Is the imaginary number i "necessary" in the pauli matrices simply because of the condition of having 3 mutually orthogonal axi?
If space were two dimensional we wouldn't need the i imaginary number?
Wonderful that makes sense, thank-you.
Ok so would it be correct to think of a density matrix as a way of packaging probabilities ( both classical probabilities in case of say someone setting up a system with classical probability or classical probability due to inefficiency of a measuring...
For the above ##|\alpha|^2 + |\beta|^2 = 1## it must mean that each sum term is less than 1 so how can it meet condition of pure state being that tr of p squared has to be equal to 1? Wouldnt squaring terms less than 1 like 1/2 give 1/4 and then eg 1/4 plus 1/4 equal half therefore not meeting...
So would it be correct to say that in case of up and down vectors that the co-efficients ( even though not explicitly stated ) could for example in the UP state vector be 1 and 0 which then when squared give the probability of measuring up in the up state 1X1 or certain 100% and in the down...
Thankyou so much for taking time to reply with all you wrote.
So it is comprised of the amplitudes squared in pure states and even could comprise of stated probabilities from classical imperfection in measurement to be included in a mixed state but is it correct to say the the probabilities will...
Thanks so much.
And then like you said, if not a pure state, you could theoretically if known even STATE probabilities which would be become the diagonal elements of the matrix. For example, if you knew somehow that there was ( for whatever reason ) a certain probability of some observable then...
Hi, I am wanting to confirm my understanding of the density matrix in quantum mechanics. Is it the wave function co-efficients squared - in other words the wave amplitudes squared which in turn are the probabilities and then these turn out to be placed into a matrix form with the squared wave...