Why isn't there a potential difference across a disconnected diode?

I think, the answer is relatively simple. Do you know the working principle of a "Schottky diode", which is based on a semiconductor-metal junction? Now - what happens if you connect a voltmeter (or any other load) across the diode? You create two Schottky junctions which exactly compensate the diffusion voltage inside the pn diode. Thus, no voltage can be measured. With other words: You cannot use the diffusion voltage to drive any current through an external load.

Err, the rest of the answers seem a little dodgy and I just stumbled upon this question so I'll take a shot at it.

I think it's because of the fact that the Fermi level becomes discontinuous under bias. I'm sure you can visualise that what the voltmeter is really measuring is how badly electrons and holes want to cross the junction. At thermal equilibrium, the electrons and holes have no intention of moving across the junction, so the voltage is 0V. IN other words, the voltmeter really only measures the difference in the Fermi levels between the 2 sides.

To understand why it does this, you have to know how a voltmeter works. Rather than literally measuring the difference in the energy level of an electron at both ends of the diode (which would be awesome), it just measures the current flowing through its high resistance. In a diode at thermal equilibrium, there's no net movement of any charge carriers and so there's no current. No current means no voltmeter reading.

It is a very nice curiosity question! Same question came up to me when I was in my second year. But until I came across the Threshold voltages in Transistors and PN junction voltage drops, the picture became little clear.

You are absolutely right (last paragraph), because there is a change in potential due to electric field in the depletion region, there is higher potential from n-type side and negative potential from p-type side, making the intrinsic potential difference built up. That is why, to allow the current to flow through diode (PN junction) you would need higher potential from P-type and n-Type such that their difference is larger than the intrinsic potential difference which is in opposite direction to applied voltage across diode. This is what we call forward biased diode! I am sure you know this basics. Now lets go to the real question ->

If you were to probe your virtual Digital voltmeter exactly at the two depletion boundaries then I am sure you would see the voltage difference there, but its quite impossible to do with the regular multimeter. I am sure there are ways that semiconductor companies have special probes to sense these voltage differences. But if you were to measure the disconnected diode from your regular multimeter (same this is taken in consideration when you simulate it in LTSPICE that the probing is done at the ends of the diode not internally). Basically, your Graph (D) has this answer it self. Graph shows that both ends of diode have no electric field present. since the Electric field is conservative, and two diode ends (ends of P and N type materials) have no charge and electric fields at the ends are cancelled due to diffusion, as a results there is no electric field present at after the diffusion region ends, that means their difference is also 0 and measured voltage difference is 0 V as well. Hope this helps!