- #1
Philip Land
- 56
- 3
Using that ##\hat{a} =a = \sqrt{\frac{mw}{2 \hbar}} \hat{x} +\frac{i}{\sqrt{2mw \hbar}} \hat{p}## and ## a \dagger = \sqrt{\frac{mw}{2 \hbar}} \hat{x} -\frac{i}{\sqrt{2mw \hbar}} \hat{p}##
We can solve for x in term of the lowering and raising operator.
Now, recently I read a derivation of ##<n| \hat{x} |m> (1)##.
Question 1: n and m were never specified, so what does the above expression actually mean?
By substitution, we can rewrite (1) to ##\sqrt{\frac{ \hbar}{2mw}} <n| (a + a \dagger )|m>##(2)
Question 2: I'm a little confused about how I can simplify the above expression. I'm not super familiar with Dirac notation. I know very well the definition of the raising and lowering operators. But can someone fill in the blanks of how they get from (2) to ## \sqrt{\frac{ \hbar}{2mw}} \cdot ( \sqrt{m} \delta_{n, m-1} + \sqrt{m+1} \delta_{n, m+1})##? No relation between n and m is defined.
That is not clear to me.
We can solve for x in term of the lowering and raising operator.
Now, recently I read a derivation of ##<n| \hat{x} |m> (1)##.
Question 1: n and m were never specified, so what does the above expression actually mean?
By substitution, we can rewrite (1) to ##\sqrt{\frac{ \hbar}{2mw}} <n| (a + a \dagger )|m>##(2)
Question 2: I'm a little confused about how I can simplify the above expression. I'm not super familiar with Dirac notation. I know very well the definition of the raising and lowering operators. But can someone fill in the blanks of how they get from (2) to ## \sqrt{\frac{ \hbar}{2mw}} \cdot ( \sqrt{m} \delta_{n, m-1} + \sqrt{m+1} \delta_{n, m+1})##? No relation between n and m is defined.
That is not clear to me.