How Does Zee Simplify Integrals in QFT Derivations?

[Cadaei]

On page 12 of Zee's QFT in a nutshell, he rewrites an integral from dirac notation to a mixture of Schrödinger/dirac

My question is, since

, what happened to the integrals, and shouldn't there be four wave functions instead of two?

I tried writing out the integrals explicitly, thinking maybe theyre integrated out with a delta function or something, but with no luck - I don't see how to get to psi(q_f).

Any help would be appreciated.

 

[samalkhaiat]

Let me rewrite the matrix element as $$\langle \Psi_{F}| e^{- i H T} | \Psi_{I} \rangle .$$ Now insert the completeness relations $$\int d X_{a} \ | X_{a} \rangle \langle X_{a} | = 1 , \ \ \mbox{for} \ \ a = I , F .$$ You find $$\langle \Psi_{F}| e^{- i H T} | \Psi_{I} \rangle = \int d X_{F} d X_{I} \langle \Psi_{F}| X_{F} \rangle \langle X_{F}| e^{ - i H T} | X_{I} \rangle \langle X_{I} | \Psi_{I} \rangle .$$ Okay, now use the definitions of the wavefunctions $$\Psi_{I} ( X_{I} ) = \langle X_{I} | \Psi_{I} \rangle , \ \ \Psi^{*}_{F}(X_{F}) = \langle \Psi_{F} | X_{F} \rangle .$$

 

[Cadaei]

I think I'm missing something extremely simple here... why is dotting ⟨X_I| into |psi_i⟩ the same as evaluating psi_I at X_I?

 

[samalkhaiat]

I think I'm missing something extremely simple here... why is dotting ⟨X_I| into |psi_i⟩ the same as evaluating psi_I at X_I?

And yes, you need to know the difference between state vector $|\Psi \rangle$ and its coordinate representation, i.e. wave-function $\Psi ( x)$. Almost all textbooks explain the Bra-Ket notations and their connection to “wave-functions”.
You seem to have no problem with $$\langle \Phi | \Psi \rangle = \int dx \ \Phi^{*} (x) \Psi (x) . \ \ \ \ (A)$$ Okay, let us work on the left-hand side by inserting the completeness relation $1=\int dx |x\rangle \langle x|$:
$$\langle \Phi | \Psi \rangle = \langle \Phi | ( \int dx |x \rangle \langle x | ) | \Psi \rangle = \int dx \ \langle \Phi | x \rangle \langle x | \Psi \rangle .$$
Now, if you compare the RHS of this equation with the RHS of Eq(A), what would you get?

Similar state of affair exists in elementary vector algebra. You know how to expand a vector $|\vec{V}\rangle$ in terms of some orthogonal unit-vectors $| e_{i}\rangle$ $$|\vec{V}\rangle = \sum_{i} V_{i} \ | e_{i} \rangle .\ \ \ \ (1)$$ You also know the ortho-normality condition $$\langle e_{i} | e_{j} \rangle = \delta_{ij} . \ \ \ \ \ \ \ (2)$$ You should also know how to calculate the components of the vector from $$V_{i} = \langle e_{i} | \vec{V}\rangle . \ \ \ \ \ \ \ (3)$$ You can now substitute (3) in (1) to obtain another familiar equation in vector algebra $$| \vec{V} \rangle = \sum_{i} | e_{i}\rangle \langle e_{i} | \vec{V} \rangle . \ \ \ \ \ \ (4)$$ This leads to the completeness relation for the unit vectors $$\sum_{i} | e_{i} \rangle \langle e_{i}\rangle = 1 . \ \ \ \ \ \ (5)$$ And finally, you know the scalar product $$\langle \vec{V}| \vec{U} \rangle = \sum_{i} V^{t}_{i} U_{i} .\ \ \ \ \ (6)$$ Now, imagine the vector space to be an infinite-dimensional complex vector space spanned by un-countable infinity of ortho-normal functions (vectors) $|e(x)\rangle \equiv | x \rangle$, i.e. just pass to the continuous limits $$i \to x , \ \ \ | e_{i}\rangle \to | x \rangle , \ \ \sum_{i} \to \int dx ,$$ and $$V_{i} \to V(x), \ \ \ V^{t}_{i} \to V^{*}(x) .$$ So, we can translate Eq(1)-Eq(6) into the continuous language as follow $$\mbox{Eq(1)} \to \ \ \ | V \rangle = \int dx \ V(x) \ | x \rangle ,$$ $$\mbox{Eq(2)} \to \ \ \ \langle x | y \rangle = \delta (x - y) .$$ $$\mbox{Eq(3)} \to \ \ \ V(x) = \langle x | V \rangle ,$$ which is what you were asking about $\Psi (x) = \langle x | \Psi \rangle$. $$\mbox{Eq(4)} \to \ \ \ | V \rangle = \int dx \ | x \rangle \langle x | V \rangle .$$ $$\mbox{Eq(5)} \to \ \ \ \int dx \ | x \rangle \langle x | = 1 .$$ And finally the equation you know $$\mbox{Eq(6)} \to \ \ \langle V | U \rangle = \int dx \ V^{*}(x) U(x) .$$

 

[Cadaei]

Then why are you reading QFT text?

Thank you for your help, I understand now.