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Let us divide the time $T$ into $N$ segments each lasting $δt = T/N$. Then we write $\langle q_F | e^{−iHT} |q_I \rangle = \langle q_F | e^{−iHδt} e^{−iHδt} . . . e^{−iHδt} |q_I \rangle $ Our states are normalized by $\langle q' | q \rangle = δ(q' − q)$ with $δ$ the Dirac delta function. ("Dirac's formulation, in the book "Quantum Field Theory" by A. Zee)

1) Why must $\langle q' | q \rangle = δ(q' − q)$ hold?

2) Is this related to $\int_{-\infty}^{\infty}\Psi^*\Psi dx = 1$?


Now use the fact that $|q \rangle$ forms a complete set of states so that $\int dq |q \rangle \langle q| = 1$.

3) Does this even make sense?

4) Shouldn't integral be $\int_{-\infty}^{\infty}$?

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1 Answer 1

1) Position is an observable. So,

i) For any two of its eigenstates $|q>$ and $|q'>$, we should have $<q'|q>=0$ whenever $q\neq q'$. This follows from that fact that position being an observable is in particular, a Hermitian operator; and that any two eigenstates of a Hermitian operator with distinct eigenvalues are orthogonal.

ii) Assuming that eigenstates $|q>$ are properly normalized, any state $|\psi>$ in the Hilbert space of states can be written as

$$|\psi>=\int_{-\infty}^{\infty} <q|\psi>|q>dq$$.

This follows from the fact that position operator is not only Hermitian but also an observable, and so (by definition of an observable) its (properly normalized) eigen vectors should form a complete set.

These two requirements can be used to find the normalization of states $|q>$ as follows :



Then from (i) we know that $f(q',q)$ should be zero when $q\neq q'$.

And from (ii) taking scalar product on both sides with $|q'>$ we get

$$<q'|\psi>=\int_{-\infty}^{\infty} <q|\psi>f(q',q)dq$$

Thus we see that $f(q',q)$ satisfies the definition of delta function $\delta(q'-q)$. So it should be equal to the delta function $\delta(q'-q)$

2) Normalization of a particular state is another issue.

3) This is just a short (and confusing) way of stating ii)

4) Yes, you are right.

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