I Bra of a multi-system ket?

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If we have a ket ##|\psi\rangle = |x\rangle \otimes |y\rangle##, then what is the corresponding bra? It seems we want, for example:

$$\langle \psi | \psi \rangle =
\langle x|x \rangle \cdot \langle y|y \rangle =
\langle x| \langle y| |x\rangle|y\rangle$$

I.e., the bra is ##\langle x| \langle y|##. But I swear I've also seen it defined as ##\langle y| \langle x|##, i.e. the order of the systems is switched. Am I crazy, or is there a use for this?

Edit: it seems I'm not the only one.
 

A. Neumaier

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Your displayed formula is incorrect. The right hand side as stated gives the absolute square of the inner product of the two kets, not the product of their squared norm.
 
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I'm sorry, it's been forever since I've touched this stuff and it seems I've forgotten how it works! Could you please be a little more explicit on where exactly I've gone wrong?

The first step says that the inner product of ##|\psi\rangle## with itself ought to equal the (scalar) product of the inner products of its subsystems.

The second step is because ##(A \otimes B)(|x\rangle \otimes |y\rangle) = A|x\rangle \otimes B|y\rangle## (where I'm treating the bras as linear functionals, and the tensor product of scalars is the scalar product).

Ultimately, is ##\langle x|\langle y|## the correct bra, or is ##\langle y|\langle x|##, or do they have different uses in different contexts?
 
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Could you please be a little more explicit on where exactly I've gone wrong?
Your equation as written claims that ##\langle x \vert x \rangle \langle y \vert y \rangle = \langle x \vert \langle y \vert \vert x \rangle \vert y \rangle##. That's not correct. It should be ##\langle x \vert x \rangle \langle y \vert y \rangle = \langle y \vert \langle x \vert \vert x \rangle \vert y \rangle##. Which says that the bra corresponding to ##\vert x \rangle \vert y \rangle## is ##\langle y \vert \langle x \vert##.
 
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Your equation as written claims that ##\langle x \vert x \rangle \langle y \vert y \rangle = \langle x \vert \langle y \vert \vert x \rangle \vert y \rangle##. That's not correct. It should be ##\langle x \vert x \rangle \langle y \vert y \rangle = \langle y \vert \langle x \vert \vert x \rangle \vert y \rangle##.
Hmm I see. I took the following equality:

$$(A \otimes B) (|x\rangle \otimes |y\rangle) = A|x\rangle \otimes B |y\rangle$$

And assumed it applied as well to linear functionals (the bras):
$$(\langle x| \otimes \langle y|) (|x\rangle \otimes |y\rangle) = \langle x|x\rangle \otimes \langle y|y\rangle$$

That's wrong? How do we arrive at your correct equality?

Which says that the bra corresponding to ##\vert x \rangle \vert y \rangle## is ##\langle y \vert \langle x \vert##.
When is this bra used? For example, when taking the inner product of ##|\psi_1\rangle |\phi_1\rangle## and ##|\psi_2\rangle |\phi_2\rangle##, we should presumably turn the first into this bra: ##\langle\phi_1| \langle\psi_1|##.

But the text I'm reading writes their inner product as ##\langle\psi_2| \langle\phi_2| |\psi_1\rangle |\phi_1\rangle## (which, aside from leaving the ##\phi## and ##\psi## in the same order, I just realized swaps the indices -- i.e., they're probably taking the other inner product).

If there's a brief reference (i.e., not a full textbook) that might help me work through the various bits of algebra I'm messing up, please point me to it!

Thanks.
 

hilbert2

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I'm not sure if the tensor products ##|x\rangle |y\rangle## and ##|y\rangle |x\rangle## are in the exact sense the same thing if the two subsystems are not identical, i.e. ##|x\rangle## and ##|y\rangle## belong to different Hilbert spaces.
 
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I took the following equality

$$
(A \otimes B) (|x\rangle \otimes |y\rangle) = A|x\rangle \otimes B |y\rangle
$$
Where are you getting that equality from? It doesn't look right to me. Unless I'm misunderstanding your notation.

the text I'm reading
What text?
 
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Where are you getting that equality from? It doesn't look right to me. Unless I'm misunderstanding your notation.
I've learned this as the definition of how the tensor product of linear operators acts on the space (where A acts on the space containing ##|x\rangle## and B acts on the space containing ##|y\rangle##). For example, Nielsen and Chuang define it that way in their Quantum Computation and Quantum Information (10th ed, p73, eq 2.45).

What text?
See p148, eq 13.45: https://mtlsites.mit.edu/Courses/6.050/2010/notes/chapter13.pdf

Might it be that there are dual definitions we're working with?
 
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See p148, eq 13.45
Ok, I see how their notation works. But their notation is easy to misunderstand when you use it to compute inner products. The question you are asking in the OP, as you ask it, is notation dependent; what the corresponding bra tensor product is to a given ket tensor product depends on how you choose your notation.

In the notation used by the reference you give, you have to write bra tensor products in the same order of Hilbert spaces as the ket tensor products; so the bra corresponding to ##| x \rangle | y \rangle## would be ##\langle x | \langle y |##. But then you have to remember to take the factors in corresponding order when taking inner products. That's complicated when you are dealing with tensor products of several Hilbert spaces.

But in the notation many other references use, the key is to make inner products look simple, and that means you need to reverse the order of Hilbert spaces when writing bra tensor products. That's the notation I was using in post #4. Note that with the order of bras reversed, the inner product is simple; it just drops right out of the formula by simply putting adjacent bras and kets together, and then factoring out the resulting inner product, which is just a number and therefore commutes with everything.
 
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That makes sense! I should have seen right from the start that both notations can be made consistent. I think I'd seen it done both ways and only subconsciously registered that different authors do it differently (but with no fundamental difference).

Thanks!
 

vanhees71

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For me the following is the standard notation
$$\hat{A} \otimes \hat{B} |\psi_1 \rangle \otimes \psi_2 \rangle=(\hat{A} |\psi_1 \rangle) \otimes (\hat{B} \psi_2 \rangle)$$
and
$$(\langle \phi_1 |\otimes \langle \phi_2|)( \psi_1 \rangle \otimes \psi_2 \rangle) = \langle \phi_1 |\psi_1 \rangle \langle \phi_2 |\psi_2 \rangle.$$
It's, of course, a matter of convention again, but I think must books and papers use the above notation.
 
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A. Neumaier

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the following is the standard notation
$$(\hat{A} \otimes \hat{B} )(|\psi_1 \rangle \otimes \psi_2 \rangle)
=\hat{A}|\psi_1\rangle\otimes \hat{B} \psi_2 \rangle$$
$$(\langle \phi_1 |\otimes \langle \phi_2)(|\psi_1 \rangle \otimes \psi_2 \rangle)
= \langle \phi_1 |\psi_1 \rangle \langle \phi_2 |\psi_2 \rangle.$$
The second is a special case of the first. You need brackets (as I added them) to make it look sensible. Then it is standard, and the mistake in post #1 is obvious.
It's, of course, a matter of convention again, but I think must books and papers use the above notation.
I don't think that there is another sensible convention.
 
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Then it is standard, and the mistake in post #1 is obvious.
Now I'm confused. Why would this be right:
$$(\langle \phi_1 |\otimes \langle \phi_2|)(|\psi_1 \rangle \otimes |\psi_2 \rangle) = \langle \phi_1|\psi_1 \rangle\langle \phi_2 |\psi_2 \rangle$$
But this be wrong?
$$(\langle x |\otimes \langle y|)(|x \rangle \otimes |y \rangle) = \langle x|x \rangle\langle y |y \rangle$$
 

A. Neumaier

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Now I'm confused. Why would this be right:

But this be wrong?
Both are right. But your final equality equation in #1 without brackets has a different meaning, since
(according to the standard conventions) the middle inner product is a number and can be moved out, giving what I stated in #2.
Moral: When writing tensor products without explicit tensor product sign, never drop essential parentheses.
 
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Both are right. But your final equality equation in #1 without brackets has a different meaning, since
(according to the standard conventions) the middle inner product is a number and can be moved out, giving what I stated in #2.
Moral: When writing tensor products without explicit tensor product sign, never drop essential parentheses.
Ah I see, thanks!
 

vanhees71

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The second is a special case of the first. You need brackets (as I added them) to make it look sensible. Then it is standard, and the mistake in post #1 is obvious.

I don't think that there is another sensible convention.
You are right, concerning the brackets. I'll correct the original posting! Thanks. Concerning the convention, I think some people indeed write it with the opposite order in the bras, but that's a rare and for me utmost confusing notation.
 

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