# Timeline for Can everything in QM be described with degrees instead of matrices and vectors?

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Oct 18 '19 at 8:34 comment |x><y| * |z> = |x> * <y|z> = |x> * c (c is the inner product) = c|x>. Remember that the inner product can be though of as how much two states are similar. And order of multiplication doesn't matter when you multiply by a scalar.
Oct 18 '19 at 2:47 comment Can you explain this |x><y| * A = (how much y is simialar to A) * |x>
Oct 17 '19 at 18:33 comment Dirac is just a way to represent vectors. At the end of the day the rules for matrix multiplication applies. if you have a scalar 'a' and a matrix/vector V. then aV = Va. the result of an inner product is a scalar, so you can write it to the left/or right of the vector. If you have 2 operators(matrices) A,B and a state (vector) x, then AB|x> means that B is acting on x, and A is acting on the resulting vector. |0><1| is the col(1,0) * row(0,1) = 2x2 matrix: [[0,1],[0,0]]. I advise you to compute some outer products and act them on different states to get the intuition for what the do/mean
Oct 16 '19 at 4:51 comment "Yes, you calculate from right to left." Why from right to left??? I find this (|φ><ψ|)|y> = |φ>(<ψ|y>) and this |ω〉〈τ|(|ψ〉) = |ω〉〈τ|ψ〉 = 〈τ|ψ〉|ω〉, it looks like doesn't matter for right-left? Look same.
Oct 10 '19 at 2:40 comment -1) "|x><y|*A" how this become (<y|A)*|x> and not |x>(<y|*A) if "Yes, you calculate from right to left"? -2) Do you know any software for that "convert Dirac to matrices and vectors" and back?
Oct 9 '19 at 8:16 comment 1- Yes, you calculate from right to left. 2- X = X = |0><1| + |1><0|, |0><1| by it self is not unitary. 3- correct except the last statement. The inner product of "<y| * A" is a scalar (number). then the the state will be that scalar * |x> (not outer product). Finally, I suggest that, if you ever struggle with the dirac notation, just convert to matrices and vectors and do the multiplication. You'll get much better intuition of what's happening and why.
Oct 9 '19 at 6:52 history edited
Oct 9 '19 at 2:31 comment In Dirac notation, we calculate from right to left? X = |0><1| this mean just what is look like "if qbit is 1 flip it to 0", or H = |+><0| "qbit from 0 to +, superposition"? You said |x><y| * A = (how much y is simialar to A) * |x>, first we calculate inner product of "<y| * A", then we do outer product of |x>PREVIOUS_INNER_PRODUCT ?
Oct 7 '19 at 13:02 comment Outer products: First it's an operator (nxn). it can be looked at in two different ways: |a><a| * x: is the projection of x into the space spanned by |a>. Second it can be used to construct unitary operators. For example the quantum equivalent to the not gate is an operation X = |0><1| + |1><0|, similarly Hadamard gate = |+><0| + |-><1| because it maps 0 to + and 1 to -. I suggest that you do the matrix multiplication yourself and you'll get a sense for how |x><y| acting on a state behaves. you'll notice that |x><y| * A = (how much y is simialar to A) * |x>
Oct 7 '19 at 13:02 comment I will try to do my best to answer your questions in simple way, doing that I'm bound to make mistakes. Yes, a ket is a column vector. Bra is the complex conjugate of a ket. this enables defining the inner product as <a|b> which looks nice. In a way bra is an operator from H -> C. It doesn't represent the system, it represents an operator that gives you how similar two states are (complex number). (outer product in next reply bcz of character limit)
Oct 6 '19 at 0:50 comment Thanks for this explanation! "kets represent the state of a system of qubits". Ket is column vector written in different form, right? That is same? Second question, what represent "bra" (row vector?) and what represent "outer product"?
Oct 5 '19 at 13:33 comment I'm afraid you've got everything wrong. kets represent the state of a system of qubits. Matrices represent an operation on a given state. You can look at a matrix and see what it does to a state by looking at it's columns. Column i is what happens to the ith bases. so an operation [c1 c2 c3..] will map the states: [1 0 0 .. ]* to c1, [0 1 0 .. ]* to c2.. etc. A state that is in a superposition of states [a b c ..]* will be mapped to ac1+bc2+... etc. For a matrix to be a valid QM operation it needs to be unitary.
Oct 5 '19 at 3:01 comment Maybe, better way to describe 10 qbits system/circuit is to use one matrix? What are Bra and Ket then? Ket represent columns of that matrix, and Bra represent rows of that 10 qbit matrix? Or I get everything wrong about qbits matrices, Bra, Ket, etc!!!
Oct 4 '19 at 6:27 comment you can represent the state of 10 independent qubits on 10 bloch spheres. However take bell state |00>+|11> There is no two points on 2 bloch spheres that represent this state. If you picked the points |0>+|1> on both of the spheres it would be the the state :|00>+|01>+|10>+|11> which is not the bell state. That is because 2 bloch spheres gives you 4 degrees of freedom, while 2 qubit system has 6 degrees of freedom. You wouldn't be able to represent the "entangled" part of the two qubits system. Back to the main question a unitary U: |00> -> |00>+|11> must be rotation on sphere with 6D surface
Oct 4 '19 at 4:52 comment I don't understand this "multi-dimensions". If we have 10 REAL ELECTRONS, how they can make multidimension surface of some sphere? They are 10 real electrons, and we should represent them with 10 bloch spheres. As you can see I'm engineer and think about real objects.
Oct 3 '19 at 13:10 review
Oct 3 '19 at 22:16
Oct 3 '19 at 13:06 history answered