Question

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

Answer 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

Answer 2

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?

Answer 3

somehow\[|e_i\rangle\langle e_j|\otimes|f_k\rangle\langle f_l|=(|e_i\rangle\otimes|f_k\rangle)(|e_j\rangle\otimes|f_l\rangle)\]where \(\otimes\) is the tensor product. I don't know if anyone here can help me with that.

Answer 4

@Jemurray3 tensors?

Answer 5

reading this:
https://www.edx.org/c4x/BerkeleyX/CS191x/asset/chap3.pdf

Answer 6

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.

Answer 7

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 \(\langle e_i|e_j\rangle\) 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?

Answer 8

No, a column vector times a row vector is a matrix.

Answer 9

for instance,
\[\left(\begin{matrix}1 \\ 0\end{matrix}\right) (1 \space 0) = \left(\begin{matrix}1 &0\\ 0 & 0\end{matrix}\right)\]

Answer 10

oh crap, that is embarrassing... I was thinking too hard

Answer 11

right, of course, I need to review my linear algebra for this stuff I suppose

Answer 12

It happens, no worries. <a | b > is a scalar (inner product) but |a><b| is an operator (outer product)

Answer 13

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?

Answer 14

I understand it mathematically, but what is the physical implication?

Answer 15

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 counter-intuitive ;P

Answer 16

I know this should be in physics now, but I've got you here, so...

Answer 17

It is indeed. I'm not tremendously familiar with quantum computation but generally speaking the easiest thing for me to visualize is a spin-1/2 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>.

Answer 18

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.

Answer 19

To the best of my understanding the primary mechanism for flipping spins and such is magnetic fields.

Answer 20

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??

Answer 21

I don't suppose you'd be able to give an example of how we might implement the Hadamard gate then?

Answer 22

http://library.thinkquest.org/07aug/01632/qubitcontrol.html

This may or may not be of help.

Answer 23

^ I realize that "may or may not" is a meaningless phrase. *This may be of help.*