Quantum mechanics/Quantum field theory on a violin string

This construction of an elementary quantum field theory will also give readers a glimpse of Fourier series expansions, Hamiltonian mechanics, and also Black-body radiation. It assumes that the reader is familiar with the solution to Schrödinger equation for a quantum harmonic oscillator.

The classical theory of transverse waves on a string

The first six modes (n= 1, 2,...,6) of a taut string clamped at both ends.

We begin with the classical theory of transverse waves on a vibrating string with length, $L$ , mass $M$ , and tension, $T$ . The dispersion relation, ω=ω(k), relates frequency to wavelength:

\omega _{n}=ck_{n}={\frac {c\pi }{L}}n

,

where $\omega =2\pi f$ is the angular frequency and $k=2\pi /\lambda$ is the wavenumber. The boundary conditions at each of the string of length $L$ imply that the wavenumber can take on only those values that cause the length to equal integral number of half wavelengths: $k_{n}L=n\pi$ , where $n$ may be taken to be a positive integer (1,2,3...). Thus we have:

k_{n}={\frac {n\pi }{L}}

.

The speed of transverse waves is,

c={\sqrt {\frac {T}{\mu }}}

,

where $\mu =M/L$ is the linear mass density.

Fourier series and wave energy

In our classical wave, the transverse displacement obeys,

y(x,t)={\sqrt {2}}\sum _{n=1}^{\infty }\xi _{n}(t)\sin {\frac {n\pi x}{L}}

While it is not customary to include the factor ${\sqrt {2}}$ in this Fourier series, the insertion of this factor redefines the coefficients $\xi _{n}$ in a way that will prove convenient for establishing that this system is equivalent to an infinite collection of simple harmonic oscillators.

The quantum mechanical version of a classical theory begins with some canonical version of the theory. We shall adopt the convention than ${\dot {\xi }}$ denotes $d\xi /dt$ .The total kinetic energy of the wave is,

KE={\frac {1}{2}}\mu \int _{0}^{L}dx\,{\dot {y}}^{2}={\frac {\mu }{2}}\sum _{n=1}^{\infty }\sum _{m=1}^{\infty }2{\dot {\xi }}_{n}{\dot {\xi }}_{m}\int _{0}^{L}\sin {\frac {n\pi x}{L}}\sin {\frac {m\pi x}{L}}\,dx

The double sum contains terms when the two indices (m,n) are equal and terms where they are not equal.

\sum _{n=1}^{\infty }\sum _{m=1}^{\infty }\{...\}=\sum _{n=m}\{...\}+\sum _{n\neq m}\{...\}

The integrals over the product of the two sine waves have simple properties because the interval of length L contains exactly an integral number of half-wavelengths (i.e., n and m are both integers). Therefore,

\int _{0}^{L}\sin {\frac {n\pi x}{L}}\sin {\frac {m\pi x}{L}}\,dx=0\ \ \ if\ n\neq m

\int _{0}^{L}\sin {\frac {n\pi x}{L}}\sin {\frac {m\pi x}{L}}\,dx={\frac {L}{2}}\ \ \ if\ n=m

Comment on inner product, orthogonal functions, and these integrals

This result is easy to remember if one notes that the average of a sinusoidal over n half wavelengths equals 1/2 if n is an integer, and that the integral of a constant over a segment equals the length of that segment:

\int \sin ^{2}(...)dx=\int {\overline {\sin ^{2}(...)}}dx={\frac {1}{2}}\int dx={\frac {1}{2}}L

This is one of many examples in physics where a class of functions (here sine functions) obeys this orthogonality condition:

\int f_{n}(x)f_{m}(x)dx=0\ \ if\ m\neq n

If this identity holds, $f_{n}$ and $f_{m}$ are said to be w:orthogonal functions because $\int f(x)g(x)dx$ is known as the inner product of the functions $f$ and $g$ (if $f$ and $g$ are real).^[1] Whenever such a collection of orthogonal functions is defined, the range of the integral must be specified (here it is from $0$ to $L$ .)

With this substitution we have for the kinetic energy of a vibrating string:

KE={\frac {1}{2}}m\sum _{n=1}^{\infty }{\dot {\xi }}_{n}^{2}

If the factor of ${\sqrt {2}}$ had not been inserted earlier, we would have redefined our amplitude so that the wave's kinetic energy would take this intuitive form.

Potential energy in a wave

The work required to stretch a string of length $L$ to a length of $L+\delta L$ is $T\delta L$ , where $T$ is the tension in the string. This work acts as a potential energy. A transverse wave with displacement $y=y(x)$ has a length given by,

\int _{x=0}^{x=L}{\sqrt {dx^{2}+dy^{2}}}=\int _{0}^{L}{\sqrt {1+\left({\frac {dy}{dx}}\right)^{2}}}dx\approx \int _{0}^{L}{\left[1+{\frac {1}{2}}\left({\frac {dy}{dx}}\right)^{2}+...\right]dx}

where we have used the approximation for small ε: (1+ε)^p≈1+pε.

If y=y(x,t) represents a wave, it is customary to replace the derivative by a partial derivative:dy/dx → ∂y/∂x. Moreover, it is convenient to express the partial derivative in terms of the wavenumber described above.

{\frac {\partial }{\partial x}}\left(\sin {\frac {n\pi x}{L}}\right)=\underbrace {\frac {n\pi }{L}} _{k_{n}}\cos {\frac {n\pi x}{L}}=k_{n}\cos {\frac {n\pi x}{L}}

where we note that $k_{n}=n\pi /L$ is the wavenumber of the n-th mode. Using the Fourier series expansion described above we have, the potential energy of the wave is

PE={\frac {1}{2}}T\int _{0}^{L}\left({\frac {\partial y}{\partial x}}\right)^{2}dx={\frac {T}{2}}\sum _{n=1}^{\infty }\sum _{m=1}^{\infty }2\xi _{n}(t)\xi _{m}(t)k_{n}k_{m}\int _{0}^{L}\cos {\frac {n\pi x}{L}}\cos {\frac {m\pi x}{L}}\,dx

As occurred previously with the kinetic energy this double sum becomes a single sum over all cases where m = n because the cosine functions are also orthogonal functions over this range of integration (provided n and m are integers). As before the integral $\int \cos ^{2}(k_{n}x)dx=L/2$ because the average value of cos² is 1/2 whenever the cosine is averaged over an integral number of half-wavelengths.

PE={\frac {1}{2}}\sum _{n=1}^{\infty }{\frac {n^{2}\pi ^{2}T}{L}}\xi _{n}^{2}={\frac {1}{2}}\sum _{n=1}^{\infty }K_{n}\xi _{n}^{2}

where the spring constant associated with the nth mode is

K_{n}={\frac {n^{2}\pi ^{2}T}{L}}=Lk_{n}^{2}T=n^{2}K_{1}

It is known for the classical wave equation for a stretched string that each mode oscillates as

\xi _{n}(t)=A\sin \omega _{n}t+B\cos \omega _{n}t

where

\omega _{n}={\sqrt {\frac {K_{n}}{M}}}=n\omega _{1}

,

and $\omega _{1}={\sqrt {\frac {K_{1}}{M}}}$ is the frequency of the lowest order standing wave in the classical vibrating string).

Quantizing the harmonic oscillators

From the known behavior of the classical violin string, we obtain equations of motion, which if cast in canonical form, will tell us how to create the quantum mechanical version of the theory. Our canonical form shall be that of Hamiltonian mechanics. Our goal is to show that the classical vibrating string is identical to an (almost?) infinite number of independent simple harmonic oscillators.

H(\xi _{1},\xi _{2},\xi _{3},...,{\mathcal {P}}_{1},{\mathcal {P}}_{2},{\mathcal {P}}_{3},...)={\frac {1}{2M}}\left({\mathcal {P}}_{1}^{2}+{\mathcal {P}}_{2}^{2}+{\mathcal {P}}_{3}^{2}++...\right)+{\frac {1}{2}}K_{1}\left(1^{2}\xi _{1}^{2}+2^{2}\xi _{2}^{2}+3^{2}\xi _{3}^{2}+...\right)

,

where ${\mathcal {P}}=M{\dot {\xi }}$ is the conjugate momentum. The wave equation for the simple harmonic oscillator is well known. The variables $\xi _{1},\xi _{2},\xi _{3},\xi _{4},...$ play the same role as $x,y,z$ in the quantum mechanics of a single particle. Schrödinger's equation is: $i\hbar {\frac {\partial }{\partial t}}\Psi ={\frac {-\hbar ^{2}}{2M}}\left({\frac {\partial ^{2}}{\partial \xi _{1}^{2}}}+{\frac {\partial ^{2}}{\partial \xi _{2}^{2}}}+{\frac {\partial ^{2}}{\partial \xi _{3}^{2}}}+...\right)\Psi +{\frac {1}{2}}K_{1}\left(1^{2}\xi _{1}^{2}+2^{2}\xi _{2}^{2}+3^{2}\xi _{3}^{2}+...\right)\Psi$

The solution is of the form, $\psi e^{-i\hbar \Omega t}$ , where $\Psi =\prod \psi _{1}({\mathcal {N}}_{1},\xi _{1})\cdot \psi _{1}({\mathcal {N}}_{2},\xi _{2})\cdot \psi _{3}({\mathcal {N}}_{3},\xi _{3})...$

and

\Omega =\omega _{1}\sum _{n=1}^{\infty }\left(n{\mathcal {N}}_{n}+{\frac {1}{2}}\right)

and

\psi _{n}({\mathcal {N}}_{1},\xi _{1})

is the energy eigenstate for the in the ${\mathcal {N}}^{\mathrm {th} }$ energy level of potential associated with a spring constant equal to $K_{n}$ .

If you really need to see these wavefunctions, here they are:

The energy eigenstates are:^[2] $\psi ({\mathcal {N}},n,\xi )={\frac {1}{\sqrt {2^{\mathcal {N}}\,{\mathcal {N}}!}}}\cdot \left({\frac {m\omega _{n}}{\pi \hbar }}\right)^{1/4}\cdot e^{-{\frac {m\omega _{n}x^{2}}{2\hbar }}}\cdot H_{\mathcal {N}}\left({\sqrt {\frac {m\omega _{n}}{\hbar }}}x\right),\qquad {\mathcal {N}}=0,1,2,\ldots .$