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

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TuringTestBest ResponseYou've already chosen the best response.1
I 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
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
according 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?
 one year ago

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

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@Jemurray3 tensors?
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
reading this: https://www.edx.org/c4x/BerkeleyX/CS191x/asset/chap3.pdf
 one year ago

Jemurray3Best ResponseYou've already chosen the best response.1
What 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.
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
I'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?
 one year ago

Jemurray3Best ResponseYou've already chosen the best response.1
No, a column vector times a row vector is a matrix.
 one year ago

Jemurray3Best ResponseYou've already chosen the best response.1
for instance, \[\left(\begin{matrix}1 \\ 0\end{matrix}\right) (1 \space 0) = \left(\begin{matrix}1 &0\\ 0 & 0\end{matrix}\right)\]
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
oh crap, that is embarrassing... I was thinking too hard
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
right, of course, I need to review my linear algebra for this stuff I suppose
 one year ago

Jemurray3Best ResponseYou've already chosen the best response.1
It happens, no worries. <a  b > is a scalar (inner product) but a><b is an operator (outer product)
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
while 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?
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
I understand it mathematically, but what is the physical implication?
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
like 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
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
I know this should be in physics now, but I've got you here, so...
 one year ago

Jemurray3Best ResponseYou've already chosen the best response.1
It 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>.
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
right, 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.
 one year ago

Jemurray3Best ResponseYou've already chosen the best response.1
To the best of my understanding the primary mechanism for flipping spins and such is magnetic fields.
 one year ago

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

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

Jemurray3Best ResponseYou've already chosen the best response.1
http://library.thinkquest.org/07aug/01632/qubitcontrol.html This may or may not be of help.
 one year ago

Jemurray3Best ResponseYou've already chosen the best response.1
^ I realize that "may or may not" is a meaningless phrase. *This may be of help.*
 one year ago
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