## TuringTest Group Title 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

1. TuringTest Group Title

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

2. TuringTest Group Title

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?

3. TuringTest Group Title

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.

4. TuringTest Group Title

@Jemurray3 tensors?

5. TuringTest Group Title
6. Jemurray3 Group Title

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.

7. TuringTest Group Title

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?

8. Jemurray3 Group Title

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

9. Jemurray3 Group Title

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

10. TuringTest Group Title

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

11. TuringTest Group Title

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

12. Jemurray3 Group Title

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

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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?

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

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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

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

17. Jemurray3 Group Title

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>.

18. TuringTest Group Title

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.

19. Jemurray3 Group Title

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

20. TuringTest Group Title

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??

21. TuringTest Group Title

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

22. Jemurray3 Group Title

http://library.thinkquest.org/07aug/01632/qubitcontrol.html This may or may not be of help.

23. Jemurray3 Group Title

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