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So let's say we have the three base states $\lvert + \rangle$, $\lvert 0\rangle$, $\lvert - \rangle$. A general state is then: $$\lvert \psi \rangle = c_+ \lvert + \rangle + c_0 \lvert 0\rangle + c_- \lvert - \rangle$$ where, for normalization, $c_+^2 + c_0^2 + c_- ^2 = 1$. You can clearly make an infinite number of choices for the $c_i$s, and so there are an infinite number of states that you can make from these base states.

It is important to note that in quantum mechanics, it is of physical significance not only which base states are combined, but exactly how they are combined (i.e. the coefficients). Apart from the information of probabilities, the coefficients also give what the system looks like in another base, and so the probabilities for different bases. And each combination of coefficients produces a completely different physical state.

So yes, instead of specifying infinite $\lvert \psi \rangle$s, we can simply specify the action of $A$ on the base states. This requires us to know only 3 vectors (or equivalently, nine numbers - the expression of each of the 3 vectors in the basis): $A\lvert + \rangle$, $A\lvert 0 \rangle$, $A\lvert - \rangle$ (correspondingly, all the $\langle j \rvert A\lvert i \rangle$).Thanks to the linearity of quantum mechanics, the action of $A$ on $\lvert \psi \rangle$ is then simply: $$A\lvert \psi \rangle = c_+ A\lvert + \rangle + c_0 A\lvert 0\rangle + c_- A\lvert - \rangle$$ There you have it - a complete description of $A$ would do to a general state (which is one among an infinite number of possible states), in terms of nine numbers, all made possible by expanding in the basis.

To recap: expanding the general state in a basis helps us to explain the action of the "appartus" on the general state, simply in terms of the action on each vector in the basis, which is much easier to keep track of.

This is a lot like writing the dot product $\mathbf{a}\cdot\mathbf{b}$ for 3D vectors $$\mathbf{a} = a_x\mathbf{e_x} + a_y\mathbf{e_y} + a_z\mathbf{e_z}$$ $$\mathbf{b} = b_x\mathbf{e_x} + b_y\mathbf{e_y} + b_z\mathbf{e_z}$$ as: $$\mathbf{a}\cdot\mathbf{b} = a_x b_x + a_y b_y + a_z b_z$$ just using linearity and the extremely simple known products: $$\mathbf{e_i} \cdot \mathbf{e_j} = \delta_{ij}$$ Without these, we'll have to specify $\mathbf{a}\cdot\mathbf{b}$ for any pair of vectors, and there is an infinite number of such pairs.

So let's say we have the three base states $\lvert + \rangle$, $\lvert 0\rangle$, $\lvert - \rangle$. A general state is then: $$\lvert \psi \rangle = c_+ \lvert + \rangle + c_0 \lvert 0\rangle + c_- \lvert - \rangle$$ where, for normalization, $\lvert c_+\rvert^2 + \lvert c_0\rvert^2 + \lvert c_-\rvert^2 = 1$. You can clearly make an infinite number of choices for the $c_i$s, and so there are an infinite number of states that you can make from these base states.

It is important to note that in quantum mechanics, it is of physical significance not only which base states are combined, but exactly how they are combined (i.e. the coefficients). Apart from the information of probabilities, the coefficients also give what the system looks like in another base, and so the probabilities for different bases. And each combination of coefficients produces a completely different physical state.

So yes, instead of specifying infinite $\lvert \psi \rangle$s, we can simply specify the action of $A$ on the base states. This requires us to know only 3 vectors (or equivalently, nine numbers - the expression of each of the 3 vectors in the basis): $A\lvert + \rangle$, $A\lvert 0 \rangle$, $A\lvert - \rangle$ (correspondingly, all the $\langle j \rvert A\lvert i \rangle$).Thanks to the linearity of quantum mechanics, the action of $A$ on $\lvert \psi \rangle$ is then simply: $$A\lvert \psi \rangle = c_+ A\lvert + \rangle + c_0 A\lvert 0\rangle + c_- A\lvert - \rangle$$ There you have it - a complete description of $A$ would do to a general state (which is one among an infinite number of possible states), in terms of nine numbers, all made possible by expanding in the basis.

To recap: expanding the general state in a basis helps us to explain the action of the "appartus" on the general state, simply in terms of the action on each vector in the basis, which is much easier to keep track of.

This is a lot like writing the dot product $\mathbf{a}\cdot\mathbf{b}$ for 3D vectors $$\mathbf{a} = a_x\mathbf{e_x} + a_y\mathbf{e_y} + a_z\mathbf{e_z}$$ $$\mathbf{b} = b_x\mathbf{e_x} + b_y\mathbf{e_y} + b_z\mathbf{e_z}$$ as: $$\mathbf{a}\cdot\mathbf{b} = a_x b_x + a_y b_y + a_z b_z$$ just using linearity and the extremely simple known products: $$\mathbf{e_i} \cdot \mathbf{e_j} = \delta_{ij}$$ Without these, we'll have to specify $\mathbf{a}\cdot\mathbf{b}$ for any pair of vectors, and there is an infinite number of such pairs.

So let's say we have the three base states $\lvert + \rangle$, $\lvert 0\rangle$, $\lvert - \rangle$. A general state is then: $$\lvert \psi \rangle = c_+ \lvert + \rangle + c_0 \lvert 0\rangle + c_- \lvert - \rangle$$ where, for normalization, $c_+^2 + c_0^2 + c_- ^2 = 1$. You can clearly make an infinite number of choices for the $c_i$s, and so there are an infinite number of states that you can make from these base states.

It is important to note that in quantum mechanics, it is of physical significance not only which base states are combined, but exactly how they are combined (i.e. the coefficients). Apart from the information of probabilities, the coefficients also give what the system looks like in another base, and so the probabilities for different bases. And each combination of coefficients produces a completely different physical state.

So yes, instead of specifying infinite $\lvert \psi \rangle$s, we can simply specify the action of $A$ on the base states. This requires us to know only 3 vectors (or equivalently, nine numbers - the expression of each of the 3 vectors in the basis): $A\lvert + \rangle$, $A\lvert 0 \rangle$, $A\lvert - \rangle$ (correspondingly, all the $\langle i \rvert A\lvert j \rangle$).Thanks to the linearity of quantum mechanics, the action of $A$ on $\lvert \psi \rangle$ is then simply: $$A\lvert \psi \rangle = c_+ A\lvert + \rangle + c_0 A\lvert 0\rangle + c_- A\lvert - \rangle$$ There you have it - a complete description of $A$ would do to a general state (which is one among an infinite number of possible states), in terms of nine numbers, all made possible by expanding in the basis.

To recap: expanding the general state in a basis helps us to explain the action of the "appartus" on the general state, simply in terms of the action on each vector in the basis, which is much easier to keep track of.

This is a lot like writing the dot product $\mathbf{a}\cdot\mathbf{b}$ for 3D vectors $$\mathbf{a} = a_x\mathbf{e_x} + a_y\mathbf{e_y} + a_z\mathbf{e_z}$$ $$\mathbf{b} = b_x\mathbf{e_x} + b_y\mathbf{e_y} + b_z\mathbf{e_z}$$ as: $$\mathbf{a}\cdot\mathbf{b} = a_x b_x + a_y b_y + a_z b_z$$ just using linearity and the extremely simple known products: $$\mathbf{e_i} \cdot \mathbf{e_j} = \delta_{ij}$$ Without these, we'll have to specify $\mathbf{a}\cdot\mathbf{b}$ for any pair of vectors, and there is an infinite number of such pairs.

So let's say we have the three base states $\lvert + \rangle$, $\lvert 0\rangle$, $\lvert - \rangle$. A general state is then: $$\lvert \psi \rangle = c_+ \lvert + \rangle + c_0 \lvert 0\rangle + c_- \lvert - \rangle$$ where, for normalization, $c_+^2 + c_0^2 + c_- ^2 = 1$. You can clearly make an infinite number of choices for the $c_i$s, and so there are an infinite number of states that you can make from these base states.

It is important to note that in quantum mechanics, it is of physical significance not only which base states are combined, but exactly how they are combined (i.e. the coefficients). Apart from the information of probabilities, the coefficients also give what the system looks like in another base, and so the probabilities for different bases. And each combination of coefficients produces a completely different physical state.

So yes, instead of specifying infinite $\lvert \psi \rangle$s, we can simply specify the action of $A$ on the base states. This requires us to know only 3 vectors (or equivalently, nine numbers - the expression of each of the 3 vectors in the basis): $A\lvert + \rangle$, $A\lvert 0 \rangle$, $A\lvert - \rangle$ (correspondingly, all the $\langle j \rvert A\lvert i \rangle$).Thanks to the linearity of quantum mechanics, the action of $A$ on $\lvert \psi \rangle$ is then simply: $$A\lvert \psi \rangle = c_+ A\lvert + \rangle + c_0 A\lvert 0\rangle + c_- A\lvert - \rangle$$ There you have it - a complete description of $A$ would do to a general state (which is one among an infinite number of possible states), in terms of nine numbers, all made possible by expanding in the basis.

To recap: expanding the general state in a basis helps us to explain the action of the "appartus" on the general state, simply in terms of the action on each vector in the basis, which is much easier to keep track of.

This is a lot like writing the dot product $\mathbf{a}\cdot\mathbf{b}$ for 3D vectors $$\mathbf{a} = a_x\mathbf{e_x} + a_y\mathbf{e_y} + a_z\mathbf{e_z}$$ $$\mathbf{b} = b_x\mathbf{e_x} + b_y\mathbf{e_y} + b_z\mathbf{e_z}$$ as: $$\mathbf{a}\cdot\mathbf{b} = a_x b_x + a_y b_y + a_z b_z$$ just using linearity and the extremely simple known products: $$\mathbf{e_i} \cdot \mathbf{e_j} = \delta_{ij}$$ Without these, we'll have to specify $\mathbf{a}\cdot\mathbf{b}$ for any pair of vectors, and there is an infinite number of such pairs.

So let's say we have the three base states $\lvert + \rangle$, $\lvert 0\rangle$, $\lvert - \rangle$. A general state is then: $$\lvert \psi \rangle = c_+ \lvert + \rangle + c_0 \lvert 0\rangle + c_- \lvert - \rangle$$ where, for normalization, $c_+^2 + c_0^2 + c_-^2 = 1$. You can clearly make an infinite number of choices for the $c_i$s, and so there are an infinite number of states that you can make from these base states.

It is important to note that in quantum mechanics, it is of physical significance not only which base states are combined, but exactly how they are combined (i.e. the coefficients). Apart from the information of probabilities, the coefficients also give what the system looks like in another base, and so the probabilities for different bases. And each combination of coefficients produces a completely different physical state.

So yes, instead of specifying infinite $\lvert \psi \rangle$s, we can simply specify the action of $A$ on the base states. This requires us to know only 3 vectors (or equivalently, nine numbers - the expression of each of the 3 vectors in the basis): $A\lvert + \rangle$, $A\lvert 0 \rangle$, $A\lvert - \rangle$ (correspondingly, all the $\langle i \rvert A\lvert j \rangle$).Thanks to the linearity of quantum mechanics, the action of $A$ on $\lvert \psi \rangle$ is then simply: $$A\lvert \psi \rangle = c_+ A\lvert + \rangle + c_0 A\lvert 0\rangle + c_- A\lvert - \rangle$$ There you have it - a complete description of $A$ would do to a general state (which is one among an infinite number of possible states), in terms of nine numbers, all made possible by expanding in the basis.

To recap: expanding the general state in a basis helps us to explain the action of the "appartus" on the general state, simply in terms of the action on each vector in the basis, which is much easier to keep track of.

This is a lot like writing the dot product $\mathbf{a}\cdot\mathbf{b}$ for 3D vectors $$\mathbf{a} = a_x\mathbf{e_x} + a_y\mathbf{e_y} + a_z\mathbf{e_z}$$ $$\mathbf{b} = b_x\mathbf{e_x} + b_y\mathbf{e_y} + b_z\mathbf{e_z}$$ as: $$\mathbf{a}\cdot\mathbf{b} = a_x b_x + a_y b_y + a_z b_z$$ just using linearity and the extremely simple known products: $$\mathbf{e_i} \cdot \mathbf{e_j} = \delta_{ij}$$ Without these, we'll have to specify $\mathbf{a}\cdot\mathbf{b}$ for any pair of vectors, and there is an infinite number of such pairs.

So let's say we have the three base states $\lvert + \rangle$, $\lvert 0\rangle$, $\lvert - \rangle$. A general state is then: $$\lvert \psi \rangle = c_+ \lvert + \rangle + c_0 \lvert 0\rangle + c_- \lvert - \rangle$$ where, for normalization, $c_+^2 + c_0^2 + c_- ^2 = 1$. You can clearly make an infinite number of choices for the $c_i$s, and so there are an infinite number of states that you can make from these base states.

It is important to note that in quantum mechanics, it is of physical significance not only which base states are combined, but exactly how they are combined (i.e. the coefficients). Apart from the information of probabilities, the coefficients also give what the system looks like in another base, and so the probabilities for different bases. And each combination of coefficients produces a completely different physical state.

So yes, instead of specifying infinite $\lvert \psi \rangle$s, we can simply specify the action of $A$ on the base states. This requires us to know only 3 vectors (or equivalently, nine numbers - the expression of each of the 3 vectors in the basis): $A\lvert + \rangle$, $A\lvert 0 \rangle$, $A\lvert - \rangle$ (correspondingly, all the $\langle i \rvert A\lvert j \rangle$).Thanks to the linearity of quantum mechanics, the action of $A$ on $\lvert \psi \rangle$ is then simply: $$A\lvert \psi \rangle = c_+ A\lvert + \rangle + c_0 A\lvert 0\rangle + c_- A\lvert - \rangle$$ There you have it - a complete description of $A$ would do to a general state (which is one among an infinite number of possible states), in terms of nine numbers, all made possible by expanding in the basis.

To recap: expanding the general state in a basis helps us to explain the action of the "appartus" on the general state, simply in terms of the action on each vector in the basis, which is much easier to keep track of.

This is a lot like writing the dot product $\mathbf{a}\cdot\mathbf{b}$ for 3D vectors $$\mathbf{a} = a_x\mathbf{e_x} + a_y\mathbf{e_y} + a_z\mathbf{e_z}$$ $$\mathbf{b} = b_x\mathbf{e_x} + b_y\mathbf{e_y} + b_z\mathbf{e_z}$$ as: $$\mathbf{a}\cdot\mathbf{b} = a_x b_x + a_y b_y + a_z b_z$$ just using linearity and the extremely simple known products: $$\mathbf{e_i} \cdot \mathbf{e_j} = \delta_{ij}$$ Without these, we'll have to specify $\mathbf{a}\cdot\mathbf{b}$ for any pair of vectors, and there is an infinite number of such pairs.