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Main class wiki page: home

Please try to include the following

• main points understood, and expand them - what is your understanding of what the points were.
  • expand these points by including many of the details the class discussed.
• main points which are not clear. - describe what you have understood and what the remain questions surrounding the point(s).
  • Other classmates can step in and clarify the points, and expand them.
• How the main points fit with the big picture of QM. Or what is not clear about how today's points fit in in a big picture.
• wonderful tricks which were used in the lecture.
  

Define Hilbert space- Hilbert space contains the set of all square-integrable functions, on a specified interval (usually ±∞, but more generally (a and b)
<math>f(x)=\int_{-\infty}^{\infty}|f(x)|^2dx < \infty</math>
where all functions that have this property make up a vector space we call Hilbert space. Hilbert space is real space that contains the inner product of two functions defined as follows:
<math> <f|g>=\int_{-\infty}^{\infty}f(x)^*g(x) dx </math>if f and g are both square-integrable the inner product is guaranteed to exist, if the inner product does not exist the then <math> <f|g>=\int_{-\infty}^{\infty}f(x)^*g(x) dx </math> diverges
*Notation relationships
We claim that the operator <math>\hat{Q}\</math>⇔<math>Q_{mn}=\int\psi_m^*\hat{Q}\psi_n dx</math>. Here we introduce new short hand notation <math><e_m|\hat{Q}e_n></math>, where <math> |e_n{>} </math>is a unit vector.
For example<math> |e_n></math> could represent the ground state wave function. That is<math>|\psi>=\psi=\sum c_n\psi_n and <\psi|=\psi^*=\sum c_n^*\psi_n </math>
Back to the claim made. Is this a sensible claim?
So, we speculate that <f|g> =<math>\sum_n F_{i}^*g_{i}</math> that is to say for example if A and B are vectors then A⋅B=<math>A_{1}B_{1}……A_{n}B_{n}</math>
We begin by inspecting that if <math>|f>=\sum_i f_{i}|e_{i}>,|g> = \sum_i g_{i}|e_{i}></math>
so <math><f|g>=\sum_i F_{i}^*|e_{i}>\sum_j g_{j}|e_{i}></math> (where i and j are independent)
Then we can say this equals <math>\sum_{ij}f_i^*g_i<e_i|e_j> </math>where<math> <e_i|e_j> =\int\psi_i^*\psi_j dx=\delta_{ij}</math> Which we know equals one if i=j and eauals zero if i≠j
∴<math>\sum f_i^*g_i\delta_{ij}= \sum f_i^*g_i</math> whooooo!

*This time we claim that <math> <Q> =\int\psi^*\hat{Q}\psi dx</math> which is the same as <math> <\psi|Q|\psi></math> here we recognize that both <math>\psi's</math> are vectors and that Q is like a matrix. We begin by examining <math>|\psi>=\sum c_n\psi_n = \sum c_n|e_n></math>
We want to check that <math> <Q»=(c_1^*…..c_n^*)(matrix)\coloum{c_1…….c_n}</math> So we speculate <math> \hat{Q} =Q_{mn} =<c_m\hat{Q} c_n>=\int\psi_m^*Q\psi_n dx</math>
With this in mind <math><Q_\psi>=\int\psi^*\hat{Q}\psi dx=\int\sum c_i^*\psi_i^*\hat{Q}</math>
So we speculate <math> \hat{Q} =Q_{mn} =<c_m\hat{Q} c_n>=\int\psi_m^*Q\psi_n dx</math>
With this in mind <math><Q_\psi>=\int\psi^*\hat{Q}\psi dx=\int\sum c_i^*\psi_i^*\hat{Q}\sum c_j\psi_jdx</math>
=<math>\sum_{ij} c_{i}^*c_{j}\int\psi_i^*\hat{Q}\psi_jdx</math>
=<math><e_i|\hat{Q}|e_j>=Q_ij</math>
= ∴ <math>\sum_{ij} c_i^*Q_{ij}c_j</math> whooooo

*There was also a question asked in class. What exactly the question was i dont remember could somebody help?
However the discussion went something like this:
In general <math>\psi(x)=1/2\pi \int\phi(k)e^ikx</math> analogous to <math>\sum c_n\psi_n(x)</math>
where <math>|c_n|^2=E</math> and <math> <E>= \sum|c_n|^2 E_n</math>
Discribe a plane wave function. If <math> f(x)=e^ikx</math> which can be thought of as a stationary state of momentum. Which implies that
<math> <f|f>=\int_{-\infty}^{\infty}|f(x)|^2dx = \infty</math>
Consider then,<math> f_k(x)=e^ikx then \int f_k(x)^*f_k(x) dk</math> which is proportional to <math>\delta(k-k')~\delta_{ij}</math> and normalizing in this fashion.\\Next, stationary states are Hamiltonian eigenstates and <math>e^ikx</math>⇔ momentum eigenstate. This means that if <math>\hat{p}e^ikx=khe^ikx</math>
So <math>|\phi(k)|^2</math> is the probability density i.e. the likely hood of finding a momentum value.
therefor if you let <math> f_k(x)=1/2\pi e^ikx</math> then <math> f_k(x)=e^ikx</math> then <math>\int f_k(x)^*f_k(x) dk</math> is now equal to <math>\delta(k-k')=\delta_{ij}</math>

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