Some Dirac notation explanations

Equation for an expectation value $\langle x \rangle$ is known to me:

\begin{align} \langle x \rangle = \int\limits_{-\infty}^{\infty} \overline{\psi}x\psi\, d x \end{align}

By the definition we say that expectation value is a sandwich: $\langle \psi|\hat{x}|\psi\rangle$. So:

\begin{align} \langle \psi|\hat{x}|\psi \rangle = \int\limits_{-\infty}^{\infty} \overline{\psi}x\psi\, d x \end{align}

Can you first confirm that these three lines are correct (I am not sure if I understand Dirac's bra-ket notation right). If they are wrong please explain:

\begin{align} \text{1st:}& & \langle \psi | \hat{x} | \psi \rangle &= | \psi\rangle \cdot \hat{x}|\psi \rangle\\ \text{2nd:}& & \langle \psi | \hat{x} | \psi \rangle &= {\langle \psi|}^\dagger \cdot \hat{x}|\psi \rangle\\ \text{3rd:}& & \langle \psi | \hat{x} | \psi \rangle &= {\langle \psi|}^\dagger \cdot \hat{x} \langle\psi |^\dagger\\ \end{align}

How do i derive relations $\langle\psi|\hat{x}|\psi\rangle = \langle \psi |\hat{x}\psi\rangle$ and $\langle\psi|\hat{x}|\psi\rangle = \langle \hat{x}^\dagger\psi |\psi\rangle$?

• Notation is supposed to be useful and follow some restricted standards. Your "identities" apparently deliberately violate what a normal physicist using the Dirac notation would write down. In the 1st, it may be OK if the "dot" is a symbol for the inner product that assumes that one of the factors is hermitian conjugated. Physicists surely don't use such a notation. The same is true for 2nd. Physicists would normally write the 2nd without the dagger. the 3rd is nonsenical in any sense, operators such as $x$ shouldn't act on the "bracket side" of a vector. Commented May 12, 2013 at 7:45
• Why don't you try to understand the notation correctly rather than trying to write down as many incorrect or marginally incorrect "identities" as you can? Also, I think it is not just notation you don't understand. The confusion in the question indicates that you don't understand the beef, the linear algebra, otherwise you wouldn't formulate the question in this way. Do you understand matrix multiplication? Ket vectors are just column thin matrices, bra vectors are the snigle rows - the hermitian conjugate to kets - and the operators are large matrices. What is so hard about it? Commented May 12, 2013 at 7:46
• Related Phys.SE questions: physics.stackexchange.com/search?q=dirac+bra+ket+notation Commented May 12, 2013 at 8:43
• Related physics.stackexchange.com/questions/57754/… see my answer. Commented May 12, 2013 at 17:54

If you are acquainted to matrices, then $|\psi\rangle$ is very much like column vector, and $\langle\psi|$ is similar to row vector. Operators correspond to square matrices. The conjugate transpose $^\dagger$ is similar to the matrix transpose $^\mathrm{T}$ in the sense that it turns columns to rows and vice versa.
Then (neither of your 1-3 lines), if we take $\cdot$ as a matrix-like product, $$\langle\psi|\hat{x}|\psi\rangle=\langle\psi|\cdot\hat{x}|\psi\rangle=|\psi\rangle^\dagger\cdot\hat{x}|\psi\rangle$$ If we group multipliers we could write $$\langle\psi|\hat{x}|\psi\rangle=\langle\psi|\cdot\Bigl(\hat{x}|\psi\rangle\Bigr)=\langle\psi|\cdot|\hat{x}\psi\rangle=\langle\psi|\hat{x}\psi\rangle$$ Other way of grouping lets us $$\langle\psi|\hat{x}|\psi\rangle=\Bigl(\langle\psi|\hat{x}\Bigr)\cdot|\psi\rangle=\Bigl(\hat{x}^\dagger\cdot\langle\psi|^\dagger\Bigr)^\dagger\cdot|\psi\rangle=\Bigl(\hat{x}^\dagger\cdot|\psi\rangle\Bigr)^\dagger\cdot|\psi\rangle=|\hat{x}^\dagger\psi\rangle^\dagger\cdot|\psi\rangle=\langle\hat{x}^\dagger\psi|\cdot|\psi\rangle=\langle\hat{x}^\dagger\psi|\psi\rangle$$
• $\dagger$ is transpose + complex conjugate. Commented May 12, 2013 at 8:40
• @user10001 Yes. But there is kind of mess of concepts resembling each other (conjugate, transpose, conjugate transpose, dual covectors, covariant vectors) and notations ($\overline{a}$,$\tilde{a}$,$a^*$,$a^\mathrm{T}$,$a^\dagger$,$a^+$), so I tried to involve it as little as possible. Commented May 12, 2013 at 9:18