Quick Braket notation question

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The discussion clarifies that \langle J_z \rangle represents the expectation value of the operator J_z, which is the average result of measurements on identically prepared quantum systems. It is expressed mathematically as \langle J_z \rangle = \langle \psi | J_z | \psi \rangle, indicating its dependence on the quantum state \vert\psi\rangle. The expectation value can also be calculated using integrals, as shown in the example with \langle x \rangle. The conversation emphasizes the importance of understanding the relationship between operators and quantum states in bra-ket notation. This information is essential for anyone learning about quantum mechanics and expectation values.
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I'm a complete noob with Braket and I've only just started getting to grips with it.

For completeness' sake though (from the book I'm currently reading), I can't seem to find a definition for:

\langle J_z \rangle

Would this just be the "magnitude" of J_z?

Thanks
 
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It's an expectation value, i.e. the average result in a (long) series of measurements of Jz on identically prepared systems.
Fredrik said:
...the average result in a series of measurements of A on identically prepared systems is

\langle A\rangle=\sum_a P(a)a=\sum_a a|\langle a|\psi\rangle|^2=\sum_a\langle a|\psi\rangle\langle\psi|A|a\rangle=\mbox{Tr}(\rho A) ...and also =\langle\psi|\Big(\sum_a|a\rangle\langle a|\Big)A|\psi\rangle=\langle\psi|A|\psi\rangle
Note that it depends on what state the system is in.

This post should be useful if you're learning bra-ket notation.
 
Actually it's the expectation value of J_z - that is,
\langle J_z \rangle = \langle \psi \vert J_z \vert \psi \rangle
In order to actually evaluate that expression, you would have to have some quantum state \vert\psi\rangle, since the expectation value of any operator depends on the quantum state.
 
Ah crud, I completely forgot that's how you write Expectation value.

so, just to confirm I have a grip on this,

\langle x \rangle = \langle \psi \vert {x} \vert \psi \rangle = \int_{-\infty}^{\infty} \psi^* x \psi dx

Is that right? Assuming the limits are from infinity to minus infinity.

Thanks
 
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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