- #1
Irishdoug
- 102
- 16
- Homework Statement
- Among the stationary states of the harmonic oscillator |n> = ##\phi_{n}(x)## only n = 0 hits the uncertainty limit ##\frac{\hbar}{2}##. But certain linear combinations also minimise the uncertainty product. They are eigenfunctions of the lowering operator. Calculate ##<x>##, ##<x^{2}>##.
- Relevant Equations
- ##x = \sqrt{\frac{\hbar}{2m\omega}}(a_{+} + a_{-})##
##x^{2} = \frac{\hbar}{2m\omega}## (##a_{+}^{2} + 2a_{+}a_{-} +1 + a_{-}^{2}##)
The hermitian conjugate of ##a_{+} = a_{-}##
Firstly, apologies for the latex as the preview option is not working for me. I will fix mistakes after posting.
So for ##<x>## = (##\sqrt{\frac{\hbar}{2m\omega}}##) ##(< \alpha | a_{+} + a_{-}| \alpha >)## = (##\sqrt{\frac{\hbar}{2m\omega}}##) ##< a_{-} \alpha | \alpha> + <\alpha | a_{-} \alpha > ##.
The final answer is then (##\sqrt{\frac{\hbar}{2m\omega}}##) ##(\alpha + \alpha^{*})##. I am unsure as to why this is i.e. why does ##< a_{-} \alpha | = 1##?
Likewise for ##<x^{2}>##:
I can get as far as (##\frac{\hbar}{2m\omega}##)(##<a_{-}^{2} \alpha | \alpha> + 2<a_{-} \alpha | a_{-} \alpha > + 1 + <\alpha |a_{-}^{2} \alpha>##
The answer is (##\frac{\hbar}{2m\omega}##) (##(\alpha^{*})^{2}) + 2\alpha^{*}\alpha + 1 + (\alpha^{*})^{2})##).
Again, I am unsure as to why the lowering operator squared leads to ##(\alpha^{*})^{2}## and ##(\alpha)^{2}##?
Can I have some help please.
So for ##<x>## = (##\sqrt{\frac{\hbar}{2m\omega}}##) ##(< \alpha | a_{+} + a_{-}| \alpha >)## = (##\sqrt{\frac{\hbar}{2m\omega}}##) ##< a_{-} \alpha | \alpha> + <\alpha | a_{-} \alpha > ##.
The final answer is then (##\sqrt{\frac{\hbar}{2m\omega}}##) ##(\alpha + \alpha^{*})##. I am unsure as to why this is i.e. why does ##< a_{-} \alpha | = 1##?
Likewise for ##<x^{2}>##:
I can get as far as (##\frac{\hbar}{2m\omega}##)(##<a_{-}^{2} \alpha | \alpha> + 2<a_{-} \alpha | a_{-} \alpha > + 1 + <\alpha |a_{-}^{2} \alpha>##
The answer is (##\frac{\hbar}{2m\omega}##) (##(\alpha^{*})^{2}) + 2\alpha^{*}\alpha + 1 + (\alpha^{*})^{2})##).
Again, I am unsure as to why the lowering operator squared leads to ##(\alpha^{*})^{2}## and ##(\alpha)^{2}##?
Can I have some help please.
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