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• 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 ofThe integral removes x dependence, but this integral still possibly have is a function depending on time. * If we set calculate the time derivative <math>\frac{\partial}{\partial t} <\hat{H}>_\Psi__n </math>, it turns out that it is zero (we can show this fairly easily) this gives us our stationary states, implying that <math><\hat{H}></math> is constant as a function of time.

• The general formAs an example of non-stationary state wave function, let's think about : <math>\Psi_{\strike 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>
• Then the complex conjugate: <math>\Psi^*_{\strike 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 still doesn't necessarily disappear. However, in the case of <math><\hat{H}></math>, these cross terms will go away if you do more calculations, which happens because<math>\psi_1(x)</math> and <math>\psi_2(x)</math> areorthogonal! .
• For example: letNote that if we define <math>\xi \equiv \frac{E_1 - E_2}{\hbar}t</math>, we might get <math>e^{i\xi} + e^{-i\xi}</math> in the above calculations of the energy expectation value .
• Real part:Knowing that <math>{\rm e}^{\pm\xi}=\cos{\xi} \pm i\sin{\xi}</math>, the imaginary part of the above expression may cancel, but the real part will remain.
• Imaginary part cancels, This will still leave 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 with no explicit time dependence is assumed. 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. From the above, we conclude that <math><\hat{\Theta}></math> is generally constant in time.

• 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 iswill not necessarily be constant with respect to time.
• Here,For example, if <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.
• Following the same logic as in the previous section above, we will conclude that the time dependence which arises from cross terms will not necessarily disappear. It turns out that if the operator represent a physical quantity which would be conserved (like angular momentum under certain condition) in Classical Mechanics, the time dependence of its expectation value in QM will also be zero.

• <math><\hat{H}>_\psi_n = E_n</math>. To show this, we can do the following
• Using this, We canoften write the Schroedinger EquationHamiltonian in this form: <math> \hat{H}\strike{\psi_n = E_n\psi_n} = (-\frac{\hbar^2}{2m}\partial^2_x + V)</math>, where <math>\partial_x</math> is short-hand for <math>\frac{\partial}{\partial x}</math>
• This short-hand form ofThen the Schroedinger Equation lets us writebecomes<math> \hat{H}\psi_n = E_n\psi_n</math>. Using this form, we can calculate the Hamiltonian expectation value in the following fashion: <math><\hat{H}>_\psi_n = \int \psi_n^*\underline{\hat{H}\psi_n}dx = \int\psi_n^*\underline{E_n\psi_n}dx</math>, where <math>E_n</math> is not an operator, but rather a number. (See lec_notes_0914.)
• Thus this becomes: <math>E_n\int\psi_n^*\psi_ndx = E_n</math>
• Since wave functions have to be normalized, we arecan assume <math>\int\psi_n^*\psi_ndx=1</math> to be normalized to 1.
• Giving If you calculate <math>\sigma^2 \equiv <\hat{H}^2> - <\hat{H}>^2 </math>you will find it to be zero. If you don't see it immediately, you should think about this until you get zero.
• Since <math>\sigma_E^2 = 0</math>, This implies that there is no fluctuation: You get the same measurement every time.

• In general, if we accept the idea that a set of stationary state wave functions forms a “complete” set, any wave functions can be expressed as <math>\psi(x) = \Sigma c_n\psi_n</math>
• A big questions may be, “How can we find out what <math>c_i</math>'s are, knowing what the function <math>\psi(x)</math> is?”
• To answer that question, Griffiths talks about “Fourier's trick.”
• I would say, this is analogous to finding the x component of some vector <math>\vec{r}</math>, which we can represent by <math>(\vec{r}\cdot\hat{x})</math>. This may sound silly because when <math>\hat{x}=\[x, y, z \]</math>, there is no need to come up with a more complex way to figure out what their x, y and z components are. They are obviously x, y and z. method works even if your have not found the right basis of the vector space such that <math>\hat{x}=\[1, 0, 0 \]</math>. This is because,
  • When you succeed in decomposing a vector to its components, the following expression will express the original vector: <math>\vec{r} = x\hat{x}+y\hat{y}+z\hat{z}</math>
• We have not learned this yet, but it turns out that in QM, doing this: <math>\int\psi_n^* ({\rm wave function})dx</math> to the wave function is essentially the dot product, where the subscript n implies the particular component (ie, x, y, z, 1, 2, 3, etc)
  • For example, the dot product between <math>\psi_n(x)</math> and <math>\psi_m(x)</math> would be <math><m|n>=\int\psi^*_n(x)\psi^_m(x)dx=0</math> unless <math>m=n</math>. Note that in QM, <math><m|n></math> is use to represent the dot product between two wave functions. Vectors are represented by the index only here. We say that the two wave functions are orthogonal. The dot product of two orthogonal vectors is zero, too.
• 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>

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