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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 $\langle\psi|\hat{x}|\psi\rangle = \langle \hat{x}^\dagger\psi |\psi\rangle$?

Qmechanic
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71GA
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1 Answers1

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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 $$\langle\psi|\hat{x}|\psi\rangle=\Bigl(\langle\psi|\hat{x}\Bigr)\cdot|\psi\rangle=\Bigl(\hat{x}^\dagger\cdot\langle\psi|^\dagger\Bigr)^\dagger\cdot|\psi\rangle=\Bigl(\hat{x}^\dagger\cdot|\psi\rangle\Bigr)^\dagger\cdot|\psi\rangle=|\hat{x}^\dagger\psi\rangle^\dagger\cdot|\psi\rangle=\langle\hat{x}^\dagger\psi|\cdot|\psi\rangle=\langle\hat{x}^\dagger\psi|\psi\rangle$$

If you are not acquainted to matrices, I strongly suggest you to get acquainted.

firtree
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