Quantum Harmonic oscillator

  • Thread starter TimeRip496
  • Start date
  • #1
254
5
upload_2016-1-31_3-3-25.png
upload_2016-1-31_3-33-26.png
upload_2016-1-31_3-3-43.png
upload_2016-1-31_3-4-23.png

When I work out $$b^+b$$, I get

$$\widehat{b^+} \widehat{b} = \frac{1}{2} (ξ - \frac{d}{dξ})(ξ + \frac{d}{dξ}) = \frac{1}{2} (ξ^2 - \frac{d^2}{dξ^2}) = \frac{mωπx^2}{h} - \frac{h}{4mωπ} \frac{d^2}{dx^2}$$

So base on what I have about, (9) should be
$$(9) = \frac{hω}{2π} (\frac{1}{2} \widehat{b^+} \widehat{b} + \frac{1}{2} ξ^2)$$
instead of (10).

Am I missing out something?
 

Attachments

  • upload_2016-1-31_3-2-57.png
    upload_2016-1-31_3-2-57.png
    18.3 KB · Views: 350
Last edited:

Answers and Replies

  • #2
blue_leaf77
Science Advisor
Homework Helper
2,629
785
##\frac{1}{2}(ξ-\frac{d}{dξ})(ξ+\frac{d}{dξ})=\frac{1}{2}(ξ^2-\frac{d^2}{dξ^2})##
The above step is not correct. Instead, it should be
$$
\frac{1}{2}(\xi-\frac{d}{d\xi})(\xi+\frac{d}{d\xi}) = \frac{1}{2}(\xi^2 +\xi\frac{d}{d\xi} - \frac{d}{d\xi}\xi - \frac{d^2}{d\xi^2}).
$$
Then simplify the second and third terms.
 
  • #3
254
5
##\frac{1}{2}(ξ-\frac{d}{dξ})(ξ+\frac{d}{dξ})=\frac{1}{2}(ξ^2-\frac{d^2}{dξ^2})##
The above step is not correct. Instead, it should be
$$
\frac{1}{2}(\xi-\frac{d}{d\xi})(\xi+\frac{d}{d\xi}) = \frac{1}{2}(\xi^2 +\xi\frac{d}{d\xi} - \frac{d}{d\xi}\xi - \frac{d^2}{d\xi^2}).
$$
Then simplify the second and third terms.
Based on yours,
$$
\frac{1}{2}(\xi-\frac{d}{d\xi})(\xi+\frac{d}{d\xi}) = \frac{1}{2}(\xi^2 +\xi\frac{d}{d\xi} - \frac{d}{d\xi}\xi - \frac{d^2}{d\xi^2}) = \frac{mωπx^2}{h} - \frac{h}{4mωπ} \frac{d^2}{dx^2} -\frac{1}{2} + x\frac{1}{2} \frac{d}{dx}
$$
This is where i am stuck again as to what do I do with $$x\frac{1}{2} \frac{d}{dx}$$ and this one( $$\frac{h}{4mωπ} \frac{d^2}{dx^2}$$) has the term 4 instead of 8 as shown in equation (9).
 
  • #4
blue_leaf77
Science Advisor
Homework Helper
2,629
785
$$
\frac{1}{2}(\xi^2 +\xi\frac{d}{d\xi} - \frac{d}{d\xi}\xi - \frac{d^2}{d\xi^2}) = \frac{mωπx^2}{h} - \frac{h}{4mωπ} \frac{d^2}{dx^2} -\frac{1}{2} + x\frac{1}{2} \frac{d}{dx}
$$
That's still incorrect.
The term ##\frac{d}{d\xi}\xi## should expand into two terms, what are they?
 
  • #5
254
5
That's still incorrect.
The term ##\frac{d}{d\xi}\xi## should expand into two terms, what are they?
##\frac{d}{d\xi}\xi = \frac{d}{d\xi}(\xi * 1) = 1 * \frac{d\xi}{d\xi} + \xi * \frac{d1}{d\xi}##
Is this correct?
 
  • #6
blue_leaf77
Science Advisor
Homework Helper
2,629
785
##\frac{d}{d\xi}\xi = \frac{d}{d\xi}(\xi * 1) = 1 * \frac{d\xi}{d\xi} + \xi * \frac{d1}{d\xi}##
Is this correct?
Yes, that one is correct.
 
  • #7
254
5
Yes, that one is correct.
What about this one( $$\frac{h}{4mωπ} \frac{d^2}{dx^2}$$) which is supposed to be in equation (9)?

The only way I can get to $$\frac{h^2}{8mωπ^2} \frac{d^2}{dx^2}$$ is by multiplying $$\frac{h}{4mωπ} \frac{d^2}{dx^2}$$ with $$\frac{h}{2π}$$, which I don't know where to get from.






How I get $$\frac{h}{4mωπ} \frac{d^2}{dx^2}$$,
First, $$dξ = \sqrt[] {\frac{2mπω}{h}} * dx$$
Squaring it, $$dξ^2 = \frac{2mπω}{h} * dx^2$$
Subbing it into $$ \frac{d^2}{dξ^2} $$ from b+*b, I get $$\frac{h}{2mωπ} \frac{d^2}{dx^2}$$ before multiplying 0.5 from the bracket outside $$\frac{1}{2}(\xi^2 +\xi\frac{d}{d\xi} - \frac{d}{d\xi}\xi - \frac{d^2}{d\xi^2})=b^+*b$$.
 
Last edited:
  • #8
blue_leaf77
Science Advisor
Homework Helper
2,629
785
upload_2016-1-31_3-3-43-png.95071.png
The coefficient in front of the second order derivative in the above equation is not correct, as you can check its dimensionality. It should be
$$
\frac{h}{4m\omega \pi}
$$
which is the result you obtained from your own calculation. This is the correct one as you can check by multiplying it with the constant outside the square bracket and it should lead to ##p^2/(2m)##.
 
  • #9
254
5
##\frac{1}{2}(ξ-\frac{d}{dξ})(ξ+\frac{d}{dξ})=\frac{1}{2}(ξ^2-\frac{d^2}{dξ^2})##
The above step is not correct. Instead, it should be
$$
\frac{1}{2}(\xi-\frac{d}{d\xi})(\xi+\frac{d}{d\xi}) = \frac{1}{2}(\xi^2 +\xi\frac{d}{d\xi} - \frac{d}{d\xi}\xi - \frac{d^2}{d\xi^2}).
$$
Then simplify the second and third terms.
The coefficient in front of the second order derivative in the above equation is not correct, as you can check its dimensionality. It should be
$$
\frac{h}{4m\omega \pi}
$$
which is the result you obtained from your own calculation. This is the correct one as you can check by multiplying it with the constant outside the square bracket and it should lead to ##p^2/(2m)##.
Thanks a lot!
 
  • #10
254
5
##\frac{1}{2}(ξ-\frac{d}{dξ})(ξ+\frac{d}{dξ})=\frac{1}{2}(ξ^2-\frac{d^2}{dξ^2})##
The above step is not correct. Instead, it should be
$$
\frac{1}{2}(\xi-\frac{d}{d\xi})(\xi+\frac{d}{d\xi}) = \frac{1}{2}(\xi^2 +\xi\frac{d}{d\xi} - \frac{d}{d\xi}\xi - \frac{d^2}{d\xi^2}).
$$
Then simplify the second and third terms.
Sorry to trouble you again.
Talking about the second and third terms, $$ \xi\frac{d}{d\xi} - \frac{d}{d\xi}\xi = \xi\frac{d}{d\xi} - (1 * \frac{d\xi}{d\xi} + \xi * \frac{d1}{d\xi}) $$
Initially I thought I can cancel out the $$\xi\frac{d}{d\xi}$$ by subtracting it with $$\xi * \frac{d1}{d\xi}$$, but in my another thread, you told me that
$$∅*(\frac{d}{dξ})\neq \frac{d∅}{dξ}$$, thats why I am stuck at this part.
 
  • #11
blue_leaf77
Science Advisor
Homework Helper
2,629
785
I should probably have corrected your post #5, instead of equal sign it might have been more appropriate to use arrow instead:
$$
\frac{d}{d\xi}\xi \rightarrow \frac{d}{d\xi}(\xi * 1) = 1 * \frac{d\xi}{d\xi} + \xi * \frac{d1}{d\xi}.
$$
with the arrow sign is intended to mean that the expression to the left is to be operated on a function, in the above case a number ##1##. The insertion of a ##1## there is supposed to be thought as an arbitrary object because remember that Hamiltonian is actually an operator, it's not yet a physical quantity. Therefore, writing ##H## only with nothing following it does not have mathematical significance. Only when you make it operate on an arbitrary state ##\psi## will the expression ##H\psi## be meaningful.
##
\xi\frac{d}{d\xi} - \frac{d}{d\xi}\xi = \xi\frac{d}{d\xi} - (1 * \frac{d\xi}{d\xi} + \xi * \frac{d1}{d\xi})##
In the above equation, if you make it as if it operates on a ##1## (or an arbitrary state ##\psi##), it should become
$$
\xi\frac{d}{d\xi}1 - (1 * \frac{d\xi}{d\xi} + \xi * \frac{d1}{d\xi}).
$$
 
  • #12
254
5
I should probably have corrected your post #5, instead of equal sign it might have been more appropriate to use arrow instead:
$$
\frac{d}{d\xi}\xi \rightarrow \frac{d}{d\xi}(\xi * 1) = 1 * \frac{d\xi}{d\xi} + \xi * \frac{d1}{d\xi}.
$$
with the arrow sign is intended to mean that the expression to the left is to be operated on a function, in the above case a number ##1##. The insertion of a ##1## there is supposed to be thought as an arbitrary object because remember that Hamiltonian is actually an operator, it's not yet a physical quantity. Therefore, writing ##H## only with nothing following it does not have mathematical significance. Only when you make it operate on an arbitrary state ##\psi## will the expression ##H\psi## be meaningful.

In the above equation, if you make it as if it operates on a ##1## (or an arbitrary state ##\psi##), it should become
$$
\xi\frac{d}{d\xi}1 - (1 * \frac{d\xi}{d\xi} + \xi * \frac{d1}{d\xi}).
$$
Thanks a lot again! One last thing if you wouldn't mind,
After converting equation (9) to equation (10),
$$\hat{b}*∅_0 = 0$$
Thus, $$\hat{b}^+\hat{b}*∅_0 = 0$$,
So equation (10) left with $$\frac{hω}{2π}(0+\frac{1}{2})∅_0=E_0∅_0$$
Shouldn't $$E_0=\frac{hω}{4π}$$ instead of $$E_0=\frac{hω}{2π}$$?
 
  • #13
blue_leaf77
Science Advisor
Homework Helper
2,629
785
Doesn't the equation ##\frac{hω}{2π}(0+\frac{1}{2})∅_0=E_0∅_0## correctly imply that ##E_0=\frac{hω}{4π}##?
 
  • #16
254
5
I think you should look for another source to study this subject, the author seems to be rather sloppy with his derivation.
Ok thanks!
 

Related Threads on Quantum Harmonic oscillator

  • Last Post
Replies
0
Views
2K
  • Last Post
Replies
3
Views
1K
  • Last Post
Replies
13
Views
2K
  • Last Post
Replies
9
Views
4K
  • Last Post
Replies
0
Views
2K
  • Last Post
Replies
8
Views
1K
  • Last Post
Replies
4
Views
1K
Replies
1
Views
1K
Replies
2
Views
4K
  • Last Post
Replies
7
Views
1K
Top