I am familiar with the representation theory of finite groups and Lie groups/algebra from the mathematical perspective, and I am wondering how quantum mechanics/quantum field theory uses concepts from representation theory. I have seen the theory of angular momentum in quantum mechanics, and I...
Thanks for your replies! I will look more into the mathematical side of L2 spaces and read the article. I believe the "physical reality" argument for why ψ approaches 0 as x approaches infinity, and I just realized that from the time independent Schrodinger equation, if we require the potential...
I am currently reading through Griffiths Quantum Mechanics textbook, and on page 14, Griffiths proves that
\frac{d}{dt}\int_{-\infty}^{\infty} |\Psi(x,t)|^2 \, dx = \left.\frac{i \hbar}{2m}\left( \Psi^* \frac{\partial \Psi}{\partial x} - \frac{\partial \Psi^*}{\partial x} \Psi \right)...
It really depends how you define addition on cartesian products. The usual definition is
(v_1+v_2,w_1+w_2)=(v_1,w_1)+(v_2,w_2)
In this case, the cartesian product is usually called a direct sum, written as V \oplus W.
If you think about it, this 'product' is more like a sum--for...
We use reduced mass since the proton would be moving too, and so the potential energy depends on both the position of the proton and electron. This is near impossible to solve, and moreover, we usually aren't interested in the proton.
If we redefine the problem in terms of the separation...
The photon momentum do have a dependency on an inertial frame.
According to relativity, E=pc. We also know that E=hf, so
p=hf/c.
The thing to know is, f is dependent on an inertial frame. In fact, f is the temporal part of a 4-vector with the wave vector k being the spatial parts.
So, the...
How do we measure momentum? As far as I know, there is no simple way of measuring momentum by itself; we need to measure the velocity of the particle with respect to a certain frame, and multiply that by m (or by \gamma m). This will give you the total momentum of the particle, not just the...
Even if a particle did have intrinsic momentum, it would be undetectable, since we can always choose an inertial reference frame where this equals zero. Intrinsic angular momentum is detectable since in every inertial frame we agree that the particle possesses angular momentum.