Can you explain to me again, why what I have drawn is not the wavefunction but the probability density? I think it might be the wavefunction.....I am still confused. I think it should be the continuation of exponential decay when the wave emerges from the barrier for E < V and for E > V, I...
Like this?
Also, just wana clarify something, I know I am probably being pedantic but anyways...Is 50 % transmission is the same as R = T = 0.5? Cause the question asks for the probability density sketches for 50% transmission.
I think 50% transmission is the same as R = T = 0.5...
This is the quantum harmonic oscillator where \hat a\dagger and \hat a are the step down and step up (SOMETIMES CALLED LADDER) operators according to your definitions.
Correction: I think you meant to say \hat N = \hat a\dagger \hat a
So would the probability density look like this? The transmitted wave and reflected waves have a reduced amplitude...i.e. they are 1/2 of the original amplitude (incident amplitude). This is the plot of the probability density. Does it make sense?
Looking forward to hearing from you soon...
Consider a Quantum Mechanical particle approaching a barrier (potential) of height V_0 and width a. What will the sketch of the probability density look like if there is a 50% chance of reflection and a 50% chance of transmission? Can you explain why cause after reading Griffith' s Quantum...
How do I plot graphs in LaTeX? Example sin(x) to begin with. :frown:
Also, how do I insert pictures in LaTeX? Example, simple circuit diagrams.
student :confused:
I have done clamping, clipping, half wave rectifier and full wave rectifier circuits. For the diagram, I think it is a cos wave with amplitude 10 V but have no idea why. I am genuinely lost here.
student
Sketch the output voltage as a function of time. The AC voltage source is V_{o}cos(\omega)t with V_{o} = 10V and \omega = 2000rad/sec.
I have posted a diode circuit question in the attachment
Ok, I think it should be a sine curve with a 10 V amplitude but am not too sure about the period...
If i multiple both sides by exp(-ik'x) the LHS gives exp(-ik'x-(x/2a)^2). I' m not sure what to do with this to simplify it further. Do i have to try to complete the square in this exponential now?
Plot distance along x axis versus peak number. calculate the slope using the method of least squares. Plot distance from 0th peak versus peak number. Calculate the slope as before. The ratio of the slopes gives the excitation potential.
There is another method using the current just before...
Is there a site where I could see what people have done in various experiments? For example, I will be doing the verlocity of light experiment and would like to know what kinds of data analysis people have done so I can get some ideas as to what I could do with my data.
Is there any such...
I know about that, but it depends on what the size of the box is. The only reason things worked out nicely to one expression was due to the fact that the box was centered at the origin and was of size L rather than what I have which is 2L.
I think there was an expression relating energy to...
I' m not very knowledgable on this issue....perhaps someone else can help ...I say excitation energy is the difference between peaks and subtract the potential at the location of the first peak to get the contact potential...WAIT for SOMEONE ELSE to back this up.
student1938
Particle in a BOX -- what are allowed momenta?
Ok I am trying to come up with the first five eigenfunctions for the particle in a box of size 2L. Now, I gave the appropriate initial conditions and get as a solution phi(x) = bcos(kx) + asin(kx). I said that phi(-L) = phi(L) = 0 which game me...
oh did you use v = omega X r where r = l and the angle is 90 degrees.....now that makes sense. So then T = (1/2)*m*(l*omega)^2 and I can now use Lagrange' s Eqs right?
the speed should just be d(lcos(theta))/dt right? I think thta lcos(theta) is the horizontal distance and we are dividing by time here to get speed right?
For some reason, I am just unable to figure it out...according to me, based on the last hint, it should be (1/2)*m*(l*cos(theta))^2...I am saying l cos theta because that is the distance to the CM from the tip of the side rods.....does that make sense? Any suggestions?
So then KE is just (1/2)*I*omega^2. The bottom rod does not rotate so there is no KE like you said. IS that all that is needed for the KE? how would I get omega? I think that I should just leave it the way it is