On Fri, Dec 7, 2018 at 2:24 PM Henry Baker <hbaker1@pipeline.com> wrote:
I don't understand.
Perhaps I have 2 different problems: the actual physical system and the mathematical abstraction of it.
Let's take the math abstraction first. (I prefer a physics-free matrix notation.)
Forget about "pure" or "mixed" states; we're talking about matrices & vectors over the standard (uncountable) complex numbers, aren't we?
Let's not even get into wave function "collapse"/"measurement"; I'm trying to understand the wave-function evolution first.
Let's consider a small set of qbits -- say 3.
If I understand what you're saying, the 3 qbits just sit there unless hit by a unitary operator; is this correct?
(By "just sit there", I mean *nothing* changes, not even any phase factors, whether observable or not.)
OK. Physically, the simplest example I can think of is the spins of three well-separated electrons, with no electric or magnetic field.
Is the "state" of the system a 3-vector of complex numbers or a 3x3 (constant?) matrix of complex numbers?
It's a 2^3=8-dimensional vector of complex numbers |ψ> = ∑ᵢ cᵢ|i> where i ranges from 000 to 111. Note that I've implicitly chosen an orthonormal basis for this vector; this amounts to picking a direction "z" in space.
By "state change", I presume that we pre- (or post-) multiply this initial "state" by a 3x3 unitary matrix; correct?
Premultiply by an 8x8 matrix U(t) = exp(iHt) where H is a Hermitian perturbation of the system.
So our computation consists of k steps -- an ordered sequence of k matrix multiplications; correct?
Yep.
The time-reversed computation is the conjugate transpose of the entire sequence; correct?
Yep.
So how can I represent a simple harmonic oscillator using one (more more) qbits? Wouldn't a harmonic oscillator "DO" something even if isolated from the rest of the universe? Wouldn't there be some time-variation of the "state" -- i.e., a periodicity?
A quantum harmonic oscillator assumes a quadratic potential V(x) = (x^2)/2. The nth eigenvector of the system has energy proportional to (n+1/2). In the Schrodinger picture, the states with energy E change phase by exp(-iEt/ℏ). In the Heisenberg picture, that phase is absorbed into the Hamiltonian operator. To get the particles in the QHO to change energy levels, you perturb H = (p^2 + x^2)/2 for some period of time, which multiplies the state by a unitary matrix as above.
At 12:54 PM 12/7/2018, Mike Stay wrote:
The pure states of a quantum system are given by the eigenvectors of a Hamiltonian operator, which is Hermitian.
The states are stationary, in the sense that only their phases change over time: |ψ_t> = exp(-iHt/℠) |ψ_0>.
If you want something to *happen*, then you need to perturb the Hamiltonian for some period of time.
If you want to measure the state, you typically start with two independent Hamiltonians (one for the quantum system and one for a pointer state) and then add a coupling perturbation for some period of time.
On Fri, Dec 7, 2018 at 12:22 PM Henry Baker <hbaker1@pipeline.com> wrote:
As I understand it, a simple quantum system undergoes transitions via unitary transformations.
Unitary transformations preserve inner products, so they are "rigid" rotations of a "configuration" in N-space (assuming that a finite dimensional quantum system can exist).
Nevertheless, from a given "state" (whatever that means), not all unitary transformations seem possible.
For example, if a system is isolated from its environment, wouldn't it "evolve" in an "inertial-like frame" manner? If the rotation analogy is correct, wouldn't this be a rigid constant-speed spin about some axis/plane? Wouldn't this spin have some "angular momentum" (resistance to slowing down) ?
Aren't there other unitary evolutions which will require some external "effort" to change its course -- i.e., axis ? And in this case, wouldn't reversibility require the re-appearance of the conjugate of this "effort" to change the course back?
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