Do we really understand quantum mechanics by Laloe F.

By Laloe F.

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The p vectors in Figure 15 don't look perpendicular. This isn't the result of drawing the three-dimensional diagram in two dimensions — rather, it's the result of drawing three complex dimensions as three real ones. In a real three-dimensional space, there's no way to find three perpendicular vectors that all have equal-sized projections onto the x-, y- and z-axes. It's only the fact that p0, p1 and p2 have the extra freedom of complex phases that allows them to be mutually perpendicular. We can write the P matrix in terms of its own eigenvectors, to get a much simpler expression for it than the one based on D and Equation (21).

The actual motion always turns out to involve a stationary point of the action: on a graph it will always be the bottom of a valley, the top of a hill, a flat mountain pass, a plain. Lagrange and Hamilton derived all their results from Newtonian mechanics, but with the advent of quantum mechanics the action principle made perfect sense. If matter is a wave too, it's subject to the same effects as light. A matter wave can't follow a perfectly narrow path, so it can only stay in phase by following a range of paths that involve more or less the same phase shift.

It's only the fact that p0, p1 and p2 have the extra freedom of complex phases that allows them to be mutually perpendicular. We can write the P matrix in terms of its own eigenvectors, to get a much simpler expression for it than the one based on D and Equation (21). Just like Equation (18) for X, we take all the possible values for the momentum, p, and multiply them by tensors that project onto the states with those values of momentum: P = (h√3/4π) (–1 p –1 ⊗p –1 + 0 p 0 ⊗p 0 + 1 p 1 ⊗p 1 ) ( 2 2 ) As with X, this matrix allows us to calculate the average momentum for any state vector ψ, as <ψ,Pψ>.

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