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Marcel
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If $u$ is uniformly distributed over the sphere, we can write it as $u=Uv$, where $U$ is a unitary transformation uniformly distributed over the unitary group. Then the quantity $|u\cdot v|^2$ is just the modulus square of a matrix element of $U$. So your question is: what is the distribution of the modulus square of a matrix element of a random unitary matrix?

Consider a column of the matrix as a random vector. The only constraint is must satisfy is that its norm must be unit, $\sum_{i=1}^N|z_i|^2=1$. In other words, the joint distribution of the entries is simply

$$\delta(1-\sum_{i=1}^N|z_i|^2)=\int ds e^{is(1-\sum_{i=1}^N|z_i|^2)}$$

The distribution of a single element, $z_1$ say, is obtained integrating over the remaining ones. Indeed, you get something proportional to $(1-|z_1|^2)^{N-1}$$(1-|z_1|^2)^{N-2}$ in dimension $N$, which is your Beta distribution.

This is independent of $v$, so the distribution of $v$ is irrelevant.

A very similar argument applies to the real case and random orthogonal matrices.

This is discussed in several articles, such as

Kus, Mostowski and Haake, J. Phys. A 21, L1073 (1988)

Haake and Zyczkowski, Phys. Rev. A 42, 1013 (1990)

K. Zyczkowski, H.-J. Sommers, J. Phys. A 33, 2045 (2000)

If $u$ is uniformly distributed over the sphere, we can write it as $u=Uv$, where $U$ is a unitary transformation uniformly distributed over the unitary group. Then the quantity $|u\cdot v|^2$ is just the modulus square of a matrix element of $U$. So your question is: what is the distribution of the modulus square of a matrix element of a random unitary matrix?

Consider a column of the matrix as a random vector. The only constraint is must satisfy is that its norm must be unit, $\sum_{i=1}^N|z_i|^2=1$. In other words, the joint distribution of the entries is simply

$$\delta(1-\sum_{i=1}^N|z_i|^2)=\int ds e^{is(1-\sum_{i=1}^N|z_i|^2)}$$

The distribution of a single element, $z_1$ say, is obtained integrating over the remaining ones. Indeed, you get something proportional to $(1-|z_1|^2)^{N-1}$ in dimension $N$, which is your Beta distribution.

This is independent of $v$, so the distribution of $v$ is irrelevant.

A very similar argument applies to the real case and random orthogonal matrices.

If $u$ is uniformly distributed over the sphere, we can write it as $u=Uv$, where $U$ is a unitary transformation uniformly distributed over the unitary group. Then the quantity $|u\cdot v|^2$ is just the modulus square of a matrix element of $U$. So your question is: what is the distribution of the modulus square of a matrix element of a random unitary matrix?

Consider a column of the matrix as a random vector. The only constraint is must satisfy is that its norm must be unit, $\sum_{i=1}^N|z_i|^2=1$. In other words, the joint distribution of the entries is simply

$$\delta(1-\sum_{i=1}^N|z_i|^2)=\int ds e^{is(1-\sum_{i=1}^N|z_i|^2)}$$

The distribution of a single element, $z_1$ say, is obtained integrating over the remaining ones. Indeed, you get something proportional to $(1-|z_1|^2)^{N-2}$ in dimension $N$, which is your Beta distribution.

This is independent of $v$, so the distribution of $v$ is irrelevant.

A very similar argument applies to the real case and random orthogonal matrices.

This is discussed in several articles, such as

Kus, Mostowski and Haake, J. Phys. A 21, L1073 (1988)

Haake and Zyczkowski, Phys. Rev. A 42, 1013 (1990)

K. Zyczkowski, H.-J. Sommers, J. Phys. A 33, 2045 (2000)

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Marcel
  • 2.6k
  • 19
  • 35

If $u$ is uniformly distributed over the sphere, we can write it as $u=Uv$, where $U$ is a unitary transformation uniformly distributed over the unitary group. Then the quantity $|u\cdot v|^2$ is just the modulus square of a matrix element of $U$. So your question is: what is the distribution of the modulus square of a matrix element of a random unitary matrix?

Consider a column of the matrix as a random vector. The only constraint is must satisfy is that its norm must be unit, $\sum_{i=1}^N|z_i|^2=1$. In other words, the joint distribution of the entries is simply

$$\delta(1-\sum_{i=1}^N|z_i|^2)=\int ds e^{is(1-\sum_{i=1}^N|z_i|^2)}$$

The distribution of a single element, $z_1$ say, is obtained integrating over the remaining ones. Indeed, you get something proportional to $(1-|z_1|^2)^{N-1}$ in dimension $N$, which is your Beta distribution.

This is independent of $v$, so the distribution of $v$ is irrelevant.

A very similar argument applies to the real case and random orthogonal matrices.