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TuringTest

  • 2 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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  1. TuringTest
    • 2 years ago
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    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
    • 2 years ago
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    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
    • 2 years ago
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    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
    • 2 years ago
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    @Jemurray3 tensors?

  5. TuringTest
    • 2 years ago
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    reading this: https://www.edx.org/c4x/BerkeleyX/CS191x/asset/chap3.pdf

  6. Jemurray3
    • 2 years ago
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    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
    • 2 years ago
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    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
    • 2 years ago
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    No, a column vector times a row vector is a matrix.

  9. Jemurray3
    • 2 years ago
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    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
    • 2 years ago
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    oh crap, that is embarrassing... I was thinking too hard

  11. TuringTest
    • 2 years ago
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    right, of course, I need to review my linear algebra for this stuff I suppose

  12. Jemurray3
    • 2 years ago
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    It happens, no worries. <a | b > is a scalar (inner product) but |a><b| is an operator (outer product)

  13. TuringTest
    • 2 years ago
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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?

  14. TuringTest
    • 2 years ago
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    I understand it mathematically, but what is the physical implication?

  15. TuringTest
    • 2 years ago
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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

  16. TuringTest
    • 2 years ago
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    I know this should be in physics now, but I've got you here, so...

  17. Jemurray3
    • 2 years ago
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    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
    • 2 years ago
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    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
    • 2 years ago
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    To the best of my understanding the primary mechanism for flipping spins and such is magnetic fields.

  20. TuringTest
    • 2 years ago
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    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
    • 2 years ago
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    I don't suppose you'd be able to give an example of how we might implement the Hadamard gate then?

  22. Jemurray3
    • 2 years ago
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    http://library.thinkquest.org/07aug/01632/qubitcontrol.html This may or may not be of help.

  23. Jemurray3
    • 2 years ago
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    ^ I realize that "may or may not" is a meaningless phrase. *This may be of help.*

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