To find the probability distribution for an observable $A$ in a given quantum state $|\psi\rangle$, you need to find the eigenvalues and eigenvectors of $A$. It should actually be obvious that you need to find the eigenvalues, since those are exactly the allowed values of $A$. However, you also need the eigenvectors (or at least the eigenspaces) associated with each eigenvalue to find the probability.

For a (nondegenerate, discrete) eigenvalue $\lambda$ of $A$ with corresponding eigenstate $|\lambda\rangle$, meaning $A|\lambda\rangle=\lambda|\lambda\rangle$, the probability of observing $\lambda$ is $P(\lambda)=\left|\langle\lambda|\psi\rangle\right|^{2}$. This occurs because we can expand the state $|\psi\rangle$ as a sum over the complete set of normalized eigenstates $$|\psi\rangle=\sum_{\lambda_{j}}c_{j}|\lambda_{j}\rangle;$$ the probability that, if $A$ is measured, it will be found in a particular eigenstate $\lambda_{j}$ is $|c_{j}|^{2}$, and since the $|\lambda_{j}\rangle$ are orthonormal (meaning $\langle\lambda_{j}|\lambda_{k}\rangle=\delta_{jk}$), $c_{j}=\langle\lambda_{j}|\psi\rangle$.

For an operator $A$ with a continuous spectrum of eigenvalues (like $x$ or $p$), the expansion in eigenstates becomes an integral, and $\left|\langle\lambda|\psi\rangle\right|^{2}$ is the probability density $\wp(\lambda)$ in $\lambda$-space. For example, the probability density in space is simply $$\wp(x)=\left|\langle x|\psi\rangle\right|^{2}=\psi^{*}(x)\psi(x),$$
so that $1=\int dx\,\wp(x)$ and $\langle x\rangle=\int dx\,x\wp(x)$. For the probability density in the wave number $k=p/\hbar$, we similarly have $\wp(k)=\left|\langle k|\psi\rangle\right|^{2}$. In that case, the normalized state $|k\rangle$ is $$|k\rangle=\frac{1}{\sqrt{2\pi}}e^{ikx}.$$ The normalization constant is chosen so that $$\langle k'|k\rangle=\int dx\,\left(\frac{1}{\sqrt{2\pi}}e^{-ikx'}\right) \left(\frac{1}{\sqrt{2\pi}}e^{ikx}\right)=\delta(k-k'),$$ is normalized in the same way as $\langle x'|x\rangle=\delta(x-x')$. This means that $$\langle k|\psi\rangle=\int dx\,\left(\frac{1}{\sqrt{2\pi}}e^{-ikx}\right) \psi(x)$$ is just the Fourier transform $\tilde{\psi}(k)$, and $\wp(k)=\left|\tilde{\psi}(k)\right|^{2}$, as expected.

The other complication is that when $A$ has degenerate eigenvalues, you have to sum the overlap probabilities (or probability densities) of finding $|\psi\rangle$ in each orthogonal eigenstate in the eigenvalue-$\lambda$ subspace.

To find the probability distribution for an observable $A$ in a given quantum state $|\psi\rangle$, you need to find the eigenvalues and eigenvectors of $A$. It should actually be obvious that you need to find the eigenvalues, since those are exactly the allowed values of $A$. However, you also need the eigenvectors (or at least the eigenspaces) associated with each eigenvalue to find the probability.

For a (nondegenerate, discrete) eigenvalue $\lambda$ of $A$ with corresponding eigenstate $|\lambda\rangle$, meaning $A|\lambda\rangle=\lambda|\lambda\rangle$, the probability of observing $\lambda$ is $P(\lambda)=\left|\langle\lambda|\psi\rangle\right|^{2}$. This occurs because we can expand the state $|\psi\rangle$ as a sum over the complete set of normalized eigenstates $$|\psi\rangle=\sum_{\lambda_{j}}c_{j}|\lambda_{j}\rangle;$$ the probability that, if $A$ is measured, it will be found in a particular eigenstate $\lambda_{j}$ is $|c_{j}|^{2}$, and since the $|\lambda_{j}\rangle$ are orthonormal (meaning $\langle\lambda_{j}|\lambda_{k}\rangle=\delta_{jk}$), $c_{j}=\langle\lambda_{j}|\psi\rangle$.

For an operator $A$ with a continuous spectrum of eigenvalues (like $x$ or $p$), the expansion in eigenstates becomes an integral, and $\left|\langle\lambda|\psi\rangle\right|^{2}$ is the probability density $\wp(\lambda)$ in $\lambda$-space. For example, the probability density in space is simply $$\wp(x)=\left|\langle x|\psi\rangle\right|^{2}=\psi^{*}(x)\psi(x),$$
so that $1=\int dx\,\wp(x)$ and $\langle x\rangle=\int dx\,x\wp(x)$. For the probability density in the wave number $k=p/\hbar$, we similarly have $\wp(k)=\left|\langle k|\psi\rangle\right|^{2}$. In that case, the normalized state $|k\rangle$ is $$\langle x|k\rangle=\frac{1}{\sqrt{2\pi}}e^{ikx}.$$ The normalization constant is chosen so that $$\langle k'|k\rangle=\int dx\,\left(\frac{1}{\sqrt{2\pi}}e^{-ikx'}\right) \left(\frac{1}{\sqrt{2\pi}}e^{ikx}\right)=\delta(k-k'),$$ is normalized in the same way as $\langle x'|x\rangle=\delta(x-x')$. This means that $$\langle k|\psi\rangle=\int dx\,\left(\frac{1}{\sqrt{2\pi}}e^{-ikx}\right) \psi(x)$$ is just the Fourier transform $\tilde{\psi}(k)$, and $\wp(k)=\left|\tilde{\psi}(k)\right|^{2}$, as expected.

The other complication is that when $A$ has degenerate eigenvalues, you have to sum the overlap probabilities (or probability densities) of finding $|\psi\rangle$ in each orthogonal eigenstate in the eigenvalue-$\lambda$ subspace.

To find the probability distribution for an observable $A$ in a given quantum state $|\psi\rangle$, you need to find the eigenvalues and eigenvectors of $A$. It should actually be obvious that you need to find the eigenvalues, since those are exactly the allowed values of $A$. However, you also need the eigenvectors (or at least the eigenspaces) associated with each eigenvalue to find the probability.

For a (nondegenerate, discrete) eigenvalue $\lambda$ of $A$ with corresponding eigenstate $|\lambda\rangle$, meaning $A|\lambda\rangle=\lambda|\lambda\rangle$, the probability of observing $\lambda$ is $P(\lambda)=\left|\langle\lambda|\psi\rangle\right|^{2}$. This occurs because we can expand the state $|\psi\rangle$ as a sum over the complete set of normalized eigenstates $$|\psi\rangle=\sum_{\lambda_{j}}c_{j}|\lambda_{j}\rangle;$$ the probability that, if $A$ is measured, it will be found in a particular eigenstate $\lambda_{j}$ is $|c_{j}|^{2}$, and since the $|\lambda_{j}\rangle$ are orthonormal (meaning $\langle\lambda_{j}|\lambda_{k}\rangle=\delta_{jk}$), $c_{j}=\langle\lambda_{j}|\psi\rangle$.

For an operator $A$ with a continuous spectrum of eigenvalues (like $x$ or $p$), the expansion in eigenstates becomes an integral, and $\left|\langle\lambda|\psi\rangle\right|^{2}$ is the probability density $\wp(\lambda)$ in $\lambda$-space. The other complication is that when $A$ has degenerate eigenvalues, you have to sum the overlap probabilities (or probability densities) of finding $|\psi\rangle$ in each orthogonal eigenstate in the eigenvalue-$\lambda$ subspace.

To find the probability distribution for an observable $A$ in a given quantum state $|\psi\rangle$, you need to find the eigenvalues and eigenvectors of $A$. It should actually be obvious that you need to find the eigenvalues, since those are exactly the allowed values of $A$. However, you also need the eigenvectors (or at least the eigenspaces) associated with each eigenvalue to find the probability.

For a (nondegenerate, discrete) eigenvalue $\lambda$ of $A$ with corresponding eigenstate $|\lambda\rangle$, meaning $A|\lambda\rangle=\lambda|\lambda\rangle$, the probability of observing $\lambda$ is $P(\lambda)=\left|\langle\lambda|\psi\rangle\right|^{2}$. This occurs because we can expand the state $|\psi\rangle$ as a sum over the complete set of normalized eigenstates $$|\psi\rangle=\sum_{\lambda_{j}}c_{j}|\lambda_{j}\rangle;$$ the probability that, if $A$ is measured, it will be found in a particular eigenstate $\lambda_{j}$ is $|c_{j}|^{2}$, and since the $|\lambda_{j}\rangle$ are orthonormal (meaning $\langle\lambda_{j}|\lambda_{k}\rangle=\delta_{jk}$), $c_{j}=\langle\lambda_{j}|\psi\rangle$.

For an operator $A$ with a continuous spectrum of eigenvalues (like $x$ or $p$), the expansion in eigenstates becomes an integral, and $\left|\langle\lambda|\psi\rangle\right|^{2}$ is the probability density $\wp(\lambda)$ in $\lambda$-space. For example, the probability density in space is simply $$\wp(x)=\left|\langle x|\psi\rangle\right|^{2}=\psi^{*}(x)\psi(x),$$
so that $1=\int dx\,\wp(x)$ and $\langle x\rangle=\int dx\,x\wp(x)$. For the probability density in the wave number $k=p/\hbar$, we similarly have $\wp(k)=\left|\langle k|\psi\rangle\right|^{2}$. In that case, the normalized state $|k\rangle$ is $$|k\rangle=\frac{1}{\sqrt{2\pi}}e^{ikx}.$$ The normalization constant is chosen so that $$\langle k'|k\rangle=\int dx\,\left(\frac{1}{\sqrt{2\pi}}e^{-ikx'}\right) \left(\frac{1}{\sqrt{2\pi}}e^{ikx}\right)=\delta(k-k'),$$ is normalized in the same way as $\langle x'|x\rangle=\delta(x-x')$. This means that $$\langle k|\psi\rangle=\int dx\,\left(\frac{1}{\sqrt{2\pi}}e^{-ikx}\right) \psi(x)$$ is just the Fourier transform $\tilde{\psi}(k)$, and $\wp(k)=\left|\tilde{\psi}(k)\right|^{2}$, as expected.

The other complication is that when $A$ has degenerate eigenvalues, you have to sum the overlap probabilities (or probability densities) of finding $|\psi\rangle$ in each orthogonal eigenstate in the eigenvalue-$\lambda$ subspace.