Dirac notation and conjugate transpose in Sakurai

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Sakurai's Modern Quantum Mechanics introduces Dirac notation, where bras and kets are related through conjugate transposes. The discussion clarifies that while bras are the conjugate transposes of kets, the notation used distinguishes between complex conjugates of numbers and the Hermitian adjoint of operators. It emphasizes that the complex conjugate of a bra is indeed a ket, and vice versa, but the notation serves to keep these concepts distinct. The conversation also touches on the mathematical representation of complex numbers and matrices, illustrating how the transpose and conjugate transpose operations apply in quantum mechanics. Overall, the thread highlights the importance of understanding these notational conventions in quantum mechanics.
Master J
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In Sakurai's Modern Quantum Mechanics, he develops the Dirac notation of bras and kets. In one part, he states (page 17):

<B|X|A>

= (<A|X^|B>)*

= <A|X^|B>*

where X^ denotes the Hermitian adjoint (the conjugate transpose) of the operator X.

My question is, since a bra is the conjugate transpose of a ket, could we write
<B|X|A>^ = <A|X|B>

(since of course X^ = X (ie. X is Hermitian) for real, measurable quantities).

What I'm trying to ask is, does Sakurai define a separate conjugate transpose of the bras and kets, whereby he just takes the complex conjugate and implicitly also transposes it (ie.
|A>* = <A| ). Normally, I would expect to see |A>^ = <A|.
 
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I think you're misunderstanding what the notation means. Stars denote the complex conjugates of numbers: i.e. a complex number a has complex conjugate a^*. Daggers (I assume it's a dagger and you just don't know LaTex; don't have my copy of Sakurai around) denote the adjoint of an operator: an operator \hat A has as its adjoint \hat A^\dagger.

I don't remember in QM kets and bras being conjugated. It's meaningful to talk about their representations being conjugated: \langle x | \psi \rangle = \langle \psi | x \rangle^*, but that's because this is a number (a function is still a number).

At any rate, I think Sakurai is just trying to keep the complex conjugate of a number distinct from the hermitian adjoint of an operator, which is what I've seen in books other than his, too.
 
Sorry, I should really learn LaTeX, but yes, that's what I meant ...

I have from Wikipedia ( http://en.wikipedia.org/wiki/Bra-ket_notation ) that the complex conjugate of a bra is a ket, and vice versa.

So, in the equation <B|X|A> = ( <A|X|B> )* , the * would some unnecessary? Again here, X is Hermitian.
 
Basically what is done is always a transpose. The conjugate operation comes in
just because of the complex notation. You can equally well describe the mathematics
in an entirely real form using only the transpose operation by substituting the complex
numbers with real valued 2x2 matrices:<br /> (a+ib) ~~\longrightarrow~~<br /> \left(\begin{array}{cc} a &amp; -b \\ b &amp; ~~a\end{array}\right) <br />

The conjugate transpose comes up if you need to describe the transpose of the
2x2 real valued matrix in the complex notation.

<br /> (a+ib)^* ~~=~~<br /> \left(\begin{array}{cc} a &amp; -b \\ b &amp; ~~a\end{array}\right)^\intercal ~~=~~<br /> (a-ib)<br />

This substitution works just as well with complex matrices, for instance the
Pauli matrix \sigma^2 which has an equivalent real valued 4x4 matrix.

<br /> \sigma^2 ~~=~~<br /> \left(\begin{array}{cc} 0 &amp; -i \\ i &amp; ~~0\end{array}\right) ~~\longrightarrow~~<br /> \left(\begin{array}{cccc} 0 &amp; ~~0 &amp; ~~0 &amp; ~~1 \\<br /> 0 &amp; ~~0 &amp; - 1 &amp; ~~0 \\<br /> 0 &amp; - 1 &amp; ~~0 &amp; ~~0 \\<br /> 1 &amp; ~~0 &amp; ~~0 &amp; ~~0<br /> \end{array}\right)<br />

The transpose of the 4x4 real matrix is equivalent to the conjugate transpose of
the 2x2 complex Pauli matrix.

<br /> (\sigma^2)^* ~~=~~<br /> \left(\begin{array}{cc} 0 &amp; -i \\ i &amp; ~~0\end{array}\right)^* ~~\longrightarrow~~<br /> \left(\begin{array}{cccc} 0 &amp; ~~0 &amp; ~~0 &amp; ~~1 \\<br /> 0 &amp; ~~0 &amp; - 1 &amp; ~~0 \\<br /> 0 &amp; - 1 &amp; ~~0 &amp; ~~0 \\<br /> 1 &amp; ~~0 &amp; ~~0 &amp; ~~0<br /> \end{array}\right)^\intercal ~~=~~ \sigma^2<br />

Another example:

<br /> (i\sigma^2)^* ~~=~~<br /> \left(\begin{array}{cc} ~~0 &amp; ~~1 \\ -1 &amp; ~~0\end{array}\right)^* ~~\longrightarrow~~<br /> \left(\begin{array}{cccc} ~~0 &amp; ~~0 &amp; ~~1 &amp; ~~0 \\<br /> ~~0 &amp; ~~0 &amp; ~~0 &amp; ~~1 \\<br /> - 1 &amp; ~~0 &amp; ~~0 &amp; ~~0 \\<br /> ~~0 &amp; -1 &amp; ~~0 &amp; ~~0<br /> \end{array}\right)^\intercal ~~=~~ -i\sigma^2<br />

Bras and kets are just row and column vectors due to the underlying matrix algebra,
and therefor they are related to each other by a transpose operation in the real
valued case and a conjugate transpose in the complex valued case.

http://en.wikipedia.org/wiki/Bra-ket_notation#Bras_and_kets_as_row_and_column_vectors

Hans
 
Last edited:
Time reversal invariant Hamiltonians must satisfy ##[H,\Theta]=0## where ##\Theta## is time reversal operator. However, in some texts (for example see Many-body Quantum Theory in Condensed Matter Physics an introduction, HENRIK BRUUS and KARSTEN FLENSBERG, Corrected version: 14 January 2016, section 7.1.4) the time reversal invariant condition is introduced as ##H=H^*##. How these two conditions are identical?

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