Momentum of Stationary States
About points...
We associate a certain number of points with each exercise.
When you click an exercise into a collection, this number will be taken as points for the exercise, kind of "by default".
But once the exercise is on the collection, you can edit the number of points for the exercise in the collection independently, without any effect on "points by default" as represented by the number here.
That being said... How many "default points" should you associate with an exercise upon creation?
As with difficulty, there is no straight forward and generally accepted way.
But as a guideline, we tend to give as many points by default as there are mathematical steps to do in the exercise.
Again, very vague... But the number should kind of represent the "work" required.
When you click an exercise into a collection, this number will be taken as points for the exercise, kind of "by default".
But once the exercise is on the collection, you can edit the number of points for the exercise in the collection independently, without any effect on "points by default" as represented by the number here.
That being said... How many "default points" should you associate with an exercise upon creation?
As with difficulty, there is no straight forward and generally accepted way.
But as a guideline, we tend to give as many points by default as there are mathematical steps to do in the exercise.
Again, very vague... But the number should kind of represent the "work" required.
About difficulty...
We associate a certain difficulty with each exercise.
When you click an exercise into a collection, this number will be taken as difficulty for the exercise, kind of "by default".
But once the exercise is on the collection, you can edit its difficulty in the collection independently, without any effect on the "difficulty by default" here.
Why we use chess pieces? Well... we like chess, we like playing around with \(\LaTeX\)-fonts, we wanted symbols that need less space than six stars in a table-column... But in your layouts, you are of course free to indicate the difficulty of the exercise the way you want.
That being said... How "difficult" is an exercise? It depends on many factors, like what was being taught etc.
In physics exercises, we try to follow this pattern:
Level 1 - One formula (one you would find in a reference book) is enough to solve the exercise. Example exercise
Level 2 - Two formulas are needed, it's possible to compute an "in-between" solution, i.e. no algebraic equation needed. Example exercise
Level 3 - "Chain-computations" like on level 2, but 3+ calculations. Still, no equations, i.e. you are not forced to solve it in an algebraic manner. Example exercise
Level 4 - Exercise needs to be solved by algebraic equations, not possible to calculate numerical "in-between" results. Example exercise
Level 5 -
Level 6 -
When you click an exercise into a collection, this number will be taken as difficulty for the exercise, kind of "by default".
But once the exercise is on the collection, you can edit its difficulty in the collection independently, without any effect on the "difficulty by default" here.
Why we use chess pieces? Well... we like chess, we like playing around with \(\LaTeX\)-fonts, we wanted symbols that need less space than six stars in a table-column... But in your layouts, you are of course free to indicate the difficulty of the exercise the way you want.
That being said... How "difficult" is an exercise? It depends on many factors, like what was being taught etc.
In physics exercises, we try to follow this pattern:
Level 1 - One formula (one you would find in a reference book) is enough to solve the exercise. Example exercise
Level 2 - Two formulas are needed, it's possible to compute an "in-between" solution, i.e. no algebraic equation needed. Example exercise
Level 3 - "Chain-computations" like on level 2, but 3+ calculations. Still, no equations, i.e. you are not forced to solve it in an algebraic manner. Example exercise
Level 4 - Exercise needs to be solved by algebraic equations, not possible to calculate numerical "in-between" results. Example exercise
Level 5 -
Level 6 -
Question
Solution
Short
Video
\(\LaTeX\)
No explanation / solution video to this exercise has yet been created.
Visit our YouTube-Channel to see solutions to other exercises.
Don't forget to subscribe to our channel, like the videos and leave comments!
Visit our YouTube-Channel to see solutions to other exercises.
Don't forget to subscribe to our channel, like the videos and leave comments!
Exercise:
For the infinite potential well show that the expectation value for the momentum of a stationary state is equal to zero and that the uncertay is sigma_p fracnpi hbarL
Solution:
The expectation value for the stationary state n is langle p rangle_n _^L psi_n^*xt hat p psi_nxt textrmdx _^L psi_n^*xt fractextrmdtextrmdxpsi_nxt textrmdx _^L A^* sink_n x -ihbarfractextrmdtextrmdxAsink_n x textrmdx |A|^_^L sink_n xcosk_n x k_n textrmd x k_n |A|^ _^L sink_n xcosk_n x textrmdx With the trigonometric identity sinalphacosbeta fracleftsinalpha-beta+sinalpha+betaright the egrand can be written as sinkn xcosk_n x fracleftsin+sin k_n xright fracsin k_n x The egral of sin k_n x is _^L sin k_n x textrmdx -fraccos k_n x k_n Big|_^L -fraccos k_n L-cos k_n -fraccos pi n-cos k_n -frac - k_n This proves the first statement. vspacemm The expectation value for p^ is langle p^ rangle_n _^L psi_n^*xt hat p^ psi_nxt textrmdx _^L psi_n^*xt -hbar^fractextrmd^textrmdx^A sink_n xe^iomega t textrmdx -hbar^_^L psi_n^*xt-Asink_n xk_n^ e^iomega t textrmdx hbar^ k^ _^L psi_n^*xt psi_nxt textrmdx hbar k_n^ where we have used the normalisation condition for the state psi_n. vspacemm It follows for the uncertay sigma_pn sqrtlangle p^ rangle_n - langle p rangle_n^ sqrthbar k_n^ - hbar k_n hbar fracnpiL fracnpihbarL
For the infinite potential well show that the expectation value for the momentum of a stationary state is equal to zero and that the uncertay is sigma_p fracnpi hbarL
Solution:
The expectation value for the stationary state n is langle p rangle_n _^L psi_n^*xt hat p psi_nxt textrmdx _^L psi_n^*xt fractextrmdtextrmdxpsi_nxt textrmdx _^L A^* sink_n x -ihbarfractextrmdtextrmdxAsink_n x textrmdx |A|^_^L sink_n xcosk_n x k_n textrmd x k_n |A|^ _^L sink_n xcosk_n x textrmdx With the trigonometric identity sinalphacosbeta fracleftsinalpha-beta+sinalpha+betaright the egrand can be written as sinkn xcosk_n x fracleftsin+sin k_n xright fracsin k_n x The egral of sin k_n x is _^L sin k_n x textrmdx -fraccos k_n x k_n Big|_^L -fraccos k_n L-cos k_n -fraccos pi n-cos k_n -frac - k_n This proves the first statement. vspacemm The expectation value for p^ is langle p^ rangle_n _^L psi_n^*xt hat p^ psi_nxt textrmdx _^L psi_n^*xt -hbar^fractextrmd^textrmdx^A sink_n xe^iomega t textrmdx -hbar^_^L psi_n^*xt-Asink_n xk_n^ e^iomega t textrmdx hbar^ k^ _^L psi_n^*xt psi_nxt textrmdx hbar k_n^ where we have used the normalisation condition for the state psi_n. vspacemm It follows for the uncertay sigma_pn sqrtlangle p^ rangle_n - langle p rangle_n^ sqrthbar k_n^ - hbar k_n hbar fracnpiL fracnpihbarL
Meta Information
Exercise:
For the infinite potential well show that the expectation value for the momentum of a stationary state is equal to zero and that the uncertay is sigma_p fracnpi hbarL
Solution:
The expectation value for the stationary state n is langle p rangle_n _^L psi_n^*xt hat p psi_nxt textrmdx _^L psi_n^*xt fractextrmdtextrmdxpsi_nxt textrmdx _^L A^* sink_n x -ihbarfractextrmdtextrmdxAsink_n x textrmdx |A|^_^L sink_n xcosk_n x k_n textrmd x k_n |A|^ _^L sink_n xcosk_n x textrmdx With the trigonometric identity sinalphacosbeta fracleftsinalpha-beta+sinalpha+betaright the egrand can be written as sinkn xcosk_n x fracleftsin+sin k_n xright fracsin k_n x The egral of sin k_n x is _^L sin k_n x textrmdx -fraccos k_n x k_n Big|_^L -fraccos k_n L-cos k_n -fraccos pi n-cos k_n -frac - k_n This proves the first statement. vspacemm The expectation value for p^ is langle p^ rangle_n _^L psi_n^*xt hat p^ psi_nxt textrmdx _^L psi_n^*xt -hbar^fractextrmd^textrmdx^A sink_n xe^iomega t textrmdx -hbar^_^L psi_n^*xt-Asink_n xk_n^ e^iomega t textrmdx hbar^ k^ _^L psi_n^*xt psi_nxt textrmdx hbar k_n^ where we have used the normalisation condition for the state psi_n. vspacemm It follows for the uncertay sigma_pn sqrtlangle p^ rangle_n - langle p rangle_n^ sqrthbar k_n^ - hbar k_n hbar fracnpiL fracnpihbarL
For the infinite potential well show that the expectation value for the momentum of a stationary state is equal to zero and that the uncertay is sigma_p fracnpi hbarL
Solution:
The expectation value for the stationary state n is langle p rangle_n _^L psi_n^*xt hat p psi_nxt textrmdx _^L psi_n^*xt fractextrmdtextrmdxpsi_nxt textrmdx _^L A^* sink_n x -ihbarfractextrmdtextrmdxAsink_n x textrmdx |A|^_^L sink_n xcosk_n x k_n textrmd x k_n |A|^ _^L sink_n xcosk_n x textrmdx With the trigonometric identity sinalphacosbeta fracleftsinalpha-beta+sinalpha+betaright the egrand can be written as sinkn xcosk_n x fracleftsin+sin k_n xright fracsin k_n x The egral of sin k_n x is _^L sin k_n x textrmdx -fraccos k_n x k_n Big|_^L -fraccos k_n L-cos k_n -fraccos pi n-cos k_n -frac - k_n This proves the first statement. vspacemm The expectation value for p^ is langle p^ rangle_n _^L psi_n^*xt hat p^ psi_nxt textrmdx _^L psi_n^*xt -hbar^fractextrmd^textrmdx^A sink_n xe^iomega t textrmdx -hbar^_^L psi_n^*xt-Asink_n xk_n^ e^iomega t textrmdx hbar^ k^ _^L psi_n^*xt psi_nxt textrmdx hbar k_n^ where we have used the normalisation condition for the state psi_n. vspacemm It follows for the uncertay sigma_pn sqrtlangle p^ rangle_n - langle p rangle_n^ sqrthbar k_n^ - hbar k_n hbar fracnpiL fracnpihbarL
Contained in these collections:
-
Schrödinger Equation by by