Responsible party: joh04684, Spherical Chicken

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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.

• Stationary States
• Expectation Values of Operators
• Hamiltonions and Energy

<math> <\hat{H}>_\Psi_n(t) = \int_{-\infty}^{\infty}\Psi^*(x,t)\hat{H}\Psi(x,t)dx</math>

• This lets us get rid of x, but still possibly have a function depending on time.
• If we set the time derivative <math>\frac{\partial}{\partial t} <\hat{H}>_\Psi__n = 0</math>, this gives us our stationary states, implying that <\hat{H}> is constant as a function of time.

• The general form: <math>\Psi_n (x,t) = \frac{1}{\sqrt{2}}\psi_1(x)e^{\frac{-i E_1 t}{\hbar}} + \frac{1}{\sqrt{2}}\psi_2(x)e^{\frac{-i E_2 t}{\hbar}}</math>
• And the complex conjugate: <math>\Psi^*_n (x,t) = \frac{1}{\sqrt{2}}\psi^*_1(x)e^{\frac{+i E_1 t}{\hbar}} + \frac{1}{\sqrt{2}}\psi^*_2(x)e^{\frac{+i E_2 t}{\hbar}}</math>
• The cross terms of <math>\Psi^*(x,t)\Psi(x,t)</math> become <math>e^{i(\frac{E_1-E_2}{\hbar})t}</math> or <math>e^{-i(\frac{E_1-E_2}{\hbar})t}</math>
• If you add these two, the time dependence doesn't necessarily disappear. However, in the case of <\hat{H}>, these cross terms will go away.
• For example: let <math>\xi = \frac{E_1 - E_2}{\hbar}t</math>, we get <math>e^{i\xi} + e^{-i\xi}</math>
• Real part: <math>\cos{\xi} \pm \sin{\xi}</math>
• Imaginary part cancels, leaving us with a time dependent portion

• <math> <\hat{\theta}>_\Psi_n = \int_{-\infty}^{\infty} \Psi_n^*(x,t)\hat{\theta}\Psi_n(x,t) dx</math>, where <math>\Psi_n^*(x,t)</math> has an <math>e^{\frac{iE_nt}{\hbar}}</math> component, and <math>\Psi_n(x,t)</math> has an <math>e^{\frac{-iE_nt}{\hbar}</math> component.
• Time dependent part cancels, resulting in <math><\hat{\theta}></math> being constant with respect to time.
• Note: <math>\hat{\theta}</math> is just a general operator, and has no explicit time dependence. Also, the subscript n appended to the two <math>\Psi</math> values implies we are dealing with stationary states.
• We are trying to see if operators other than <math>\hat{H}</math> are constant, as well.

• For an arbitrary <math>\Psi(x,t)</math>, we have: <math><\hat{\theta}>_\Psi = \int_{-\infty}^{\infty} \Psi^*(x,t)\hat{\theta}\Psi(x,t)dx</math>, which is not necessarily constant with respect to time.
• Here, <math>\Psi^*(x,t) = (c_1\psi_1(x)e^{\frac{-iE_1t}{\hbar}}+c_2\psi_2(x)e^{\frac{-iE_2t}{\hbar}})^*</math>
• If you have only one term (i.e. ground state), you have stationary states.
• If you have multiple energy states, it's possible some <math><\hat{\theta}></math> are not stationary.

• <math><\hat{H}>_\psi_n = E_n</math>
• Giving <math>\sigma^2 = <\hat{H}^2> - <\hat{H}>^2 = 0</math>
• Since <math>\sigma_E^2 = 0</math>, this implies that there is no fluctuation: You get the same measurement every time.
• Using this, we can write the Schroedinger Equation in short hand: <math> \hat{H}\psi_n = E_n\psi_n = (-\frac{\hbar^2}{2m}\partial^2x + V)</math>, where <math>\partial x</math> is short-hand for <math>\frac{\partial}{\partial x}</math>
• This short-hand form of the Schroedinger Equation lets us write: <math><\hat{H}>_\psi_n = \int \psi_n^*\hat{H}\psi_ndx = \int\psi_n^*E_n\psi_ndx</math>, where <math>E_n</math> is not an operator, but rather a number.
• Thus this becomes: <math>E_n\int\psi_n^*\psi_ndx = E_n</math>
• Since we are dealing with stationary states, we are assuming <math>\int\psi_n^*\psi_ndx</math> to be normalized to 1.

• In general: <math>\psi(x) = \Sigma c_n\psi_n</math>
• Thus, we can say: <math>\int\psi_m^*(\psi_n(x))dx = \int\psi_m^*(\Sigma c_n\psi_n)dx</math>
• Since <math>\Sigma c_n</math> is a set of constants, we can pull it out such that:
  • For <math>m \neq n</math>, we have <math>\Sigma c_n\int\psi_m^*\psi_ndx = 0</math> for different indices
  • For <math>m = n</math>, we have <math>\Sigma c_n\int\psi_{m=n}^*\psi_ndx = 1</math>
  • This is due to orthogonality
• The only terms to survive are those where <math> m = n</math>, leaving us <math> \Sigma_n c_n\int\psi_m^*\psi_ndx = c_m</math>
• Thus, our left hand side of the equation becomes: <math>LHS = \int\psi_m^*\psi(x)dx = c_m</math>
• With: <math>\psi(x) = \Sigma c_n\psi_n</math>
• This is similar to finding the x component of <math>\vec{r}</math>, which we can represent by <math>(\vec{r}\cdot\hat{x})</math>
• Here, <math>\int\psi_n^* (~)dx</math> is essentially the dot product, where the subscript /n implies the particular component (ie, x, y, z, 1, 2, 3, etc)