Is There an Error in the Derivation of Quantum Harmonic Oscillator?

[TimeRip496]

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?

 

[blue_leaf77]

##\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.

 

[TimeRip496]

##\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).

 

[blue_leaf77]

$$
\frac{1}{2}(\xi^2 +\xi\frac{d}{d\xi} - \frac{d}{d\xi}\xi - \frac{d^2}{d\xi^2}) = \frac{mωπx^2}{h} - \frac{h}{4mωπ} \frac{d^2}{dx^2} -\frac{1}{2} + x\frac{1}{2} \frac{d}{dx}
$$

That's still incorrect.
The term ##\frac{d}{d\xi}\xi## should expand into two terms, what are they?

 

[TimeRip496]

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?

 

[blue_leaf77]

##\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.

 

[TimeRip496]

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 $$\frac{1}{2}(\xi^2 +\xi\frac{d}{d\xi} - \frac{d}{d\xi}\xi - \frac{d^2}{d\xi^2})=b^+*b$$.

 

[blue_leaf77]

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)##.

 

[TimeRip496]

##\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!

 

[TimeRip496]

##\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
$$∅*(\frac{d}{dξ})\neq \frac{d∅}{dξ}$$, that's why I am stuck at this part.

 

[blue_leaf77]

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
$$
\xi\frac{d}{d\xi}1 - (1 * \frac{d\xi}{d\xi} + \xi * \frac{d1}{d\xi}).
$$

 

[TimeRip496]

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 $$E_0=\frac{hω}{2π}$$?

 

[blue_leaf77]

Doesn't the equation ##\frac{hω}{2π}(0+\frac{1}{2})∅_0=E_0∅_0## correctly imply that ##E_0=\frac{hω}{4π}##?

 

[TimeRip496]

Doesn't the equation ##\frac{hω}{2π}(0+\frac{1}{2})∅_0=E_0∅_0## correctly imply that ##E_0=\frac{hω}{4π}##?

Yes, but from the source ##E_0=\frac{hω}{2π}##.
Source: http://vixra.org/pdf/1307.0007v1.pdf

 

[blue_leaf77]

Yes, but from the source ##E_0=\frac{hω}{2π}##.
Source: http://vixra.org/pdf/1307.0007v1.pdf

I think you should look for another source to study this subject, the author seems to be rather sloppy with his derivation.

 

[TimeRip496]

I think you should look for another source to study this subject, the author seems to be rather sloppy with his derivation.

Ok thanks!