 '12 at 23:08
1  
Related: physics.stackexchange.com/a/23033/2451 –  Qmechanic Sep 29 '12 at 23:50
1  
Why are you sticking to this book? it is confusing and unclear. (For one, it seems to be trying to justify the number basis which it takes for granted in the beginning.) There are plenty of far better resources to learn the quantum harmonic oscillator –  Emilio Pisanty Oct 3 '12 at 13:27
show 5 more comments

2 Answers

up vote 3 down vote accepted
+100

The likely interpretation:

we assume there is a continuous range of possible values, then we are able to measure it so that we get some particular value of $\langle N \rangle$ and particular uncertainty, but still then $N$ is not exactly determined. In this context, for some states, $\langle (N - \langle N \rangle^2 \rangle \leq [\frac{1}{2}(n-\langle N \rangle)]^2$, where the left part is uncertainty. And I think the aforementioned part is where you got confused. The next part will come as obvious.

share|improve this answer
add comment

I) It seems that the question(v4) is essentially

Prove the that the point spectrum of the infinite-dimensional matrix $$ N~:= \left(\begin{array}{ccccc} 0 & 0 & 0 & 0 & \ldots \\ 0 & 1 & 0 & 0 & \\ 0 & 0 & 2 & 0 & \\ 0 & 0 & 0 & 3 & \\ \vdots & & & & \ddots\\ \end{array}\right) $$ is the non-negative numbers $${\rm Spec}_{\rm pt}(N)~=~ \mathbb{N}_0~:=~\{0,1,2,3,\ldots\}.$$

II) Now the (linear) operator $N$ is an unbounded operator, so let us for simplicity (in order to avoid various technical issues related to unbounded operators) instead ask

Prove the that the point spectrum of the bounded operator $$ e^{-N}~:= \left(\begin{array}{ccccc} e^0 & 0 & 0 & 0 & \ldots \\ 0 & e^{-1} & 0 & 0 & \\ 0 & 0 & e^{-2} & 0 & \\ 0 & 0 & 0 & e^{-3} & \\ \vdots & & & & \ddots\\ \end{array}\right) $$ is the numbers $${\rm Spec}_{\rm pt}(e^{-N})~=~\{e^{-n}\mid n \in\mathbb{N}_0 \} ~=~\{e^{0},e^{-1},e^{-2},e^{-3},\ldots\}.$$

III) Or more generally

If a bounded operator $A$ is diagonal in an orthonormal basis $(|i\rangle)_{i\in I}$ for a Hilbert space $$H~=~{\rm span}\{|i\rangle | i\in I\}$$ with eigenvalues $\lambda_i\in \mathbb{C}$, show that the point spectrum is $$ {\rm Spec}_{\rm pt}(A)~=~ \{ \lambda_i | i\in I\}~\in \mathbb{C}. $$

This is almost a triviality. A sketched proof could be as follows:

  1. Since $A$ is a normal operator, the eigenspaces for different eigenvalues are pairwise orthogonal $$ \forall i,j\in I:~~\lambda_i~\neq~\lambda_j\qquad \Rightarrow \qquad {\rm ker}(A-\lambda_i{\bf 1}) ~\perp~ {\rm ker}(A-\lambda_j{\bf 1}).$$

  2. Since $A$ is diagonal, the eigenspaces corresponding to the diagonal elements $\{ \lambda_i | i\in I\}$ span the full Hilbert space $$H~=~\oplus_{i\in I} {\rm ker}(A-\lambda_i{\bf 1}). $$

  3. Hence there are no room for eigenvalues $\lambda$ that are not diagonal elements $\{ \lambda_i | i\in I\}$, $${\rm ker}(A-\lambda{\bf 1}) ~\subseteq~ H^{\perp}~=~ \{0\}. $$

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.