This post is part of our journey to explore Quantum Field Theory. Last week I introduced the idea of a Probability Distribution (PD).
“https://platitudes.home.blog/2024/06/08/the-remarkable-theory-of-quantum-fields/“
This week I’m going to take this a little bit further and start using the notation used in Quantum Mechanics (QM).
Quantum Mechanics is the theory of probability, with one small, but extremely significant twist.
Take a very simple example of the roll of a six sided die. There are six possible outcomes. While the die is still rolling, there is an equal probability that any of the six sides will appear face up. Each has a 1 in 6 chance, or a probability of one sixth, 1/6. So while in this state, the Probability Distribution (PD) looks like this.
We can represent this as a sequence of numbers where we just list the probabilities for each possible outcome, (1/6,1/6,1/6,1/6,1/6,1/6). And to save some ink, I’ll give this a shorthand name. I’ll write it as |e>, where the letter “e” stands for “equal probability”. The funny brackets are conventional in QM.
|e> = (1/6,1/6,1/6,1/6,1/6,1/6) “e” for equal
Now suppose the die comes to rest with the number 4 facing up. The PD has now changed.
A 4 is now certain, it has a probability of 1. If we don’t do anything to the die it will just sit there, continuing to display a 4. Again, we can write this as a bunch of numbers, and give that a name.
|4> = (0,0,0,1,0,0)
This means there is zero probability for the faces 1,2,3,5 and 6, and a probability if 1, certainty, for the value 4. Similarly, if the number 6 ends up face up, then we can represent it’s PD as
|6> = (0,0,0,0,0,1)
If we want to represent any one of the above six then we write
|n> = ( 1 in the nth position and zero everywhere else).
How to Compare States
If we look at the two states |4> and |6>, we see that they have nothing in common.
|4> = (0,0,0,1,0,0)
|6> = (0,0,0,0,0,1)
The only number that is non-zero in |4> is in the fourth position, but the fourth position in |6> is a zero.
Similarly, the only number that is non-zero in |6> is in the sixth position, but the sixth position in |4> is a zero.
We say that the “overlap” between the two states is zero. As a shorthand, this is written as.
<4|6> = 0
As these are PDs, what we’re really saying is that, if the die is in the state |6>, then the probability of it being in the state |4> is zero.
Any PD always perfectly overlaps with itself < i | i > = 1. The probability of being in state i, given that it’s already in state i, is exactly 1, certainty.
Now compare the state |4> with the equal probability state |e>
|4> = (0,0,0,1,0,0)
|e> = (1/6,1/6,1/6,1/6,1/6,1/6)
Now there is some overlap between the two states. The “1” in position four in |4> matches a “1/6” in position four in |e>. If the die is in the |e> state then there is a non-zero probability that it will end up in |4>.
The probability that we’ll end up in the |4> state, given that we’re in the equal probabilty state is written as <4|e> and has the value 1/6.
(For those with some maths, this is just the inner product of two vectors |i> and |j>. It is P( i | j ), sort of, see below.)
And the twist? Well that’s just a little bit more mathematical, so I’ve added it as a comment.
Platitude of the Day 
This is just a little bit more mathematical than the main posts, so I’ve done it as an optional comment. You don’t have to read this bit. If you do read it, you don’t necessarily have to understand all of it. Most of the discussion on Quantum Field Theory won’t rely on it. However if you want to see why QM is so weird then you might find it interesting.
Trust me, this is as complicated as the maths gets. If you can get through this bit, then you’ll be able to get through the whole thing no problem.
The way to calculate the overlap is to mutliply corresponding components and add the results. So
<4|4> = 0x0 + 0x0 + 0x0 + 1×1 + 0x0 + 0x0 = 1
<4|6> = 0x0 + 0x0 + 0x0 + 1×0 + 0x0 + 0x1 = 0
But
<e | e> = 1/6×1/6 + 1/6×1/6 + 1/6×1/6 + 1/6×1/6 + 1/6×1/6 + 1/6×1/6
= 6/36
= 1/6.
Now I’ve just said that <e | e>= 1, whereas the formula I just gave gives 1/6. The solution to this is to deal with the square roots of PDs rather than the PDs themselves. In QM these are called Probability Amplitudes (PAs).
PDs like |6> = (0,0,0,0,0,1) remain as they are as the square root of 1 is 1. However our equal probability state changes. It is now
|e> = (1/√6,1/√6,1/√6,1/√6,1/√6,1/√6).
This now gives the correct overlap with itself
<e | e> = 1/√6×1/√6 + 1/√6×1/√6 + 1/√6×1/√6 + 1/√6×1/√6 + 1/√6×1/√6 + 1/√6×1/√6
= 6/6
= 1
If we want to calculate the probability of the “4” side coming face up, given that the die is still rolling, then according to our comparison rule, this is
<4|e> = 0 x 1/√6 + 0 x 1/√6 + 0 x 1/√6 + 1 x 1/√6 + 0 x 1/√6 + 0 x 1/√6
= 1/√6
Which is clearly wrong. The probability of throwing a “1” is 1/6, not 1/√6. That’s because we took the square root of all the probabilities in the PD. To get the right result we have to take the overlap and then square it.
<1|e>² = (1/(√6))² = 1/6, which is finally correct.
It is the need to use Probability Amplitudes, i.e. the square roots of Probability Distributions, that leeds to all of the weirdness of QM. For example, consider this modified version of |4>, where the amplitude is a negative number.
|-4> = (0,0,0,-1,0,0)
This has the same probability as |4>. <-4|e>² = (-1 x 1/√6 )² = 1/6.
It’s got just as good a claim to represent the state of “4” face up as |4> has. When you get two alternatives like this, QM allows you to add the Probability Amplitudes.
But |4> + |-4> = (0,0,0,1,0,0) + (0,0,0,-1,0,0) = (0,0,0,0,0,0)
The die has vanished!
It is the addition of Probability Amplitudes that leads to interference effects in QM.
For example, in the famous double slit experiment, a photon of light has a choice of going through two different slits. The Probability Amplitude of the photon varies with time, and the time to travel through one hole and on to the screen, differs from the time taken to travel through the other hole. So the amplitudes at a given point on the screen can vary. At some points they reinforce one another, but at others they completely cancel and no photons ever arrive there.
Most of the time I’m not going to bother making the distinction between amplitudes and probabilities, unless it’s important to the argument. I just thought you’d like to see where all the weirdness in QM comes from.
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Thanks for the post and the supplementary comment! Bits of my A-level physics are starting to swim back into view. Looking forward very much to the next instalment.
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