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I am currently reading Quantum Mechanics The Theoretical Minimum by Leonard Susskind. In the second lecture he says that for a given state of a spin $|A\rangle = a|u\rangle + b|d\rangle$: The probability for the spin to be in the $|u\rangle$ state is: $a^*a$ (where $a^*$ is the complex conjugate of $a$), and $a^*a = \langle A|u\rangle\langle u|A\rangle$. My question is, can we rewrite the last equation like this: $\langle A|A\rangle\langle u|u\rangle$? And if the answer is no, then why?

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    $\begingroup$ Note that both $\langle u|u\rangle$ and $\langle A|A\rangle$ are equal to $1$ as $|u\rangle$ and $|A\rangle$ are unit vectors, so the quantity you wrote is always equal to $1$. $\endgroup$
    – Tristan Nemoz
    Aug 25, 2023 at 13:32

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We can see that $\langle A|u\rangle\langle u|A\rangle \neq \langle A|A\rangle\langle u|u\rangle$ with a simple example.

Let $|A\rangle = 1/\sqrt{2}|u\rangle + 1/\sqrt{2}|d\rangle$. With quick matrix multiplication we see: \begin{equation} \langle A|u\rangle\langle u|A\rangle = (1\sqrt{2})^2 = 1/2 = |a|^2 \end{equation} as expected. Now, we can just see by inspection that $\langle A|A\rangle\langle u|u\rangle = 1$.

Intuitively, we can see this is not equivalent because the inner product between two states $|\psi\rangle$ and $|\phi\rangle$ is defined as $\langle \psi | \phi \rangle$, and this represents the probability amplitude for the state $|\phi\rangle$ to be measured as the state $|\psi\rangle$ (the reason as to why we can see just from inspection that $\langle u|u\rangle = 1$).

Mathematically, we know that this is not equivalent is because matrix multiplication is non-commutative. This means that when multiplying matrices, we can't just switch their order as we do with numbers. Hence, these expressions are not the same.

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  • $\begingroup$ Another question: Can we multiply two ket vectors or it needs to be a ket with a bra vector? $\endgroup$
    – zizaaooo
    Aug 25, 2023 at 14:09
  • $\begingroup$ You take the tensor product of ket vectors, which is effectively multiplication but subtly different. See this question on the physics SE for a full answer. $\endgroup$
    – banercat
    Aug 25, 2023 at 14:19
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The inner product is not commutative because in general, $\langle u | v \rangle = \langle v | u \rangle^*$, where * denotes the complex conjugate.

It doesn't even really make sense to discuss whether or not it is associative, because the associative property is only defined for (closed) binary operations on a magma. But the inner product operation maps elements from one set (the Hilbert space of state vectors) to a completely different set (the complex numbers $\mathbb{C}$), so it isn't a binary operation. The notion of associativity simply doesn't apply.

(Strictly speaking, the commutative property is only defined for binary operations as well, so it doesn't make sense to talk about whether or not the inner product is commutative either. But unlike for the associative property, there isn't really a need to restrict the notion of commutativity to binary operations: there's a natural generalization to functions $S \times S \rightarrow T$ with $S \neq T$.)

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