| Both sides previous revisionPrevious revisionNext revision | Previous revision |
| classes:2009:fall:phys4101.001:lec_notes_1028 [2009/10/28 20:02] – x500_voukx002 | classes:2009:fall:phys4101.001:lec_notes_1028 [2009/10/31 15:45] (current) – yk |
|---|
| ===== Oct 28 (Wed) ===== | ===== Oct 28 (Wed) 3.4 generalized probability, delta-func normalization ===== |
| ** Responsible party: poit0009, Hydra ** | ** Responsible party: poit0009, Hydra ** |
| |
| \\ | \\ |
| |
| | ==== Main points discussed ==== |
| | * Normalization of non-normalizable wave function: why <math> < f_{p}|f_{p'}> =\delta (p-p^{'})</math> makes sense? |
| | * generalized probability interpretation. |
| | * momentum-space based wave function, energy-space based wave function, real-space based wave function (what we have been dealing with) |
| |
| | ===Normalization of non-normalizable wave function=== |
| First, we start off with the question of why <math> < f_{p}|f_{p'}> =\delta (p-p^{'})</math> | First, we start off with the question of why <math> < f_{p}|f_{p'}> =\delta (p-p^{'})</math> |
| And what exactly does this mean? | And what exactly does this mean? |
| |
| To begin, we know that we can operate on an eigenfunction as follows: | To begin, we know that we can operate on an eigenfunction as follows: |
| <math>\widehat{p}|f_{p}>=p|f_{p}></math> where we have simply multiplied “p” onto each part of the vector | <math>\widehat{p}|f_{p}>=p|f_{p}></math> where we have simply multiplied “p” onto each part of the vector. |
| |
| So for any “Q” we have <math>\widehat{Q}|q_{n}>=q_{n}|q_{n}></math> | >In the past, for some operator “Q,” its eigenvalue and eigenvector, we can express this relation as <math>\widehat{Q}|q_{n}>=q_{n}|q_{n}></math> |
| | > |
| | >and we tried to normalize the eigenvector so that <math><q_{n}|q_{n}>=1 </math> |
| | > |
| | >or more in general, for two eigenvectors <math> q_{n}</math> and <math>q_{m} </math> we will have <math><q_{m}|q_{n}>=\delta _{m,n}</math>. |
| |
| and we know that <math><q_{n}|q_{n}>=1 </math> | For our current case: <math> f_{p}</math>, we know <math> f_{p}=Ae^{i\frac{p}{\hbar}x}</math> where the usual "k" has been replaced with <math>\frac{p}{\hbar}</math> |
| |
| so for two vectors <math> q_{n} , q_{m} </math> we will have <math><q_{m}|q_{n}>=\delta _{m,n}</math> | Then can we normalize this wave function by setting |A| properly so that <math> \int_{-\infty}^{\infty}f_{p}^{*}\: f_{p^{'}}\; dx = 1</math>? |
| |
| So to solve the case for <math> f_{p}</math>, we know <math> f_{p}=Ae^{i\frac{p}{\hbar}x}</math> where the usual "k" has been replaced with <math>\frac{p}{\hbar}</math> | Let's try. |
| |
| Then | <math>1 = |A|^{2} \int_{-\infty}^{\infty}e^{-i\frac{p}{\hbar}x}\: e^{i\frac{p'}{\hbar}x}\; dx </math> |
| <math>A^{2} \int_{-\infty}^{\infty}f_{p}^{*}\: f_{p^{'}}\; dx </math> | |
| |
| <math>= A^{2} \int_{-\infty }^{\infty}e^{-i\frac{p}{\hbar}x}\: \; e^{i\frac{p'}{\hbar}x}\; dx </math> | <math>=|A|^{2} \int_{-\infty}^{\infty}e^{-i\frac{p-p^{'}}{\hbar}x}\; dx</math> |
| |
| <math>=A^{2} \int_{-\infty}^{\infty}e^{-i\frac{p-p^{'}}{\hbar}x}\; dx</math> | This is, as you may remember, one way to express the delta function (times 2pi), where <math>\frac{p-p^{'}}{\hbar}</math> should be the argument of the delta function. |
| |
| <math>=A^{2} 2\pi \int_{-\infty}^{\infty}\delta \left ( \frac{p-p^{'}}{\hbar} \right )dp | <math>=|A|^{2} 2\pi \delta \left ( \frac{p-p^{'}}{\hbar} \right ) |
| </math> | </math> |
| |
| and then we use a u-substitution <math>p=\hbar k </math>so <math>dp=\hbar dk</math> so we find | Since this form of delta function is not very familiar to us, we need to use substitution to bring it to a more familiar form, where the argument should look like (k-k') form. Also delta function mean something only when it goes inside an integral, and that integral should be over //p// originally, but when we do the substitution to //k//, dp has to be converted to dk in a proper way. |
| |
| <math>=2\pi \hbar A^{2}\int \delta (k-k^{'})dk</math> | Following this plan, we use a substitution <math>p=\hbar k </math>so <math>dp=\hbar dk</math> so we find |
| |
| so we find <math>A=\frac{1}{\sqrt{2\pi \hbar}}</math> | <math>=2\pi\hbar |A|^{2} \delta (k-k^{'})</math> where the implied integral variable has changed from //p// to //k//. |
| | |
| | This suggests that there is no way to accomplish <math> \int_{-\infty}^{\infty}f_{p}^{*}\: f_{p^{'}}\; dx = 1</math> no matter what value we choose for //A//. |
| | |
| | If we give up this idea, and normalize <math>f_p</math> to the delta function, then <math>A=\frac{1}{2\pi\hbar}</math>, but at this point, this is an arbitrary decision. We could have chosen so that <math> \int_{-\infty}^{\infty}f_{p}^{*}\: f_{p^{'}}\; dx = 10000\delta(k-k')</math> equally justifiably as <math> \int_{-\infty}^{\infty}f_{p}^{*}\: f_{p^{'}}\; dx = \delta(k-k')</math>. |
| | |
| | In any case, for now, we find <math>A=\frac{1}{\sqrt{2\pi \hbar}}</math> |
| |
| so <math> f_{p}=\frac{1}{\sqrt{2\pi \hbar}}e^{i\frac{p}{\hbar}x} </math> | so <math> f_{p}=\frac{1}{\sqrt{2\pi \hbar}}e^{i\frac{p}{\hbar}x} </math> |
| The above expression is the all important equation representing momentum space, <math>\Phi(p,t)</math> | The above expression is the all important equation representing momentum space, <math>\Phi(p,t)</math> |
| |
| | >Let's remember that we expanded wave functions in terms of stationary-state wave functions (linear combination), <math>|\psi>=\sum C_n \psi_n(x)</math>, where C_n's can be calculated by <math>C_n=<\psi_n|\psi></math>. |
| | > |
| | >Furthermore, we learned that |C_n|^2 can be INTERPRETED to be the probability that the particle would be observed to be in the state with energy E_n. To make sure this interpretation makes sense, we also checked that |
| | ><math>\sum C_{n}=1 </math> and |
| | ><math>\sum |C_{n}|^{2}E_{n}=<E>_{\Psi}</math>. |
| | >To prove that this is possible, we start with <math> \hat{H}\Psi=E\Psi</math> where <math>\hat{H}=\left( -\frac{\hbar^{2}}{2m}dx+V \right )</math> is the Hamiltonian. |
| | > |
| | >With the definition of expectation value, <math> <E>=\int \Psi^{*}(x,t) \: \hat{H} \; \Psi(x,t)\; dx</math> |
| | > |
| | ><math>=\int \sum C_{n}\Psi_{n}^{*}\; \hat{H}\; \sum C_{m}\Psi_{m}\; dx</math> |
| | > |
| | >=<math> E_{n}\sum C_{n}^{*}C_{m}\int \Psi_{n}^{*}\Psi_{m}\; dx</math> |
| | > |
| | >Here <math>\Psi_{m}^{*}\Psi_{n}=\delta_{m,n}</math> |
| | > |
| | >So it is proven that <math>\sum |C_{n}|^{2}E_{n}=<E></math> |
| | |
| | |
| | Instead of <math>\psi_n</math>'s, can we use <math>f_p</math>'s to do the same (except the summation will be integral)? |
| |
| The equation for <math>\Psi(p,t)</math> is a result of a Fourier transform, =><math> \sum C_{n}\Psi_{n}</math> | Namely, <math>\Psi(x,t)=\int \Phi(p,t)f_p dp</math> where <math>\Phi(p,t)</math> plays the role of C_n's. It should be calculable by <math>\Phi(p,t)=<f_p|Psi(x,t)</math>. Furthermore, <math>|Phi(p,t)|^2</math> should be interpreted as probability (density). For the last part, we need to be able to show that |
| Here, the sum of all coefficients are equal to 1 <math>\sum C_{n}=1 </math> which holds true when the wavefunction is normalized. | <math>\int |\Phi(p,t)|^2 dp = 1</math> and |
| | <math>\int p |\Phi(p,t)|^2 dp = <p></math>. |
| |
| | It turns out that if you choose so that <math> \int_{-\infty}^{\infty}f_{p}^{*}\: f_{p^{'}}\; dx = 1</math> everything works out. (Students are strongly encouraged to show this claim is true. Some of them may appear in the next quiz! (//Yuichi//) |
| |
| |