Recent content by knut-o

So, does that mean that the Z, in 1983 was produced with annihilation of the proton and anti-proton? Because that gives two photons, but considering the energies, isn't there possible that some of them (enough to measure) was Z's? And the W+'s were then produced when a anti-down "collides" with...

I am familiar with Feynman-diagrams, at least to the degree where it sort of makes somewhat sense. The problem is that I have large issues understanding the Z-boson, works. I only found a few pictures where it's a Feynman-diagram of two leptons (electrons) that become a Z that after that...

Homework Statement So, I got this task, which I feel either really dumb, or nobody is able to explain: I got this question: Z and W was discovered in 1973 at CERN, how were they produced? Include Feynman diagram and a short description. Then there is this question: How is the Z and W produced...

So I end up with something like integrating dS from Si to S(T,P) which gives me that S(T,P) - Si=CPln(T/Ti)+C and just say that C=0 and move Si over? And how would it work to calculate \Delta G and \Delta S for difference between liquid water and water vapoir? I know I have CP, Ti, but what...

Not impossible, :p . So confusing! But I am supposed to show that (1) is true, And if I put dP to 0, then the integration becomes the integral of ? Makes no sense :s . I assume then it becomes an integration constant that I somehow need to show is S(Ti,P)?

Homework Statement I know that Gibbs free energy: G=H-TS, and therefor \frac{dG}{dT}=-S , and that dS=C_P\frac{dT}{T} Now, I am to show that more generally, dS=C_P\frac{dT}{T}-\frac{dV}{dT}dP(1) (assuming that the difference between delta and d is mostly the same (symbolwise). The hint I...

Oh, that's not to bad. I really feel like a physisist when it comes to having a mess in my papers and writing. And since it's the squared of k, the positive and negative version will provide the same energi. And, aaarhh, linearly independency, I hate that stuff. If you mean you can express the...

There is, however, a small problem with b) . It's the calculations, I am just assuming that the task is easy, which is why I haven't really bothered with the derivations to a big degree (the mathematical problems isn't to hard for me). Is it fair to assume that V(\psi)=0? Or am I supposed to...

Aha! It makes sense :p . I thought the integral was from -inf to inf like normal. That N=\sqrt{\frac{1}{2\pi}} makes more sence, it even fits what my book says, I think. Then I guess there is only c) left, and the problem is that I am pretty certain the solution is that psi is 2pi, and k is...

Quantum physics, I believe Lz is the term for angular momentum. I have a one dimensional system with a particle with mass M and it's moving a long a circle with radius R. a)Use Lz=MvR to express the particle kinetic energi with help of Lz. Then use the substituion that L_z\rightarrow...

\psi _2=a_+\psi _1=B1(m\omega x -\hbar\frac{d}{dx})\cdot e^{-\frac{m\omega}{2\hbar}x^2}=B1(m\omega x^2e^{-\frac{m\omega}{2\hbar}x^2} - \hbar\frac{dxe^{-\frac{m\omega}{2\hbar}x^2}}{dx})? If I do this derivation, sort out the constants etc, I will end up with \psi _2? Which should give me...

Does that mean I multiply \psi _1 with a_+? And sorting out the constants, and deriving psi _1 where it's needed for x, or simply multiply it with mwx ? So I get something along the lines of: Where B1=(\frac{m\omega}{\pi\hbar})^{\frac{1}{4}}*\frac{1}{\hbar} \psi _2=a_+\psi _1=B1(m\omega x...

Now, I've done some real attempts. Now, it's hard to write the many integrals in latex. The integrals still goes from -inf to inf. For psi 0: <x^2>=\sqrt{\frac{\hbar ^3}{(m\omega)^3}}\alpha ^2\int \xi ^2e^{-\xi ^2}=\frac{\hbar}{2m\omega} <p^2>=\alpha ^2\hbar\sqrt{m\omega}(\int e^{-\xi...

I assume I am going to end up with meter, and I don't do that now, end up wit Js2 or something.. Odd function? Odd function over symetric limits? I must admit I am not quite sure about this. Part of the problem is doing it correcltly since I most likely going to screw it up when it comes to...

Oh, no wonder I had some severe mistakes in my calculations. I calculated <x^2>=\int x\psi (x,t)^4dx\\ <p^2>=\int (\psi\cdot \frac{\partial\psi}{\partial x})^2dx But let us do some calculations, or rather. I got osme answers that I don't know if are correct or not. For <x2> I got...