Bra-ket notation to Matrix for entangled pairs

[nomadreid]

I am confused about the the notation |ab> for an entangled pair. Isn't this the same as the tensor product |a> \otimes |b>? If so, I run into another confusion when using the corresponding matrices. I read that I should apply a Hadamard operator H twice to the input state |01>. Does this mean (H|0>)\otimes (H|1>)? I don't see how it could mean H(H(|01>), since |01> is represented by a four-by-one matrix, whereas H is represented by a two-by-two matrix. So what does this mean? Thanks

 

[Greg Bernhardt]

@nomadreid did you find any more insight on this topic?

 

[nomadreid]

No, unfortunately.

 

[DrClaude]

I am confused about the the notation |ab> for an entangled pair. Isn't this the same as the tensor product |a> \otimes |b>?

Yes.

If so, I run into another confusion when using the corresponding matrices. I read that I should apply a Hadamard operator H twice to the input state |01>. Does this mean (H|0>)\otimes (H|1>)? I don't see how it could mean H(H(|01>), since |01> is represented by a four-by-one matrix, whereas H is represented by a two-by-two matrix. So what does this mean?

The statement is ambiguous. The use of the word "twice" mean for me that the same thing has to be done two times. If you were to apply a Hadamard transformation to both qubits, the language would be different, I think. However, what is not said is which qubit the Hadamard transformation must be applied to.

To apply the transformation to a single qubit in a two-qubit system, one must use $H_1 \otimes 1_2$ where the subscript refer to which qubit is operated on. In matrix form, this is a 4x4 matrix.

That said, the Hadamard matrix is symmetric and unitary. What happens when it is applied twice in row?