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I have problem in understanding equation (1.23), I croped this image from Mark_Srednicki "Quantum field theory". Can anyone show me the reason for the equation (1.23)?

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The first term is the inner product of the state $|\psi,t>$ with itself. In the second he has inserted a complete set of states $|x>$ using the identity $ 1 = \int\mathrm{d}x |x><x|$. In the third he has used the fact that $ <x|\psi,t> = \psi(x,t)$ is the position-space representation of the wavefunction. I'm not sure where the $t$ argument in the wavefunction has gone, unless he's using that the norm of the state is conserved. –  Michael Brown Feb 15 '13 at 16:00
You posted your awnser while I was writing mine, not nice ;-) –  Noldig Feb 15 '13 at 16:03

2 Answers 2

up vote 1 down vote accepted

The inner product of two wavefunctions $\phi$ and $\psi$ (on 1D, for simplicity) is defined to be the integral $\int_{-\infty}^\infty \phi^\ast(x)\psi(x)dx$, and Dirac notation assigns the symbol $\langle\phi|\psi\rangle$ to this object, in analogy to the mathematician's usual notation $\langle u,v\rangle$ for the inner product of two vectors $u$ and $v$. Thus, you have $$\langle\phi|\psi\rangle=\int_{-\infty}^\infty \phi^\ast(x)\psi(x)dx.$$

Now, the definite-position states are defined in such a way that their inner product with some state $|\psi\rangle$ gives you the wavefunction: $\langle x|\psi\rangle=\psi(x)$; alternatively, you can say that the wavefunction of a definite-position state (call it $|x_0\rangle$) is a delta function centered at $x_0$: $\langle x|x_0\rangle=\delta(x-x_0)$. (You can and should check, using the integral above, that my two previous statements are equivalent.)

Putting these two puzzle pieces together you can get an alternative version of the integral: $$\langle\phi|\psi\rangle=\int_{-\infty}^\infty \phi^\ast(x)\psi(x)dx=\int_{-\infty}^\infty \langle\phi|x\rangle\langle x|\psi\rangle dx.$$ (Here I have used the conjugate symmetry of the inner product: $\langle\phi|x\rangle=\langle x|\phi\rangle^\ast=\phi^\ast(x)$.) Now, in this second integral the objects $|\psi\rangle$ (a Hilbert space vector) and $\langle\phi|$ (a vector in the dual space, i.e. a linear functional on the Hilbert space) no longer depend on the integration variable $x$, so we can pull them out: $$\langle\phi|\psi\rangle=\langle\phi|\left[\int_{-\infty}^\infty dx|x\rangle\langle x| \right]|\psi\rangle.$$ This is what is known as a resolution of the identity: each object $|x\rangle\langle x|$ is a Hilbert space operator (it acts linearly on vectors $|\psi\rangle$ to give other vectors $\psi(x) |x\rangle$), and when you integrate over them you get the identity operator: $$\int_{-\infty}^\infty dx|x\rangle\langle x|={1}.$$

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Because of the completeness of the eigenfunctions of the position operator you can add a factor of one in the middle of the bracket, which is the integral in the middle. The projection of a bra to the position eigenstate is simply the wave function in position space and if you change the bra with the ket in a braket you have to take the complex conjugate.

I think that's all information you need.

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