IPhO 2006 · Theory problem 1

Gravity in a Neutron Interferometer

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Figure 1

Figure 1a

Figure 2

Figure 1b

Physical situation We consider the situation of the famous neutron-interferometer experiment by Collela, Overhauser and Werner, but idealize the set-up inasmuch as we shall assume perfect beam splitters and mirrors within the interferometer. The experiment studies the effect of the gravitational pull on the de Broglie waves of neutrons.

The symbolic representation of this interferometer in analogy to an optical interferometer is shown in Figure 1a. The neutrons enter the interferometer through the IN port and follow the two paths shown. The neutrons are detected at either one of the two output ports, OUT1 or OUT2. The two paths enclose a diamond-shaped area, which is typically a few \(\mathrm{cm^2}\) in size.

The neutron de Broglie waves (of typical wavelength of \(10^{-10}\ \mathrm{m}\)) interfere such that all neutrons emerge from the output port OUT1 if the interferometer plane is horizontal. But when the interferometer is tilted around the axis of the incoming neutron beam by angle \(\phi\) (Figure 1b), one observes a \(\phi\) dependent redistribution of the neutrons between the two output ports OUT1 and OUT2.

Geometry For \(\phi=0^\circ\) the interferometer plane is horizontal; for \(\phi=90^\circ\) the plane is vertical with the output ports above the tilt axis.

1.1 (1.0) How large is the diamond-shaped area \(A\) enclosed by the two paths of the interferometer?

1.2 (1.0) What is the height \(H\) of output port OUT1 above the horizontal plane of the tilt axis?

Express \(A\) and \(H\) in terms of \(a\), \(\theta\), and \(\phi\).

Optical path length The optical path length \(N_{\mathrm{opt}}\) (a number) is the ratio of the geometrical path length (a distance) and the wavelength \(\lambda\). If \(\lambda\) changes along the path, \(N_{\mathrm{opt}}\) is obtained by integrating \(\lambda^{-1}\) along the path.

1.3 (3.0) What is the difference \(\Delta N_{\mathrm{opt}}\) in the optical path lengths of the two paths when the interferometer has been tilted by angle \(\phi\)? Express your answer in terms of \(a\), \(\theta\), and \(\phi\) as well as the neutron mass \(M\), the de Broglie wavelength \(\lambda_0\) of the incoming neutrons, the gravitational acceleration \(g\), and Planck’s constant \(h\).

1.4 (1.0) Introduce the volume parameter

\[ V=\frac{h^2}{gM^2} \]

and express \(\Delta N_{\mathrm{opt}}\) solely in terms of \(A\), \(V\), \(\lambda_0\), and \(\phi\). State the value of \(V\) for \(M=1.675\times10^{-27}\ \mathrm{kg}\), \(g=9.800\ \mathrm{m\,s^{-2}}\), and \(h=6.626\times10^{-34}\ \mathrm{J\,s}\).

1.5 (2.0) How many cycles — from high intensity to low intensity and back to high intensity — are completed by output port OUT1 when \(\phi\) is increased from \(\phi=-90^\circ\) to \(\phi=90^\circ\)?

Experimental data The interferometer of an actual experiment was characterized by \(a=3.600\ \mathrm{cm}\) and \(\theta=22.10^\circ\), and 19.00 full cycles were observed.

1.6 (1.0) How large was \(\lambda_0\) in this experiment?

1.7 (1.0) If one observed 30.00 full cycles in another experiment of the same kind that uses neutrons with \(\lambda_0=0.2000\ \mathrm{nm}\), how large would be the area \(A\)?

Hint: If \(|\alpha x|\ll1\), it is permissible to replace \((1+x)^\alpha\) by \(1+\alpha x\).

Official English text of the problem.

Answers

Define the signed difference as \(\Delta N_{\mathrm{opt}}=N_{\mathrm{lower}}-N_{\mathrm{upper}}\), with the paths named according to Figure 1a. The phase results retain first order in the gravitational energy change.

  • 1.1: \(\displaystyle A=2a^2\tan\theta.\)
  • 1.2: \(\displaystyle H=2a\sin\theta\sin\phi.\)
  • 1.3: \(\displaystyle \Delta N_{\mathrm{opt}}\simeq\frac{2M^2ga^2\lambda_0}{h^2}\tan\theta\sin\phi.\)
  • 1.4: \(\displaystyle \Delta N_{\mathrm{opt}}\simeq\frac{A\lambda_0}{V}\sin\phi,\qquad V=1.597\times10^{-14}\ \mathrm{m^3}.\)
  • 1.5: \(\displaystyle C=\frac{2A\lambda_0}{V}=\frac{4M^2ga^2\lambda_0}{h^2}\tan\theta\) cycles.
  • 1.6: \(\displaystyle \lambda_0=1.441\times10^{-10}\ \mathrm{m}=0.1441\ \mathrm{nm}.\)
  • 1.7: \(\displaystyle A=1.198\times10^{-3}\ \mathrm{m^2}=11.98\ \mathrm{cm^2}.\)

Solution

1.1 — Enclosed area

Label the entrance beam splitter \(O\), the upper mirror \(U\), the lower mirror \(D\), and the exit beam splitter \(R\), according to Figure 1a.

Each side of the diamond has length

\[ \ell=\frac{a}{\cos\theta}. \]

The horizontal diagonal \(OR\) is \(2a\), while the perpendicular diagonal \(UD\) is \(2a\tan\theta\). Therefore,

\[ \boxed{A=\frac12(2a)(2a\tan\theta)=2a^2\tan\theta.} \]

Tilting the apparatus does not change this area.

1.2 — Height of OUT1

The incoming beam is collinear with \(OD\), so the tilt axis lies along \(OD\). The opposite side \(UR\) is parallel to this axis.

The perpendicular distance, within the interferometer plane, from \(R\) to the tilt axis is

\[ d=\ell\sin(2\theta) =\frac{a}{\cos\theta}\,2\sin\theta\cos\theta =2a\sin\theta. \]

Rotating the plane by \(\phi\) gives this distance a vertical component \(d\sin\phi\). Hence

\[ \boxed{H=2a\sin\theta\sin\phi.} \]

Both \(U\) and \(R\) have height \(H\), while \(O\) and \(D\) remain at height zero. The OUT1 beam is parallel to the tilt axis, so it stays at height \(H\). For negative \(\phi\), \(H\) is negative.

1.3 — Difference in optical path lengths

Use nonrelativistic mechanics, a uniform gravitational field, and elastic interactions with the stationary ideal optical elements. Set the gravitational potential energy to zero on the tilt axis.

The incoming momentum and kinetic energy are

\[ p_0=\frac{h}{\lambda_0}, \qquad K_0=\frac{h^2}{2M\lambda_0^2}. \]

Energy conservation at height \(z\) gives

\[ \frac{p(z)^2}{2M}+Mgz=\frac{p_0^2}{2M}. \]

Using the de Broglie relation, the reciprocal wavelength is therefore

\[ \frac{1}{\lambda(z)} =\frac{p(z)}{h} =\frac{1}{\lambda_0} \sqrt{1-\frac{2M^2g\lambda_0^2z}{h^2}}. \]

We retain first order in the small ratio \(Mg|z|/K_0\). The hint then gives

\[ \frac{1}{\lambda(z)} \simeq \frac{1}{\lambda_0} -\frac{M^2g\lambda_0}{h^2}z. \]

At this accuracy we use the straight segments shown in the figure. For fixed endpoints, the extra geometrical length caused by a small gravitational deflection is second order in that deflection.

The two sloping segments \(OU\) and \(DR\) both have length \(\ell\), and their heights vary identically from \(0\) to \(H\). Thus their optical path lengths are equal:

\[ J=\int_0^\ell \frac{ds}{\lambda(Hs/\ell)}. \]

The remaining segments are at constant heights: \(OD\) is at zero and \(UR\) is at \(H\). Consequently,

\[ N_{\mathrm{lower}}=\frac{\ell}{\lambda_0}+J, \qquad N_{\mathrm{upper}}=J+\frac{\ell}{\lambda(H)}. \]

Subtracting,

\[ \begin{aligned} \Delta N_{\mathrm{opt}} &=\ell\left(\frac{1}{\lambda_0}-\frac{1}{\lambda(H)}\right)\\ &=\frac{\ell}{\lambda_0} \left[ 1-\sqrt{1-\frac{2M^2g\lambda_0^2H}{h^2}} \right]\\ &\simeq\frac{M^2g\lambda_0}{h^2}\ell H. \end{aligned} \]

Since

\[ \ell H =\frac{a}{\cos\theta}\,2a\sin\theta\sin\phi =2a^2\tan\theta\sin\phi, \]

the requested result is

\[ \boxed{ \Delta N_{\mathrm{opt}} \simeq \frac{2M^2ga^2\lambda_0}{h^2} \tan\theta\sin\phi. } \]

For positive \(\phi\), the upper path has the smaller optical length because its elevated segment has a smaller neutron momentum and hence a longer wavelength. Reversing the subtraction convention reverses the sign without changing the interference predictions.

1.4 — Volume parameter

Using \(A=2a^2\tan\theta\) and \(V=h^2/(gM^2)\),

\[ \boxed{\Delta N_{\mathrm{opt}}\simeq\frac{A\lambda_0}{V}\sin\phi.} \]

The dimensions of \(V\) are

\[ [V]= \frac{(\mathrm{kg\,m^2\,s^{-1}})^2} {(\mathrm{m\,s^{-2}})\mathrm{kg^2}} =\mathrm{m^3}. \]

Numerically,

\[ V= \frac{(6.626\times10^{-34})^2} {9.800(1.675\times10^{-27})^2} =1.5967876\ldots\times10^{-14}\ \mathrm{m^3}. \]

Thus

\[ \boxed{V=1.597\times10^{-14}\ \mathrm{m^3}.} \]

1.5 — Number of intensity cycles

An optical path difference of one corresponds to a relative phase change of \(2\pi\):

\[ \delta=2\pi\Delta N_{\mathrm{opt}}. \]

The given constructive interference at OUT1 when \(\phi=0\) fixes the constant phase offset. For the ideal equal-amplitude paths,

\[ \frac{I_{\mathrm{OUT1}}}{I_{\mathrm{in}}} =\frac{1+\cos\delta}{2} =\cos^2\!\left(\pi\frac{A\lambda_0}{V}\sin\phi\right). \]

Maxima occur at integer \(\Delta N_{\mathrm{opt}}\), and minima at half-integer values. Each increase of one in \(\Delta N_{\mathrm{opt}}\) therefore produces one intensity cycle.

As \(\phi\) increases from \(-90^\circ\) to \(90^\circ\), \(\sin\phi\) increases monotonically from \(-1\) to \(1\). Hence the total number of cycles is

\[ \begin{aligned} C &=\Delta N_{\mathrm{opt}}(90^\circ) -\Delta N_{\mathrm{opt}}(-90^\circ)\\ &=\frac{A\lambda_0}{V} -\left(-\frac{A\lambda_0}{V}\right). \end{aligned} \]

Therefore, to the same first-order accuracy,

\[ \boxed{ C=\frac{2A\lambda_0}{V} =\frac{4M^2ga^2\lambda_0}{h^2}\tan\theta. } \]

1.6 — Incoming wavelength

The experimental area is

\[ A=2(0.03600\ \mathrm{m})^2\tan(22.10^\circ) =1.052502\ldots\times10^{-3}\ \mathrm{m^2}. \]

Solving the cycle formula for the wavelength,

\[ \lambda_0=\frac{CV}{2A}. \]

Using \(C=19.00\) and retaining unrounded intermediate values,

\[ \lambda_0 = \frac{19.00(1.5967876\ldots\times10^{-14}\ \mathrm{m^3})} {2(1.052502\ldots\times10^{-3}\ \mathrm{m^2})} =1.441278\ldots\times10^{-10}\ \mathrm{m}. \]

Thus

\[ \boxed{\lambda_0=1.441\times10^{-10}\ \mathrm{m} =0.1441\ \mathrm{nm}.} \]

The approximation used above is well justified. These neutrons have speed

\[ v_0=\frac{h}{M\lambda_0}\simeq2.745\times10^3\ \mathrm{m\,s^{-1}}, \]

and the largest height is \(H_{\max}=2a\sin\theta\simeq0.02709\ \mathrm{m}\). Therefore,

\[ \frac{MgH_{\max}}{K_0} =\frac{2gH_{\max}}{v_0^2} \simeq7.05\times10^{-8}\ll1. \]

1.7 — Area for the second experiment

Rearranging the same cycle formula,

\[ A=\frac{CV}{2\lambda_0}. \]

With \(C=30.00\) and \(\lambda_0=0.2000\ \mathrm{nm}=2.000\times10^{-10}\ \mathrm{m}\),

\[ A= \frac{30.00(1.5967876\ldots\times10^{-14}\ \mathrm{m^3})} {2(2.000\times10^{-10}\ \mathrm{m})} =1.1975907\ldots\times10^{-3}\ \mathrm{m^2}. \]

Hence

\[ \boxed{A=1.198\times10^{-3}\ \mathrm{m^2}=11.98\ \mathrm{cm^2}.} \]

The solution was checked against the official solution.