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TuringTest
 3 years ago
How can you multiply a ket by a bra in that order? (no this is not a stupid joke)
TuringTest
 3 years ago
How can you multiply a ket by a bra in that order? (no this is not a stupid joke)

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TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.1I have an expression for a way to write a matrix in the for\[A=\sum_{i,j=1}^ma_{ij}e_i \rangle\langle e_j\]and I don't understand how to multiply the ket by the bra. @Luis_Rivera no, it comes from a quantum physics thing

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.1according to my understanding, a ket can be thought of as a vertical vector, and a bra a horizontal one. Matrix multiplication is not commutative, and #rows /=#colums for each, so... how does that work?

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.1somehow\[e_i\rangle\langle e_j\otimesf_k\rangle\langle f_l=(e_i\rangle\otimesf_k\rangle)(e_j\rangle\otimesf_l\rangle)\]where \(\otimes\) is the tensor product. I don't know if anyone here can help me with that.

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.1reading this: https://www.edx.org/c4x/BerkeleyX/CS191x/asset/chap3.pdf

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0What are you trying to do? In \[ A = \sum a_{ij} \mid e_i \rangle \langle e_j \mid \] the e_i and e_j vectors are the basis vectors.

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.1I'm trying to understand what \(e_i\rangle\langle e_j\) are doing next to each other I understand that \(\langle e_j\) is the Hermitian conjugate of \(e_j\rangle\), and so the inner product is \(\langle e_ie_j\rangle\) which makes sense because e_i is a row vector and e_j is a column vector, but the other way around it would be a column vector times a row vector, which is undefined, right? Or am I just hopelessly confused?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0No, a column vector times a row vector is a matrix.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0for instance, \[\left(\begin{matrix}1 \\ 0\end{matrix}\right) (1 \space 0) = \left(\begin{matrix}1 &0\\ 0 & 0\end{matrix}\right)\]

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.1oh crap, that is embarrassing... I was thinking too hard

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.1right, of course, I need to review my linear algebra for this stuff I suppose

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0It happens, no worries. <a  b > is a scalar (inner product) but a><b is an operator (outer product)

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.1while we're on the subject, can you help me visualize the implementation of a quantum gate on a system, like a photon or hydrogen atom? I'm having trouble understanding what a unitary transformation actually does, is it just a change in basis of sorts?

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.1I understand it mathematically, but what is the physical implication?

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.1like a NOT gate, if a photon is in the 1> state, what does it mean, or how does it pass through this gate? and this gate will then return the photon in a 0> state? could you clarify? I don't know about you, but I find QM a bit counterintuitive ;P

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.1I know this should be in physics now, but I've got you here, so...

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0It is indeed. I'm not tremendously familiar with quantum computation but generally speaking the easiest thing for me to visualize is a spin1/2 system. An electron may be either spin up or down which you can denote 0> and 1> respectively. A "not" transformation just flips the spin, i.e. takes 0> to 1> and 1> to 0>.

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.1right, but *how* do we flip it? quantum gates seem to be no more than mathematical constructs. Is it similar to the measuring affect where we are collapsing the wave function? I figure not, since it always flips the outcome, as opposed to just changing the probability of measuring that state.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0To the best of my understanding the primary mechanism for flipping spins and such is magnetic fields.

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.1So as far as you know quantum gates are implemented physically, and are not just like a change of basis which we arbitrarily construct mathematically??

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.1I don't suppose you'd be able to give an example of how we might implement the Hadamard gate then?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0http://library.thinkquest.org/07aug/01632/qubitcontrol.html This may or may not be of help.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0^ I realize that "may or may not" is a meaningless phrase. *This may be of help.*
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