Lancaster and Blundell Chapter 23

(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 ,

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

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(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

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

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(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.