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I am self-studying QFT by following Mark Srednicki and trying to derive some results in the text. I am not sure whether my steps are correct in a precise mathematical sense, so I quote the problem and list my steps together with their justifications in the paragraphs below. May I invite you guys to proofread my steps and point out the errors (especially conceptual mistakes)? Thank you.

Notation: All three position vectors are denoted by $x$ and the delta function $\delta(x-x_0)$ is the three dimensional one.

Motivation: Show the equivalence between the second quantization formalism (to be defined below) and the Schrodinger time-dependent equation (TDSE) in coordinate representation for a system of N (fixed) spin-zero bosons.

For simplicity, let us consider a system consisting of two non-interacting spin-zero bosons confined in a volume by an infinite potential well.

The TDSE in the coordinate representation for this system reads

$i\hbar\partial_t\Psi = -\frac{\hbar^2}{2m}\nabla^2_1\Psi-\frac{\hbar^2}{2m}\nabla^2_2\Psi$

$\Psi=\Psi(x_1,x_2;t)$ where the indices refer to the particles

Now we introduce a set of bosonic destroyers and creators satisfying the SHO-like commutation relations: $[a(x),a(x^\prime)]=[a(x)^\dagger,a^\dagger(x^\prime)]=0;$ and $[a(x),a^\dagger(x^\prime)]=\delta(x-x^\prime)$ (These field operators are the ones in the Schrodinger picture with no time dependence.)

In terms of $a(x),a^\dagger(x)$, the Hamiltonian operator for this system reads

$H=\int dx a^\dagger(x)(-\frac{\hbar^2}{2m}\nabla^2)a(x) \tag{1}$

and the most general form of a time-dependent state for this system in the Schrodinger picture reads

$|\Psi(t)\rangle =\int dx_1dx_2\Psi(x_1,x_2;t)a^\dagger(x_1)a^\dagger(x_2)|0\rangle \tag{2}$

I wish to prove the assertion in the text that $i\hbar|\Psi(t)\rangle=H|\Psi(t)\rangle$ if and only if the c-number function $\Psi(x_1,x_2;t)$ in equation (2) satisfies the TDSE in coordiante representation as in the case of the QM of a single particle.

Before proceeding further, I wish to ask some questions about equation $(1)$ and $(2)$

In equation $(1)$, may I understand $H=\int dx a^\dagger(x)(-\frac{\hbar^2}{2m}\nabla^2)a(x)$ as $\int dx a^\dagger(x)b(x)$

i.e The differentiation of the field operator $a(x)$ (not c-number) with respect to coordinate results in another field operator $b(x)$?

For equation $(2)$, I can interpret it as either

$[\int dx_1dx_2\Psi(x_1,x_2;t)a^\dagger(x_1)a^\dagger(x_2)]|0\rangle$ (We integrate the field operator first and apply the resultant operator to the vacuum state)

$\int dx_1dx_2[\Psi(x_1,x_2;t)a^\dagger(x_1)a^\dagger(x_2)|0\rangle]$ (We let the integrand act on the vaucum state first and then integrate the resultant state)

I think both interpretations are correct because $(A+B)|\psi\rangle=A|\psi\rangle+B|\psi\rangle$ where A and B are linear operators. In particular, the ideas of integrating an operator over some continuous variables are familiar from ordinary quantum mechanics. e.g $\int dx |x\rangle\langle x|$ and any single particle state $|\alpha\rangle = I|\alpha\rangle=( \int dx |x\rangle\langle x|)|\alpha\rangle = \int dx (|x\rangle\langle x|\alpha\rangle) = \int dx |x\rangle\langle x|\alpha\rangle $

Now I present my steps of establishing the equivalence between the second quantization formalism and the TDSE in coordinate representation for this simple system. According to equation $(2)$

$i\hbar\partial_t|\Psi(t)\rangle = \int dx_1dx_2 i\hbar\partial_t\Psi(x_1,x_2;t)a^\dagger(x_1)a^\dagger(x_2)|0\rangle$

This step is quite trivial

Next, $H|\Psi(t)\rangle =(\int dx a^\dagger(x)(-\frac{\hbar^2}{2m}\nabla^2)a(x))(\int dx_1dx_2 \Psi a^\dagger(x_1)a^\dagger(x_2)|0\rangle)= \int dxdx_1dx_2 a^\dagger(x)((-\frac{\hbar^2}{2m}\nabla^2)a(x))\Psi a^\dagger(x_1)a^\dagger(x_2)|0\rangle$

In the second line, the product in the first line becomes an integral over a space 3 higher than the configuration space in dimensions. Now come to the step which I am not pretty sure its validity. My argument goes like this: Since the Laplacian acts on the variable $x$ only, we can absorb all the creation operators and the c-number function $\Psi$ into the operand of the Laplacian. i.e $H|\Psi(t)\rangle=\int dxdx_1dx_2 a^\dagger(x)(-\frac{\hbar^2}{2m}\nabla^2(a(x)a^\dagger(x_1)a^\dagger(x_2)\Psi))|0\rangle \tag{3}$

i.e The Laplacian inside the integral now acts on the product $a(x)a^\dagger(x_1)a^\dagger(x_2)\Psi$

Assuming the validity of steps above and write the field operator product $a(x)a^\dagger(x_1)a^\dagger(x_2)$ as $[a(x),a^\dagger(x_1)]a^\dagger(x_2)+a^\dagger(x_1)a(x)a^\dagger(x_2)=[a(x),a^\dagger(x_1)]a^\dagger(x_2)+a^\dagger(x_1)[a(x),a^\dagger(x_2)]+a^\dagger(x_1)a^\dagger(x_2)a(x) \tag{4}$

The last term in equation $(4)$ can be dropped as it will kill the vacuum state (this point becomes more explicit if we interpret the vacuum state as living inside the integral) Using the commutation relations for the creators and destroyers, equation $(4)$ becomes $\delta(x-x_1)a^\dagger(x_2)+a^\dagger(x_1)\delta(x-x_2)$

Substitute this back into equation $(3)$, we have

$H|\Psi(t)\rangle=\int dxdx_1dx_2 a^\dagger(x)(-\frac{\hbar^2}{2m}\nabla^2_{x}(\delta(x-x_1)a^\dagger(x_2)\Psi+a^\dagger(x_1)\delta(x-x_2)\Psi))|0\rangle$

Integrating with respect to $x$ first which can always be done as addition of operators commute, the order of integration is not material. Now come to another step I am not pretty sure its validity. After integrating with respect to $x$, $H|\Psi(t)\rangle = \int dx_1dx_2 [a^\dagger(x_1)(-\frac{\hbar^2}{2m}\nabla^2_1(a^\dagger(x_2)\Psi))+a^\dagger(x_2)(-\frac{\hbar^2}{2m}\nabla^2_2(a^\dagger(x_1)\Psi)]|0\rangle$

Since the field operators being differentiated inside the integral depend on the respective coordinate via the c-number function $\Psi$ only, all the creation operators can be pulled out of the respective Laplacian. Moreover, since the bosonic creation operators at different position communte, we can arrange operator product in the second term as $a^\dagger(x_1)a^\dagger(x_2)$ and the final result is

$H|\Psi(t)\rangle = \int dx_1dx_2 (-\frac{\hbar^2}{2m}\nabla^2_1\Psi-\frac{\hbar^2}{2m}\nabla^2_2\Psi)a^\dagger(x_1)a^\dagger(x_2)|0\rangle$

Recalling that $i\hbar\partial_t|\Psi(t)\rangle = \int dx_1dx_2 i\hbar\partial_t\Psi(x_1,x_2;t)a^\dagger(x_1)a^\dagger(x_2)|0\rangle$

We see that the TDSE in its abstract form $i\hbar|\Psi(t)\rangle = H|\Psi(t)\rangle$ is satisfied if and only if the c-number function $\Psi $ satisfy the TDSE in the coordinate representation as in the case of the quantum mechanics of a single particle. Thank you for reading and welcome for comments.

I am new to the subject and wish to make sure that my understanding and derivation are correct. Your comments and critism are greatly appreciated.

Appendix

I was wondering whether there exists an identity $ \int f(x)\delta(x-a)dx = f(a)$ for an operator-valued function f. To explore this, Let us consider the expectation value of $\int f(x)\delta(x-a)dx$ over some test state $|\Psi\rangle$ $\langle\Psi|(\int f(x)\delta(x-a)dx)|\Psi\rangle = \int \langle\Psi|f|\Psi\rangle\delta(x-a)dx = \langle\Psi|f(a)|\Psi\rangle$ and this holds for any test state $|\Psi\rangle$, so $ \int f(x)\delta(x-a)dx = f(a)$.

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  • $\begingroup$ Further to my post above, I would like to raise one more question. Is this identity holds regardless whether $f$ is c-number-valued or operator-valued? $\int dx \nabla^2_{x}(\delta(x-y)f(y)) = \nabla^2_{y}f(y)$ $\endgroup$ – Heng Fai Chang Nov 21 '17 at 19:00

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