10 ms·

I like that quote myself, but one thing I don't understand is - are the amplitudes always of a magnitude 1? Because if qbits are really always the same magnitud

I like that quote myself, but one thing I don't understand is - are the amplitudes always of a magnitude 1? Because if qbits are really always the same magnitude, why represent them as two-dimensional? Wouldn't the math be easier if we converted it to polar form and just had one measurement per superposition? Sorry if this is a dumb question, I'm not formally trained in any of this stuff...

drdeca 8y ago
There are 2 complex numbers, one of them the coefficient of |0> and |1>. The sum of the square of the magnitudes of these numbers is 1. Alternatively/equivalently , the magnitude of the vector taken as a whole, is 1. So if you have (a+bi)|0> + (c+di)|1>, the magnitude of this vector is a^2 + b^2 + c^2 + d^2 = 1 Edit: however, the vector only has to have magnitude 1 if it is describing the entire system, I think?
millstone 8y ago
It's a good question. The cartoon says that qubits are "unit vectors in 2D Hilbert space." However because the length is 1, you can think of it as a vector in 2D projective Hilbert space. Projective space is what you get when you throw away a vector's length and only talk about its angles. Some intuition: if you have a particle and you look everywhere for it, your chance of finding it must be 100%. It can't be 50% (where could it be?) and it can't be 200% (duh). QM in one dimension is formulated as "a 1d complex unit vector", i.e. a complex number of magnitude 1, which is routinely represented in polar form (a complex exponential). QM in two dimensions is a "2d complex unit vector" which is really 3d: you get three linearly independent components, and then the fourth one is decided for you. You can think of this as spherical coordinates, but complex numbers are much easier to manipulate, so that's what we do.

10 ms·

I like that quote myself, but one thing I don't understand is - are the amplitudes always of a magnitude 1? Because if qbits are really always the same magnitud

by SuoDuanDao 8y ago

drdeca 8y ago
There are 2 complex numbers, one of them the coefficient of |0> and |1>. The sum of the square of the magnitudes of these numbers is 1. Alternatively/equivalently , the magnitude of the vector taken as a whole, is 1. So if you have (a+bi)|0> + (c+di)|1>, the magnitude of this vector is a^2 + b^2 + c^2 + d^2 = 1 Edit: however, the vector only has to have magnitude 1 if it is describing the entire system, I think?
millstone 8y ago
It's a good question. The cartoon says that qubits are "unit vectors in 2D Hilbert space." However because the length is 1, you can think of it as a vector in 2D projective Hilbert space. Projective space is what you get when you throw away a vector's length and only talk about its angles. Some intuition: if you have a particle and you look everywhere for it, your chance of finding it must be 100%. It can't be 50% (where could it be?) and it can't be 200% (duh). QM in one dimension is formulated as "a 1d complex unit vector", i.e. a complex number of magnitude 1, which is routinely represented in polar form (a complex exponential). QM in two dimensions is a "2d complex unit vector" which is really 3d: you get three linearly independent components, and then the fourth one is decided for you. You can think of this as spherical coordinates, but complex numbers are much easier to manipulate, so that's what we do.