Finding the wavelength of particles in the centre of mass frame I'm having a bit of trouble understanding this question.
If I have two identical non-relativistic particles that are travelling perpendicularly to each other with their respective De Broglie wavelengths as $\lambda_{1}$ and $\lambda_{2}$, find the wavelength of both particles in the frame where their centre of mass is at rest. 
Surely $\lambda_{1}$ and $\lambda_{2}$ are those wavelengths?
EDIT: I've attempted the question again and have come to answer but I am not too sure whether or not it's correct.
First, I supposed that particle one is travelling vertically (parallel to j) with momentum $\vec{p_{1}}$ and that particle two is travelling horizontally (parallel to i) with momentum $\vec{p_{2}}$.
It follows that using $\lambda = \frac{h}{|p|}$ that $$\vec{p_{1}} = (0, \frac{h}{\lambda_{1}})$$
$$\vec{p_{2}} = (\frac{h}{\lambda_{2}},0)$$
And so it follows that the speed of the centre of mass and of the individual particles are given by:
$$V_{cm}=\frac{\vec{p_{1}}+\vec{p_{2}}}{2m}$$ 
$$V_{1}=(0,\frac{h}{\lambda_{1}m})$$ 
$$V_{2}=(\frac{h}{\lambda_{2}m,},0)$$
where $m$ is the mass of each of the particles, so then the velocity of each of the particles in centre of mass frame are given by:
$$V_{1,cm}= V_{1} - V_{cm}$$ $$V_{2,cm}= V_{2} - V_{cm}$$
so in the centre of mass frame particles 1 and 2 have the respective momenta:
$$\vec{q_{1}}=(\frac{-h}{2\lambda_{2}}\frac{h}{2\lambda_{1}})$$ with $$|\vec{q_{1}}|=\frac{h}{2}\sqrt{\frac{1}{\lambda_{2}^{2}}+\frac{1}{\lambda_{1}^{2}}}$$
and 
$$\vec{q_{2}}=(\frac{h}{2\lambda_{2}}\frac{-h}{2\lambda_{1}})$$ with $$|\vec{q_{2}}|=\frac{h}{2}\sqrt{\frac{1}{\lambda_{2}^{2}}+\frac{1}{\lambda_{1}^{2}}}$$
and so then it follows that $$\lambda_{1,cm}=\lambda_{2,cm}=\frac{2}{\sqrt{\frac{1}{\lambda_{2}^{2}}+\frac{1}{\lambda_{1}^{2}}}}$$
Am I supposed to get that the wavelengths are equal? Would it be better to use the relativistic equation for momentum or is that not necessary for the question?
 A: This is a very easy question that raises a very difficult point.
The easy bit: use $\lambda_1$ and $\lambda_2$ to get the momenta $p_1$ and $p_2$. Add them vectorially to find the centre of momentum frame and its velocity. Subtract that velocity from the particles' velocities to get their velocities in the centre of momentum frame, hence their momenta in hat frame, and hence (using $\lambda=h/p$ ) their wavelengths.
I guess you could have done all that if you weren't thinking about the difficult point: these are wavelengths, this is a non-relavistic problems so lengths don't change, so why do they?
The de Broglie wavelength is trickier than it looks. You can't measure it with a ruler. It has no absolute meaning. The only way to measure it is by interference. For example, if you have a beam of similar particles you could do a Young's slits experiment with slits and a screen a distance $D$ away, and see peaks at $d sin\theta = n \lambda$. You measure $\theta$ as $\tan^{-1}(x/D)$ and that tells you $\lambda$. 
Now consider the experiment from the point of view of an observer in a frame moving in the same direction as the particle beam (and at some fraction of their speed). The particles have lower momentum in this frame. They go through the slits as before but the screen - which was, of course, stationary - is now perceived as moving towards the particles, so they will (according to the second observer) have travelled a distance less than $D$, and the angles will appear bigger, so the second observer will measure the wavelength to be longer, in agreement with the lower momentum.   
So: particle wavelengths (and frequencies) should be treated with care. They change when the observer changes - even non-relativistically - but in such a way that the measurable consequences stay the same. 
A: If you have 2 De Broglie wavelengths, you can associate momenta with each of these particles. From these momenta, find the center of mass frame, and their momenta in this frame. Now, find the wavelengths associated to these momenta.
