Dirac_Clarity_GOAT
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- TL;DR Summary
- ##\sum_{E} c_E \exp(- \frac{-i}{\hbar} Et) u_{E}(x) ## where ## c_E = \int_{-\infty}{+\infty} u_{E}^{*}(x) \psi_{0}##
Note: My Intro QM course is structured such that we work with Fourier Transformations etc. Before Bra-Ket Notation. Sorry if the notation is cumbersome/exoteric
Hi, in my Homework I was tasked to "derive" the time dependent wavefunction ## \psi(x,t) ## from the initial wave function:
I have attempted the derivation:
Please help me see if I make sense:
$$
\psi(x,0) = \frac{1}{\sqrt{ \sqrt{ 2 \pi } \sigma_{0} }} \exp\left( \frac{i}{\hbar} p_{0} x \right) \exp\left( -\frac{1}{4 \sigma_{0}^{2}} (x- x_{0})^{2} \right)
$$
I derive the energy eigenfunctions from the Potential free Schrodinger equation to get:
$$
u_{E_{n}}(x) = A_{1} \exp(i k_{n} x) + A_{2} \exp(- i k_{n} x)
$$
Where ## k_{n} = \pm \sqrt{ \frac{2mE_{n}}{\hbar^{2}} } ## is the n-th wave number of the free-particle.
And ## A_1 , A_2 ## are Complex numbers to be determined.
Then substituting into:
$$
\psi(x,t) = \sum_{n = 1}^{\infty} c_{E_n} \exp\left( -\frac{i}{\hbar} E_nt \right) u_{E_n}(x)
$$
Although the expression is sum over the different energy eigenvalues, for the next step I have Chosen to integrate over the wave-vector. I do not know the reason. I have done so because my energy eigenfunction are expressed in k. I also have the vague notion that in the Fourier transform of from k -> x involves taking a superposition of different standing waves. For high wave-number => small wave-length => tends to be out of phase => destructive interference => 0 amplitude => localised wave packet which is what I want.
Here is the expression:
$$
\psi(x,t) = \int_{-\infty}^{\infty} [C_1 e^{i(k_0-k)x_0} e^{-\frac{(k_0-k)^2\sigma_0^2}{\hbar^2}} + C_2 e^{i(k_0+k)x_0} e^{-\frac{(k_0+k)^2\sigma_0^2}{\hbar^2}}] e^{-i\frac{\hbar k^2}{2m}t} [A_1 e^{ikx} + A_2 e^{-ikx}] dk
$$
After some tedious calculations I get the wave-function as a sum of 4-components:
1. $$C_1A_1 \int_{-\infty}^{\infty} e^{i(k_0-k)x_0} e^{-\frac{(k_0-k)^2\sigma_0^2}{\hbar^2}} e^{-i\frac{\hbar k^2}{2m}t} e^{ikx} dk$$
2. $$C_1A_2 \int_{-\infty}^{\infty} e^{i(k_0-k)x_0} e^{-\frac{(k_0-k)^2\sigma_0^2}{\hbar^2}} e^{-i\frac{\hbar k^2}{2m}t} e^{-ikx} dk$$
3. $$C_2A_1 \int_{-\infty}^{\infty} e^{i(k_0+k)x_0} e^{-\frac{(k_0+k)^2\sigma_0^2}{\hbar^2}} e^{-i\frac{\hbar k^2}{2m}t} e^{ikx} dk$$
4. $$C_2A_2 \int_{-\infty}^{\infty} e^{i(k_0+k)x_0} e^{-\frac{(k_0+k)^2\sigma_0^2}{\hbar^2}} e^{-i\frac{\hbar k^2}{2m}t} e^{-ikx} dk$$
From this helpful video
http://Phase Velocity versus Group velocity
And Ehrenfest Theorem: I deduce the Group velocity of the Gaussian Packet is:
$$
v_{g} = \frac{\hbar k_{0}}{m} = \frac{p_{0}}{m}
$$
But what was the initial phase velocity of the energy eigenfunctions?
Hi, in my Homework I was tasked to "derive" the time dependent wavefunction ## \psi(x,t) ## from the initial wave function:
I have attempted the derivation:
Please help me see if I make sense:
$$
\psi(x,0) = \frac{1}{\sqrt{ \sqrt{ 2 \pi } \sigma_{0} }} \exp\left( \frac{i}{\hbar} p_{0} x \right) \exp\left( -\frac{1}{4 \sigma_{0}^{2}} (x- x_{0})^{2} \right)
$$
I derive the energy eigenfunctions from the Potential free Schrodinger equation to get:
$$
u_{E_{n}}(x) = A_{1} \exp(i k_{n} x) + A_{2} \exp(- i k_{n} x)
$$
Where ## k_{n} = \pm \sqrt{ \frac{2mE_{n}}{\hbar^{2}} } ## is the n-th wave number of the free-particle.
And ## A_1 , A_2 ## are Complex numbers to be determined.
Then substituting into:
$$
\psi(x,t) = \sum_{n = 1}^{\infty} c_{E_n} \exp\left( -\frac{i}{\hbar} E_nt \right) u_{E_n}(x)
$$
Although the expression is sum over the different energy eigenvalues, for the next step I have Chosen to integrate over the wave-vector. I do not know the reason. I have done so because my energy eigenfunction are expressed in k. I also have the vague notion that in the Fourier transform of from k -> x involves taking a superposition of different standing waves. For high wave-number => small wave-length => tends to be out of phase => destructive interference => 0 amplitude => localised wave packet which is what I want.
Here is the expression:
$$
\psi(x,t) = \int_{-\infty}^{\infty} [C_1 e^{i(k_0-k)x_0} e^{-\frac{(k_0-k)^2\sigma_0^2}{\hbar^2}} + C_2 e^{i(k_0+k)x_0} e^{-\frac{(k_0+k)^2\sigma_0^2}{\hbar^2}}] e^{-i\frac{\hbar k^2}{2m}t} [A_1 e^{ikx} + A_2 e^{-ikx}] dk
$$
After some tedious calculations I get the wave-function as a sum of 4-components:
1. $$C_1A_1 \int_{-\infty}^{\infty} e^{i(k_0-k)x_0} e^{-\frac{(k_0-k)^2\sigma_0^2}{\hbar^2}} e^{-i\frac{\hbar k^2}{2m}t} e^{ikx} dk$$
2. $$C_1A_2 \int_{-\infty}^{\infty} e^{i(k_0-k)x_0} e^{-\frac{(k_0-k)^2\sigma_0^2}{\hbar^2}} e^{-i\frac{\hbar k^2}{2m}t} e^{-ikx} dk$$
3. $$C_2A_1 \int_{-\infty}^{\infty} e^{i(k_0+k)x_0} e^{-\frac{(k_0+k)^2\sigma_0^2}{\hbar^2}} e^{-i\frac{\hbar k^2}{2m}t} e^{ikx} dk$$
4. $$C_2A_2 \int_{-\infty}^{\infty} e^{i(k_0+k)x_0} e^{-\frac{(k_0+k)^2\sigma_0^2}{\hbar^2}} e^{-i\frac{\hbar k^2}{2m}t} e^{-ikx} dk$$
From this helpful video
http://Phase Velocity versus Group velocity
And Ehrenfest Theorem: I deduce the Group velocity of the Gaussian Packet is:
$$
v_{g} = \frac{\hbar k_{0}}{m} = \frac{p_{0}}{m}
$$
But what was the initial phase velocity of the energy eigenfunctions?