Question

Algebra

Answer 1

\[a _{+}= \frac{ 1 }{ \sqrt{2hmw} }(-h \frac{ d }{ dx }+mwx)\]

Answer 2

\[\psi_{0}= (\frac{ mw }{ \pi h })^{1/4} e^{-\frac{ mwx^2 }{ 2h }}\]

Answer 3

what is most efficient way to calculate \[(a_{+})^2 \psi_{0}\]

Answer 4

This cannot be algebra LOL

Answer 5

well stuck at algebra part

Answer 6

@hartnn

Answer 7

a+ is an operator? i see d/dx there

Answer 8

yes

Answer 9

do it term by term,

like square the \(\dfrac{1}{\sqrt{2hmw}}\) part and multiply it with \(\dfrac{mw}{\pi h}\)

Answer 10

I thought to evaluate (a+)^2 first

Answer 11

\[(a_{+})^2= \frac{ 1 }{ 2hmw}(-h^2 \frac{ d^2 }{ dx^2 }-2hmw+m^2w^2x^2)\]

Answer 12

ok, then you'll have to do it one by one only,

calculate (-h d/dx + mwx) e^(...)

m,w,h, pi are constants...you can put them aside initially

also, calculating (a+)^2 might give you wrong answer

Answer 13

nope

for middle term you did d/dx(x) right?
i am not sure that is even valid

Answer 14

you'll have to keep it as (-2hmxw d/dx )

Answer 15

are those operator rules?

Answer 16

not sure...but i am thinking logically

(a+)^2 is an operator too,
it should not operate on the dependent variable x

Answer 17

the safest way to get
\((a_{+})^2 \psi_{0}\) is
\((a_{+})(a_{+}) \psi_{0}\)
only

Answer 18

\((a_{+})[(a_{+}) \psi_{0}]\)

Answer 19

although
\((a_{+})^2= \frac{ 1 }{ 2hmw}(-h^2 \frac{ d^2 }{ dx^2 }\color{red}{-2xhmw\frac{d}{dx}}+m^2w^2x^2)\)
can work,
need to check.

Answer 20

\[(a_{+}\psi_{0}= (\frac{ mw }{ \pi h })^{1/4} (\frac{ 2mwx^2 }{ h }-1)\]

Answer 21

*e^()

Answer 22

doesn't seem right?

used the chain rule, right?

Answer 23

thats given in the textbook

Answer 24

wait, let me rewrite

Answer 25

wait, let me rewrite \[\frac{ 1 }{ \sqrt{2hmw} } (\frac{ mw }{ \pi h})^{1/4} 2mwxe^{\frac{ -mwx^2 }{ 2h }}\]

Answer 26

I can handle from here thanks

Answer 27

We can't take derivative with in the operator right? took a note of it thanks

Answer 28

that seems correct now.
and yes.

operator should not operate on itself :P

welcome ^_^