2

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
  • 201,751
71GA
  • 2,555
  • 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. – Luboš Motl May 12 '13 at 07:45
  • 1
    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? – Luboš Motl May 12 '13 at 07:46
  • Related Phys.SE questions: http://physics.stackexchange.com/search?q=dirac+bra+ket+notation – Qmechanic May 12 '13 at 08:43
  • Related http://physics.stackexchange.com/questions/57754/what-is-a-dual-cotangent-space/57793#57793 see my answer. – joshphysics May 12 '13 at 17:54

1 Answers1

3

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
  • 2,043
  • $\dagger$ is transpose + complex conjugate. – user10001 May 12 '13 at 08:40
  • @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. – firtree May 12 '13 at 09:18
  • So the general importance here is that matrix multiplication has to be defined and thie is why it is important if we use operator from the left or the right :) – 71GA May 21 '13 at 05:43