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(Maybe this should go in philosophy? Feel free to move it.)

I'm confused.

Well, that's probably because I haven't really learned QM properly, except for the basics of Quantum Computing.

But still, I'm confused.

Why is measurement modeled as a projection, or a "collapse" of a wavefunction? Wouldn't it make more sense to incorporate the measuring device into the quantum state?

My perspective is probably limited because of how little I know... but when we went over Quantum Computing in class, it struck me that many quantum gates act like measurements. For simplicity, let's look at a CNOT gate specifically:

As a reminder, a CNOT gate is an operation on two bits that acts as follows on the basis states:

[tex]\mathrm{CNOT}|x>\otimes|y> = |x>\otimes|x \oplus y>[/tex]

In particular, if we prepare |y> to be |0>, then we have the following action on the basis states:

[tex]\mathrm{CNOT}|x>\otimes|0> = |x>\otimes|x>[/tex]

This strikes me as looking very much like a measurement -- we took the value of one qubit and "stored" it in another qubit.

The CNOT gate, incidentally, can also be used as a comparator: observe that:

[tex]\mathrm{CNOT}|0>\otimes|0> = |0>\otimes|0>[/tex]

[tex]\mathrm{CNOT}|1>\otimes|1> = |1>\otimes|0>[/tex]

[tex]\mathrm{CNOT}|0>\otimes|1> = |0>\otimes|1>[/tex]

[tex]\mathrm{CNOT}|1>\otimes|0> = |0>\otimes|1>[/tex]

So we take two bits in, and when we apply CNOT, the second bit tells us if they were equal or not.

We can even observe it multiple times, and compare observations. For example, take the following quantum program:

Take the qubit x as input.

Let y = |0>

Let z = |0>

"measure" x by applying CNOT to (x,y)

"measure" x by applying CNOT to (x,z)

"compare" y to z by applying CNOT to (y, z)

Return z

Of course, by this, I mean you start with [itex]|x>\otimes|0>\otimes|0>[/itex], then apply CNOT to the first two, then to the outer two, then to the last two. After applying these operations, the final qubit will always be zero... the two measurements are always the same!

If we model real-world measurements in this way, it seems weirdness goes away. Sure, we'd be doomed to live in a world where Schrödinger's cat is perpetually both dead and alive, but when we compare any observations, they will always be consistent.

Other weirdness goes away too -- I did some scratching and it would appear that there's no problem measuring the spin state in both the x and y directions. Incidentally, what little I've done feels awfully classical in this perspective, which is nice.

I know I have work to do... the next thing I guess is try to really understand Bell's theorem, and show that this model of measurement behaves properly -- I guess I need to find a program that looks like the thought experiment and show that the "right" answer falls out? I have a sinking feeling that it's not going to, though... which means I'm going to be tormented until I understand what's going wrong.

EDIT: Fixed a mistake in the definition of CNOT

I'm confused.

Well, that's probably because I haven't really learned QM properly, except for the basics of Quantum Computing.

But still, I'm confused.

Why is measurement modeled as a projection, or a "collapse" of a wavefunction? Wouldn't it make more sense to incorporate the measuring device into the quantum state?

My perspective is probably limited because of how little I know... but when we went over Quantum Computing in class, it struck me that many quantum gates act like measurements. For simplicity, let's look at a CNOT gate specifically:

As a reminder, a CNOT gate is an operation on two bits that acts as follows on the basis states:

[tex]\mathrm{CNOT}|x>\otimes|y> = |x>\otimes|x \oplus y>[/tex]

In particular, if we prepare |y> to be |0>, then we have the following action on the basis states:

[tex]\mathrm{CNOT}|x>\otimes|0> = |x>\otimes|x>[/tex]

This strikes me as looking very much like a measurement -- we took the value of one qubit and "stored" it in another qubit.

The CNOT gate, incidentally, can also be used as a comparator: observe that:

[tex]\mathrm{CNOT}|0>\otimes|0> = |0>\otimes|0>[/tex]

[tex]\mathrm{CNOT}|1>\otimes|1> = |1>\otimes|0>[/tex]

[tex]\mathrm{CNOT}|0>\otimes|1> = |0>\otimes|1>[/tex]

[tex]\mathrm{CNOT}|1>\otimes|0> = |0>\otimes|1>[/tex]

So we take two bits in, and when we apply CNOT, the second bit tells us if they were equal or not.

We can even observe it multiple times, and compare observations. For example, take the following quantum program:

Take the qubit x as input.

Let y = |0>

Let z = |0>

"measure" x by applying CNOT to (x,y)

"measure" x by applying CNOT to (x,z)

"compare" y to z by applying CNOT to (y, z)

Return z

Of course, by this, I mean you start with [itex]|x>\otimes|0>\otimes|0>[/itex], then apply CNOT to the first two, then to the outer two, then to the last two. After applying these operations, the final qubit will always be zero... the two measurements are always the same!

If we model real-world measurements in this way, it seems weirdness goes away. Sure, we'd be doomed to live in a world where Schrödinger's cat is perpetually both dead and alive, but when we compare any observations, they will always be consistent.

Other weirdness goes away too -- I did some scratching and it would appear that there's no problem measuring the spin state in both the x and y directions. Incidentally, what little I've done feels awfully classical in this perspective, which is nice.

I know I have work to do... the next thing I guess is try to really understand Bell's theorem, and show that this model of measurement behaves properly -- I guess I need to find a program that looks like the thought experiment and show that the "right" answer falls out? I have a sinking feeling that it's not going to, though... which means I'm going to be tormented until I understand what's going wrong.

EDIT: Fixed a mistake in the definition of CNOT

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