biology, math, and physics

(24.1) Verify eqn 24.25 and show that eqn 24.29 solves eqn 24.28.

Integrating the term by parts gives as promised, while the term gives . Therefore,

$$-\frac12(\partial_\mu A_\nu\partial^\mu A^\nu-\partial_\mu A_\nu\partial^\nu A^\mu)+\frac12m^2 A^\mu A_\mu=\frac 12A_\mu\left([\partial^2+m^2]\eta_{\mu\nu}-\partial_\mu\partial_\nu\right)A^\nu$$

Evaluating

$$[-(p^2-m^2)g^{\mu\nu}+p^\mu p^\nu]\frac{-i(g_{\nu\lambda}-p_\nu p_\lambda/m^2)}{p^2-m^2}$$

$$=ig^{\mu\nu}(g_{\nu\lambda}-p_\nu p_\lambda/m^2)-i\frac{p^\mu p_\lambda-p^\mu p^2 p_\lambda/m^2}{p^2-m^2}\\=ig^\mu_\lambda – i\frac{p^\mu p_\lambda}{m^2}+i\frac{(p^2-m^2) p^\mu p_\lambda}{m^2(p^2-m^2)}=ig^\mu_\lambda$$

---

(24.2) Consider the Lagrangian with a shift

$$\mathcal L=\frac12(\partial_\mu\phi)^2-\frac{m^2}2\phi^2-\frac g8\phi^4+\frac{1}{2g}\left(\sigma-\frac g2\phi^2\right)^2$$

By performing a functional integral over the field , show that doesn’t change the dynamics of the theory.

The Lagrangian density becomes

$$\frac12(\partial_\mu\phi)^2-\frac{m^2}2\phi^2-\frac g8\phi^4+\frac g8\phi^4+\frac 1{4g^2}\sigma^2-\frac12\sigma \phi^2\\=\frac12(\partial_\mu\phi)^2-\frac{m^2}2\phi^2+\frac1{4g^2}\sigma^2-\frac12\sigma\phi^2$$

Identifying and , The path integral over gives a factor of

$$B[\det a]^{-\frac 12}e^{-\frac i2\int d^4x d^4y \left(-\frac{\phi^2}{2}\right)2g\left(-\frac{\phi^2}2\right)}$$

Here, we realize that is diagonal, i.e. a Dirac delta . Thus we recover the term in the Lagrangian, and the overall constant out in front cancels.

The Euler-Lagrange equations for are

$$\frac{\partial \mathcal L}{\partial\sigma}=0$$

since there is no dependence on the derivatives of .

$$\frac1{2g^2}\sigma+\frac{\phi^2}2=0$$

Thus is entirely determined by , and thus there are no dynamical degrees of freedom in . The Feynman diagrams have vertices with 2 particles and 1 particle.

---

(24.3) We want to do the integral

$$Z(J)=\int dx\ e^{-\frac12 Ax^2-\frac\lambda{4!}x^4+Jx}$$

$$=\int dx\ e^{-\frac12 Ax^2+Jx}e^{-\frac\lambda{4!}x^4}=\sum_n \frac1{n!}\left(-\frac\lambda{4!}x^4\right)^n\int dx\ e^{-\frac12 Ax^2+Jx}\\=\left[\sum_n\frac1{n!}\left(-\frac\lambda{4!}\frac{\partial^4}{\partial J^4}\right)^n\right]\int dx\ e^{-\frac12 Ax^2+Jx}\\=\left[e^{-\frac\lambda{4!}\frac{\partial^4}{\partial J^4}}\right]\left[\left(\frac{2\pi}A\right)^{\frac12}e^{\frac{J^2}{2A}}\right]$$

---

(24.4) By analogy, the generating functional for theory is

$$Z[J]=\left[e^{-\frac\lambda{4!}\int d^4z\ \frac{\delta^4}{\delta J(z)^4}}\right]\mathcal Z_0[J]$$

where

$$\mathcal Z_0[J]=e^{-\frac12\int d^4xd^4y\ J(x)\Delta(x-y)J(y)}$$

Act on with the functional derivative four times.

Applying the definition of the functional derivative,

$$\frac{\delta \mathcal Z_0[J]}{\delta J(z)}=\lim_{\epsilon\rightarrow 0}\frac1\epsilon(Z[J(x)+\epsilon\delta(z-x)]-Z[J(x)])$$

$$=\lim_{\epsilon\rightarrow 0}\frac1\epsilon \left[e^{-\frac12\int d^4xd^4y\left[\left(J(x)+\epsilon\delta(x-z)\right)\Delta(x-y)\left(J(y)+\epsilon\delta(x-y)\right)\right]}-e^{-\frac12\int d^4xd^4y\ J(x)\Delta(x-y)J(y)}\right]$$

Expanding the exponentials and keeping only the terms to first order in , we indeed find

$$=\left[-\int d^4y\ \Delta(z-y)J(x)\right]\mathcal Z_0[J]$$

The rest of this calculation is very tedious, and I skip it. The functional derivative acts as expected with the product rule and chain rules.

(23.1) With Lagrangian , use the Euler-Lagrange equation to find and show that the Lagrangian may be expressed equivalently as .

Since there is no dependence on ,

$$\frac{\partial L}{\partial x}=0$$

$$\frac12(\hat Ax+x\hat A)+b=0$$

Treating as a number,

$$x=-\frac b{\hat{\mathcal A}}$$

Plugging this into ,

$$\left(-\frac b{\hat{\mathcal A}}\right)\hat{\mathcal A}\left(-\frac b{\hat{\mathcal A}}\right)+b\left(-\frac b{\hat{\mathcal A}}\right)=b\frac{1}{2\hat{\mathcal A}}b-b\frac{1}{\hat{\mathcal A}}b=-b\frac 1{2\hat{\mathcal A}}b$$

---

(23.2) The path integral derivation of Wick’s theorem.

$$\int_{-\infty}^\infty dx\ x^2e^{-\frac12ax^2}=\frac{d}{da}\int_{-\infty}^\infty dx\ e^{-\frac12ax^2}=-2\frac{d}{da}\sqrt{\frac{2\pi}{a}}=\sqrt{\frac{2\pi}{a^3}}$$

Define

$$\langle x^n\rangle=\frac{\int_{-\infty}^\infty dx\ x^ne^{-\frac12ax^2}}{\int_{-\infty}^\infty dx\ e^{-\frac12ax^2}}$$

Calculate .

$$\langle x^2\rangle=\frac{\sqrt{\frac{2\pi}{a^3}}}{\sqrt{\frac{2\pi}{a}}}=\frac1a$$

$$\langle x^4\rangle=\frac{-2\frac{d}{da}\sqrt{\frac{2\pi}{a^3}}}{\sqrt{\frac{2\pi}{a}}}=\frac{3}{a^2}$$

When is odd, the numerator is by antisymmetry of the integrand, so . Otherwise when is even, acting times on the numerator with the operator gives

$$\langle x^n\rangle=\frac{(n-1)!!}{a^{\frac n2}}$$

Diagrammatically, each factor of comes from a possible contraction of the different ‘s.

Consider now the integral

$$\mathcal K=\int dx_1\cdots dx_N e^{-\frac12\mathbf x^T\mathbf A\mathbf x+\mathbf b^T\mathbf x}\\=\left(\frac{(2\pi)^N}{\det \mathbf A}\right)^{\frac12}e^{\frac12\mathbf b^T\mathbf A^{-1}\mathbf b}$$

Find the corresponding moments.

Ignoring the overall factor of which will cancel with the denominator, we need to differentiate with respect to the components of to get $$\langle x_ix_j\rangle=\frac{\int dx_1\cdots dx_N x_ix_j e^{-\frac12\mathbf x^T\mathbf A\mathbf x+\mathbf b^T\mathbf x}}{\int dx_1\cdots dx_N e^{-\frac12\mathbf x^T\mathbf A\mathbf x+\mathbf b^T\mathbf x}}$$

$$\frac{d}{db_i}\frac{d}{db_j}e^{\frac12\mathbf b_m\mathbf A^{-1}_{mn}\mathbf b_n}=\frac12\left(\delta_{im}\delta_{jn}+\delta_{in}\delta_{jm}\right)A^{-1}_{mn}e^{\frac12\mathbf b_m\mathbf A^{-1}_{mn}\mathbf b_n}$$

Assuming that are symmetric,

$$\langle x_ix_j\rangle=\left(\mathbf A^{-1}\right)_{ij}$$

It’s easy to see how this generalizes, and there will be a bunch of Kronecker deltas which give every possible contraction. Just as in the 1D case, odd moments vanish since the integral is antisymmetric. Thus

$$\langle x_ix_jx_kx_l\rangle=\left(\mathbf A^{-1}\right)_{ij}\left(\mathbf A^{-1}\right)_{kl}+\left(\mathbf A^{-1}\right)_{ik}\left(\mathbf A^{-1}\right)_{jl}+\left(\mathbf A^{-1}\right)_{il}\left(\mathbf A^{-1}\right)_{jk}$$

---

(23.3) Show that the amplitude for the forced harmonic oscillator with constant force to stay in the ground state from time to is

This is the interaction picture

$$\mathcal A=e^{-\frac12\int dt’ dt\ f(t)G(t,t’)f(t’)}$$

$$G(t,t’)=\frac{\theta(t-t’)e^{-i\omega(t-t’)}+\theta(t’-t)e^{i\omega(t-t’)}}{2m\omega}$$

This is just the value of the path integral, with time ordering as in the Feynman propagator. Performing the integral,

$$\int_{-\infty}^\infty dt\int_{-\infty}^\infty dt’\ f_0^2 G(t,t’)=-\frac{if_0}{m\omega^2}\left(T-\frac{\sin\omega T}{\omega}+i\frac2\omega\sin^2\frac{\omega T}{2}\right)$$

The imaginary part of the integral is the simple phase acquired from normal, unforced time evolution.

(19.1) Write down the momentum space amplitudes for the processes in Fig. 19.6.

a) $$(2\pi)^4\delta^{(4)}(p-q)\frac{-i\lambda}{2}\int \frac{d^4k}{(2\pi)^4}\frac{i}{k^2-m^2+i\epsilon}$$

b) $$(2\pi)^4\delta^{(4)}(p-q)\frac{-\lambda^2}{4}\left(\int \frac{d^4k}{(2\pi)^4}\frac{i}{k^2-m^2+i\epsilon}\right)^2$$

c) $$\frac{-i\lambda}{8}\int \frac{d^4k_1 d^4k_2}{(2\pi)^8}\frac{i}{k_1^2-m^2+i\epsilon}\frac{i}{k_2^2-m^2+i\epsilon}$$

d) $$(2\pi)^4\delta^{(4)}(p-q)\frac{-\lambda^2}{3!}\int\frac{d^4k_1 d^4k_2}{(2\pi)^8}\frac{i}{k_1^2-m^2+i\epsilon}\frac{i}{k_2^2-m^2+i\epsilon}\frac{i}{(q-k_1-k_2)^2-m^2+i\epsilon}$$

e) $$(2\pi)^4\delta^{(4)}(p-q)\frac{-\lambda^2}{4}\int\frac{d^4k_1d^4k_2}{(2\pi)^8}\frac{i}{k_1^2-m^2+i\epsilon}\left(\frac{i}{k_2^2-m^2+i\epsilon}\right)^2$$

f) $$(2\pi)^4\delta^{(4)}(p_1+p_2-q_1-q_2)\frac{i\lambda^3}{4}\\\int\frac{d^4k_1d^4k_2}{(2\pi)^8}\\\frac{i}{k_1^2-m^2+i\epsilon}\frac{i}{(p_1+p_2-k_1)^2-m^2+i\epsilon}\\\frac{i}{k_2^2-m^2+i\epsilon}\frac{i}{(p_1+p_2-k_2)^2-m^2+i\epsilon}$$

g) $$(2\pi)^4\delta^{(4)}(p_1+p_2-q_1-q_2)\frac{i\lambda^3}{2}\\\int\frac{d^4k_1d^4k_2}{(2\pi)^8}\\\frac{i}{k_1^2-m^2+i\epsilon}\frac{i}{k_2^2-m^2+i\epsilon}\\\frac{i}{k_1^2-m^2+i\epsilon}\\\frac{i}{(q_1+q_2-k_1)^2-m^2+i\epsilon}$$

h) $$(2\pi)^4\delta^{(4)}(p_1+p_2-q_1-q_2)\frac{-\lambda^4}{8}\\\int\frac{d^4k_1d^4k_2d^4k_3}{(2\pi)^12}\left(\frac{i}{k_1^2-m^2+i\epsilon}\right)^2\\\frac{i}{k_2^2-m^2+i\epsilon}\frac{i}{k_3^2-m^2+i\epsilon}\\\left(\frac{i}{(q_1+q_2-k_1)^2-m^2+i\epsilon}\right)^2$$

i) $$(2\pi)^4\delta^{(4)}(p_1+p_2-q_1-q_2)\frac{i\lambda}{2}\\\int\frac{d^4k_1d^4k_2}{(2\pi)^8}\\\frac{i}{k_1^2-m^2+i\epsilon}\frac{i}{k_2^2-m^2+i\epsilon}\\\frac{i}{(q_1+q_2-k_2)^2-m^2+i\epsilon}\frac{i}{(k_2-k_1-p_2)^2-m^2+i\epsilon}$$

---

(19.2) Draw the interaction vertex for . Find the contributions to the amplitude up to second order in the interaction strength. What are the symmetry factors?

If I’m ever motivated, I’ll do the remainder of this homework, which is again not very enlightening. Ditto for chapter 20.

(18.1) Consider a spin-1/2 particle in a static magnetic field subject to a perpendicular, oscillating magnetic field

$$\hat H=\gamma B_0\hat S_z+\gamma B_1(\hat S_x\cos\gamma B_0t+\hat S_y\sin\gamma B_0 t)$$

where is the gyromagnetic ratio. Write the problem in the interaction representation.

$$\hat H=\hat H_0+\hat H’$$

$$\hat H_0=\gamma B_0\hat S_z$$

$$\hat H’=\gamma B_1(\hat S_x\cos\gamma B_0t+\hat S_y\sin\gamma B_0 t)$$

Simplify the interaction Hamiltonian using the circular basis .

$$\hat S_\pm=\hat S_x\pm i\hat S_y$$

$$\hat S_\pm e^{\pm i\omega t}=e^{i\omega \hat S_z t}\hat S_{\pm}e^{-i\omega \hat S_z t}$$

$$\hat S_x\cos\gamma B_0t+\hat S_y\sin\gamma B_0t=\hat S_++\hat S_-$$

$$\hat H_I=\frac12\gamma B_1(\hat S_++\hat S_-)$$

Find the interaction picture evolution operator.

$$\hat U_I(t_2, t_1)=T\left[e^{-i\int_{t_1}^{t_2} dt\hat H_I}\right]$$

Since there is no time dependence,

$$\hat U_I(t_2,t_1)=e^{-\frac12\gamma B_1(\hat S_++\hat S_-)(t_2-t_1)}$$

What is the probability a particle initially in the state at will still be in that state at time ?

We need to compute

$$e^{-i\frac{\omega}{2} t}\langle \uparrow|U_I(t,0)|\downarrow\rangle$$

where the factor out in front comes from the non-interacting Hamiltonian which has expectation for the state. The only nonzero terms in the Taylor expansion of the exponential are when acts on and acts on . Additionally, . We find that

$$e^{-i\frac\omega2t}\langle \uparrow|e^{-\frac12\gamma B_1(\hat S_++\hat S_-)t}|\uparrow\rangle=e^{-i\frac\omega2t}\cos\frac{\gamma B_1t}{2}$$

The probability is thus $$\cos^2\frac{\gamma B_1t}{2}$$

which of course means the probability of the transition to is $$1-\cos^2\frac{\gamma B_1t}{2}=\sin^2\frac{\gamma B_1t}{2}$$

---

(18.2) Show that , where the states on the RHS are the simple-world states.

$$|\psi_I(\pm \infty)\rangle=|\psi\rangle_{\text{simple world}}=|\psi\rangle$$

By definition,

$$|\psi_I(+\infty)\rangle=\hat S|\psi_I(-\infty)\rangle=\hat S|\psi\rangle$$

Inserting a complete orthonormal basis of states,

$$|\psi_I(+\infty)\rangle=\sum_\phi|\phi\rangle\langle\phi| \hat S|\psi\rangle=\sum_\phi\langle \phi|\hat S|\psi\rangle|\phi\rangle$$

---

(18.3) Use Wick’s thoerem to express the string of Bose operators in terms of normal ordered fields and contractions.

These operators are time-independent, so

$$\hat a_{\mathbf p}\hat a^\dagger_{\mathbf q}\hat a_{\mathbf k}=T[ \hat a_{\mathbf p}\hat a^\dagger_{\mathbf q}\hat a_{\mathbf k}]=N[\text{all contractions}]$$

To compute contractions, we see that , whereas all other contractions are .

$$\overline{\hat b\hat b^\dagger}=(\hat b\hat b^\dagger-N[\hat b\hat b^\dagger])=[\hat b,\hat b^\dagger]$$

Computing,

$$\hat a_{\mathbf p}\hat a^\dagger_{\mathbf q}\hat a_{\mathbf k}=\hat a^\dagger_{\mathbf q}\hat a_{\mathbf p}\hat a_{\mathbf k}+[\hat a_{\mathbf p},\hat a^\dagger_{\mathbf q}]\hat a_{\mathbf k}\\=\hat a^\dagger_{\mathbf q}\hat a_{\mathbf p}\hat a_{\mathbf k}+\hat a_{\mathbf k}\delta^{(3)}(\mathbf{p-q})$$

---

(18.4) Find an expression for in terms of normal-ordered products.

$$\hat b\hat g\hat b\hat b^\dagger\hat b^\dagger=\hat b \hat g(\hat b^\dagger\hat b+[\hat b,\hat b^\dagger])\hat b^\dagger=\hat b\hat g\hat b^\dagger \hat b\hat b^\dagger+\hat b\hat g\hat b^\dagger\\=\hat b^\dagger\hat g\hat b\hat b\hat b^\dagger+\hat g\hat b\hat b^\dagger+\hat b\hat g\hat b^\dagger=\hat b^\dagger\hat g\hat b\hat b^\dagger\hat b+\hat b^\dagger\hat g\hat b+2(\hat g\hat b^\dagger\hat b+\hat g)\\=\hat b^\dagger\hat g\hat b^\dagger\hat b\hat b+\hat b^\dagger\hat g\hat b+3\hat g\hat b^\dagger\hat b+2\hat g\\=\hat b^\dagger\hat b^\dagger\hat b\hat b\hat g+4\hat b^\dagger\hat b\hat g+2\hat g$$

Calculating the same with Wick’s theorem,

$$\hat b\hat g\hat b\hat b^\dagger\hat b^\dagger=T[\hat b\hat g\hat b\hat b^\dagger\hat b^\dagger]=N[\text{all contractions}]$$

The contractions of are all 0.

$$=N[\hat b\hat g\hat b\hat b^\dagger\hat b^\dagger+4\hat g\hat b\hat b^\dagger+2\hat g]\\=\hat b^\dagger\hat b^\dagger\hat b\hat b\hat g+4\hat b^\dagger\hat b\hat g+2\hat g$$

---

(18.5) Use Wick’s theorem on

$$\langle 0|\hat c^\dagger_{\mathbf p_1-\mathbf q}\hat c^\dagger_{\mathbf p_2+\mathbf q}\hat c_{\mathbf p_2}\hat c_{\mathbf p_1}|0\rangle$$

$$=-\delta^{(3)}(\mathbf{p_1-q-p_2})\delta^{(3)}(\mathbf{p_2+q-p_1})+\delta^{(3)}(\mathbf{p_1-q-p_1})\delta^{(3)}(\mathbf{p_2+q-p_2})\\=\delta^{(3)}(\mathbf{q})\delta^{(3)}(\mathbf{q})-\delta^{(3)}(\mathbf{p_1-p_2-q})\delta^{(3)}(\mathbf{p_2-p_1+q})$$

(we ignored the overall normalizations).

(17.1) Calculate the retarded field propagator for a free particle in momentum space and the time domain.

$$G^+_0(\mathbf p,t_x,\mathbf q, t_y)=\theta(t_x-t_y)\langle 0|\hat a_{\mathbf p}(t_x)\hat a^\dagger_{\mathbf q}(t_y)|0\rangle$$

We go into the Schrodinger picture:

$$=\theta(t_x-t_y)\langle 0|e^{i\hat Ht_x}\hat a_{\mathbf p}e^{-i\hat Ht_x}e^{i\hat Ht_y}\hat a^\dagger_{\mathbf q}e^{-i\hat Ht_y}|0\rangle$$

Since and , the exponentials are .

$$=\theta(t_x-t_y)\langle \mathbf p|e^{-i\hat Ht_x}e^{i\hat Ht_y}|\mathbf q\rangle\\=\theta(t_x-t_y)e^{-i(E_{\mathbf p}t_x-E_{\mathbf q}t_y)}\langle \mathbf p|\mathbf q\rangle\\=\theta(t_x-t_y)e^{-i(E_{\mathbf p}t_x-E_{\mathbf q}t_y)}\delta^{(3)}(\mathbf p-\mathbf q)$$

---

(17.2) Demonstrate that the (1+1-dimensional) free scalar propagator is the Green’s function of the Klein-Gordon equation.

We write

$$\Delta(x,y)=\langle0|T\hat\phi(x)\hat\phi(y)|0\rangle=\int \frac{d^2p}{(2\pi)^2}e^{-ip\cdot(x-y)}\frac{i}{p^2-m^2+i\epsilon}$$

Operating on this with ,

$$\partial_t^2-\partial_x^2+m^2(\Delta(x,y))=\int\frac{d^2p}{(2\pi)^2}\left(-p_0^2+p_x^2+m^2\right)\frac{ie^{-ip\cdot(x-y)}}{p_0^2-p^2-m^2+i\epsilon}\\=-i\int\frac{d^2p}{(2\pi)^2}\frac{p_0^2-p_x^2-m^2}{p_0^2-p^2-m^2+i\epsilon}e^{-ip\cdot(x-y)}\\=-i\delta^{(2)}(x-y)$$

---

(17.3) Show that the action for the free scalar field may be written $$S=\frac12\int\frac{d^4p}{(2\pi)^4}\tilde\phi(-p)(p^2-m^2)\tilde\phi(p)$$

We integrate the Lagrangian density by parts to get

$$\mathcal L=\frac12\phi(x)\partial^2\phi(x)-\frac{m^2}2\phi(x)^2=\frac12\phi(x)(\partial^2-m^2)\phi(x)$$

We substitute the Fourier transform for each copy of :

$$\mathcal S=\frac12\int\frac{d^4xd^4pd^4q}{(2\pi)^8}\tilde\phi(p)e^{ip\cdot x}(\partial^2-m^2)\tilde\phi(q)e^{iq\cdot x}=\frac12\int\frac{d^4xd^4pd^4q}{(2\pi)^8}\tilde\phi(p)e^{i(p+q)\cdot x}\phi(q)(-q^2-m^2)$$

The integral over gives , so integrating over gives

$$\frac12\int\frac{d^4p}{(2\pi)^4}\tilde\phi(-p)(p^2-m^2)\tilde\phi(p)$$

We may thus identify as times the inverse of the quadratic term in the momentum-space action ().

---

(17.4) Find the Feynman propagator for the quantum harmonic oscillator with spring constant .

We write the Lagrangian as

$$L=\frac{\dot x^2}{2m}+\frac{m\omega_0^2}{2}x$$

Substituting the Fourier transform of , the same steps as before (integrating the first term by parts and identifying the free propagator) gives

$$\tilde G(\omega)=\frac1m\frac{i}{\omega^2-\omega_0^2+i\epsilon}$$

---

(17.5) Consider the Lagrangian density $$\mathcal L=\frac12(\partial_x\phi)^2+\frac{m^2}2\phi(x)^2$$ Discretize this theory () and find the momentum space expression for the action.

The derivative terms becomes a finite difference $$\frac{\phi_{j+1}-\phi_j}{a}=\frac{1}{a\sqrt{Na}}\sum_p\tilde\phi_p(e^{ip(j+1)a}-e^{ipja})$$

The integral becomes . Carrying out some tedious sums, we find that the momentum is .

$$S=\frac12\sum_p\tilde\phi_{-p}\left(\frac2{a^2}-\frac{2\cos pa}{a^2}+m^2\right)\tilde\phi_p$$

$$\tilde G(p)=\frac{i}{\frac2{a^2}-\frac{2\cos pa}{a^2}+m^2}$$

The Lagrangian for a 1D elastic string in (1+1)-Minkowski space is 

$$\mathcal L=\frac12\left((\partial_0\phi)^2-(\partial_1\phi)^2\right)$$

Discretize this theory in space only and find the propagator, with .

This is truly unenlightening.

$$\tilde G(\omega,p)=\frac{i}{\omega^2-\omega_0^2(1-\cos pa)}$$

To reason a little, the acts as the effective mass in this theory, leading to the in the denominator, whereas the comes out of the spatial derivative term. There is an extra minus sign on that term since the Lagrangian has the extra minus sign compared to before.

(16.1) Solve the Schrodinger equation for a one-dimensional infinite square well.

Separation of variables gives

$$\psi(x,t)=\sum_n e^{-iE_n t}\psi_n(x)$$

The time-independent Schrodinger equation is

$$\hat H\psi_n=E_n\psi$$

$$\frac1{2m}\partial_x^2\psi_n=-E_n\psi$$

The general solution to this homogeneous ODE is

$$\psi_n=A_n\sin \sqrt{2mE_n}x+B_n\cos\sqrt{2mE_n}x$$

Since we require , . Since we require , . Since we require the wavefunction to be nonzero, . The energy is thus ,

$$E_n=\frac{n^2\pi^2}{2ma^2}$$

Computing the retarded Green’s function,

$$\langle n|\hat U(t_2-t_1)|n\rangle=\langle n|e^{-i\hat H(t_2-t_1)}|n\rangle\\=\langle n|e^{-iE_n(t_2-t_1)}|n\rangle=e^{-i\frac{n^2\pi^2}{2ma^2}(t_2-t_1}\langle n|n\rangle=e^{-i\frac{n^2\pi^2}{2ma^2}(t_2-t_1)}$$

Finally, we enforce :

$$G^+(n, t_2, t_1)=\theta(t_2-t_1)e^{-i\frac{n^2\pi^2}{2ma^2}(t_2-t_1)}$$

Find .

$$G_0^+(n,\omega)=\int_{-\infty}^\infty dt_2 G^+(n,t_2, t_1)e^{i\omega(t_2-t_1)}\\=\int_{-\infty}^\infty dt_2 \theta(t_2-t_1) e^{i\left(\omega -\frac{n^2\pi^2}{2ma^2}+i\epsilon\right)(t_2-t_1)}$$

We add a factor of so that the integral converges.

$$=\frac{1}{i\left(\omega-\frac{n^2\pi^2}{2ma^2}+i\epsilon\right)}e^{i\left(\omega-\frac{n^2\pi^2}{2ma^2}+i\epsilon\right)(t_2-t_1)}\bigg\vert_{t_1}^\infty\\=\frac{i}{\omega-\frac{n^2\pi^2}{2ma^2}+i\epsilon}$$

---

(16.2) Derive .

I suppress the complete sets of states for .

$$G^+_0(x,y,E)=\sum_n\langle n|\int_{-\infty}^\infty \theta(t_2-t_1)e^{-iE_n (t_2-t_1)}e^{iE(t_2-t_1)}|n\rangle\\=\sum_n\langle n|\int_{t_1}^\infty e^{i\left(E-E_n+i\epsilon\right)(t_2-t_1)}|n\rangle\\=\sum_n\langle n|\frac{1}{i\left(E-E_n+i\epsilon\right)}e^{i\left(E-E_n+i\epsilon\right)(t_2-t_1)}\bigg\vert_{t_1}^\infty|n\rangle\\=\sum_n\frac{i\phi_n(x)\phi_n^*(y)}{E-E_n+i\epsilon}$$

Use the Fourier definition of the Heaviside function to derive .

$$G^+_0(p,t)=\theta(t)e^{-iE_pt}$$

$$G^+_0(p,E)=\int_{-\infty}^\infty dt G^+_0(p,t)e^{iE t}\\=\int_{-\infty}^\infty dt \int_{-\infty}^\infty\frac{dz}{2\pi}\frac{ie^{-izt}}{z+i\epsilon}e^{i(E-E_p)t}$$

Performing the integral,

$$=\int_{-\infty}^\infty dz \frac i{z+i\epsilon}\int_{-\infty}^\infty \frac{dt}{2\pi} e^{-i(z-E+E_p)t}$$

$$=\int_{-\infty}^{\infty} dz \frac{i}{z+i\epsilon}\delta(E-E_p-z)$$

$$=$\frac{i}{E-E_p+i\epsilon}}$

---

(16.3) Consider the one-dimensional simple harmonic oscillator with a forcing function described by an equation of motion

$$m\frac{\partial^2}{\partial t^2}A(t-u)+m\omega_0^2 A(t-u)=\tilde F(\omega) e^{-i\omega(t-u)}$$

Solving for A(t-u),

A specific solution is

$$-\frac{\tilde F(\omega)}{m}\frac{e^{-i\omega(t-u)}}{\omega^2-\omega_0^2}$$

Therefore the general solution is

$$A(t-u)=-\frac{\tilde F(\omega)}{m}\frac{e^{-i\omega(t-u)}}{\omega^2-\omega_0^2}+B(t)$$

where is a solution to the homogeneous equation of motion.

To find the Green’s function, we write

$$\left(m\frac{\partial^2}{\partial t^2}+m\omega_0^2\right)G(t,u)=\delta(t-u)$$

Performing a Fourier transform,

$$\left(m\frac{\partial^2}{\partial t^2}+m\omega_0^2\right)\int \frac{d\omega}{2\pi}\tilde{G}(\omega)e^{-i\omega(t-u)}=\int \frac{d\omega}{2\pi} e^{-i\omega(t-u)}$$

$$m(-\omega^2+\omega_0^2)\tilde G(\omega)=1$$

$$\tilde{G}(\omega)=-\frac{1}{m}\frac{1}{\omega^2-\omega^2_0}$$

Thus

$$G(t,u)=-\frac{1}{m}\int_{-\infty}^\infty\frac{d\omega}{2\pi}\frac{e^{-i\omega(t-u)}}{\omega^2-\omega^2_0}$$

If is subject to the boundary conditions that at we have and , show that for .

The poles of the integral are at . By the residue theorem,

$$G^+(t,u)=-\frac 1{2\pi m}\left[\frac{e^{-i\omega(t-u)}}{\omega-\omega_0}\bigg\vert_{\omega=-\omega_0}+\frac{e^{-i\omega(t-u)}}{\omega+\omega_0}\bigg\vert_{\omega=\omega_0}\right]\\=\frac{1}{m\omega_0}\frac{e^{i\omega_0(t-u)}-e^{-i\omega_0(t-u)}}{2i}\\=\frac{1}{m\omega_0}\sin\omega_0(t-u)$$

Given the forcing function , find the trajectory.

$$A(t)=\int du G^+(t,u) F_0\sin\omega_0 u\\=\frac{F_0}{m\omega_0}\int_0^t du \sin(\omega_0(t-u))\sin(\omega_0 u)$$

Using a version of the sum of angles formula for cosine, , this integral comes out to

$$A(t)=\frac{F_0}{2m\omega_0^2}\left(\sin\omega_0 t-\omega_0 t\cos\omega_0 t\right)$$

---

(16.4) Find the Green’s function for the Helmholtz equation.

$$(\nabla^2+k^2)G_k(x)=\delta^{(3)}(x)$$

Performing the Fourier transform,

$$\int\frac{d^3q}{(2\pi)^3}(\nabla^2+k^2)\tilde G_k(q)e^{-iq\cdot x}=\int\frac{d^3q}{(2\pi)^3}e^{-iq\cdot x}$$

Equating the Fourier transforms,

$$\tilde G_k(q)=\frac{1}{k^2-q^2}$$

Take the Fourier transform of for outgoing waves.

$$-\int d^3x\frac{e^{-i|k||x|}}{4\pi|x|}e^{-iq\cdot x-\epsilon|x|}\\=-\int d^3x\frac{e^{-i|k||x|}}{4\pi|x|}e^{-i|q||x|\cos\theta-\epsilon|x|}$$

$$=-\frac{1}{4\pi}\int_0^{2\pi}d\phi\int_{-1}^1d(\cos\theta)\int_0^\infty x^2 dx\frac1x e^{i(|k|+i\epsilon)|x|+i|q||x|\cos\theta}$$

The integral becomes , while the integral gives

$$=-\frac{1}{2}\int_0^\infty dx\ xe^{i(|k|+i\epsilon)|x|}\frac{1}{i|q||x|}(e^{i|q||x|}-e^{-i|q||x|})$$

$$=-\frac{1}{|q|}\frac{|q|}{|q|^2-(|k|+i\epsilon)^2}=\frac{1}{|k|^2-|q|^2+i\epsilon}$$

after we ignore the second order term, and rewrite the remaining positive infinitesimal as .

Take the inverse Fourier transform of .

This is very similar, and we perform a contour integral with the residue theorem, whose poles are already nicely separated as shown above.

What do you expect for an incoming wave solution?

We have to shift the poles the opposite way, so we will have an advanced propagator with in the denominator.