As you increase the energy of a photon it's wavelength shortens. Is it theoretically posible for light to not have a wavelength? Like a still pond?
[Physics] Can light have zero wavelength
electromagnetic-radiationphotonswavelength
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Ted's comment is correct, redshift makes the wavelength increase. But if you actually meant zero frequency (which corresponds to infinite wavelength), then it's almost possible, if you emit a photon from just outside the event horizon of a black hole.
Specifically, suppose the photon is emitted from a small coordinate distance $\epsilon R_s$ ($\epsilon \ll 1$) outside the event horizon. According to Wikipedia, the redshift of that photon once it reaches an infinite distance away from the hole would be
$$\lim_{r\to +\infty}z(r) = \frac{1}{\sqrt{1-\left(\frac{R_s}{R^*}\right)}}-1 = \frac{1}{\sqrt{1-\left(\frac{R_s}{R_s(1 + \epsilon)}\right)}}-1 \approx \sqrt{\frac{1}{\epsilon}}$$
Redshift relates to wavelength as $z = \frac{\lambda_o - \lambda_e}{\lambda_e}$ ($\lambda_o$ is the observed wavelength, $\lambda_e$ is the wavelength at emission), so given that $\epsilon$ is very small,
$$\lambda_o \approx \frac{\lambda_e}{\sqrt{\epsilon}}$$
In other words, by emitting a photon close enough to the event horizon, you can arrange for it to be redshifted to as large a positive wavelength (and thus as small a frequency) as you want. But there's no value of $\epsilon$ that will actually give you an infinite wavelength (zero frequency).
If you were to go all the way down to $\epsilon = 0$, i.e. emit the photon from right on the event horizon, then it would just be stuck there, since there are no outgoing null geodesics.
*Everything here applies only to Schwarzschild black holes, although I don't know of anything that would allow a photon to be redshifted to infinite wavelength from any other kind of black hole either.
The classical electromagnetic waves may have many forms. The simplest examples are "monochromatic" waves with a well-defined frequency $f$ – the number of periods per second. The wavelength – the distance between two maxima of the wave – is $\lambda=c/f$.
When you change something about the electromagnetic field at one point, i.e. when you try to modify an electromagnetic wave, you can't instantly modify anything at a distant place. The information and influence can never propagate faster than $c$, the speed of light in the vacuum. This is a completely general fact about Nature that follows from the special theory of relativity and it applies to quantum theory, too. If you start with a wave of a huge wavelength $L$ and modify quickly something about the wave in a small region and in a timescale much shorter than the period of the electromagnetic wave, in a hope that this allows you to "instantly" affect distant points of space, you will fail. Instead of making a long electromagnetic wave collaborate on your project, you will create some electromagnetic waves of a much higher frequency and shorter wavelength. Too long electromagnetic waves just don't allow you to make things "really quickly" – in times shorter than $T$, their period – and too locally – in regions much shorter than the wavelength. Whenever the resolution of time or space is much better, it just proves that some electromagnetic waves of much higher frequency and shorter wavelength are present.
To describe the waves in the microwave oven, it is enough to consider classical physics i.e. avoid the notion of photons. The microwaves – electromagnetic waves whose wavelength is just a little bit shorter than the size of the oven – have a negligible chance to get through the holes because these holes are much smaller than the wavelength. The mechanism is sometimes called electromagnetic shielding. How does it work?
Well, if you want to discuss waves of wavelength $\lambda$ only, their dependence on space must always have the form $A\cdot \cos(2\pi x / \lambda)$: waves with the right distance between the minima. However, the metallic cage surrounding the interior of the oven imposes the potential $\phi=0$ at a very dense network of points. When you try to write down the potential as $A\cdot \cos(2\pi x / \lambda)$, while making sure that it's zero at all points where there is a conductor, you will find out that there is no solution except for $A=0$. The wave of the given wavelength just can't get through at all. Alternatively, you could calculate the reflection from the metallic points of the mesh (not counting the holes). They would interfere with each other and guarantee that the probability of reflection is nearly 100 percent.
In other words, the microwave-frequency photon is "really large", at least as large as the wavelength, and it just "doesn't fit in". It could fit in if it "pretended" it was a shorter-wavelength photon but that would be a different one. The microwave oven prescribes a frequency $f$ and the corresponding wavelength is always $c/f$ and cannot change. You may also look at the situation with a "poor resolution" so that you "neglect" distances shorter than the wavelength. From this viewpoint, the metallic caging of the oven looks solid despite the small holes. That's ultimately why it looks solid to the electromagnetic waves.
All these things may also be phrased in terms of photons – which is needed in quantum theory – even though, as I said, it is not necessary for microwaves because they contain a huge number of coherent photons so that this large group of photons behaves classically.
Individual photons are described by wave functions that mathematically look like an electromagnetic wave. It may be monochromatic but it may be a mixture of different frequencies, too. Individual photons' wave functions propagate through the oven or anything else pretty much just like the classical electromagnetic waves. That's why photons of microwave frequencies can't escape from the microwave, either. But the interpretation is different: the wave function isn't directly measurable like the electric fields. Instead, it encodes (after squaring) the probability density that the photon is here or there.
If you try to catch a photon whose wave function is spread over a large region, you have a certain calculable chance that you will catch it "somewhere here", in a particular nearby region $V$. But when you do so, the probability is zero that the photon is simultaneously at some other, distance place $P$. So whatever you do to the photon here – if you are lucky at all and the photon "shows up" here – will not affect what is happening very far from you. Effectively, if you see the photon here, the "wave function collapses" and you're sure that the photon isn't "over there" so it can't do anything there. The photon was never over there; it only had a chance to be there but your measurement showed that the chance didn't materialize. These points are often misunderstood because people are trying to imagine that the wave function of the photon is a real wave that must leave some traces even if the photon is ultimately seen elsewhere. But it can't and it doesn't leave any traces whatsoever: the nonzero wave function only quantified a "potential" that something may be seen somewhere. When it's seen elsewhere, it's a proof that the potential was never "real".
At any rate, you will see that photons can't be used to send superluminal signals, either.
Best Answer
The relationship between energy and wavelength:
$$E = h f = \frac{h c}{\lambda}$$
As $\lambda$ goes to zero, $E$ goes to infinity.
So "no".