For example in Deutsch's Algorithm, we apply $H$ to $|0\rangle$ and $|1\rangle$ respectively. I have seen the notation this result in the state: $$\frac{1}{2}(|0\rangle+|1\rangle)(|0\rangle-|1\rangle)$$ However, it is unclear to me how exactly I should read this. From my understanding, the result should be the outer product of $\begin{pmatrix} 1/\sqrt{2} \\ 1/\sqrt{2} \end{pmatrix}$ and $\begin{pmatrix}1/\sqrt{2} \\ -1/\sqrt{2}\end{pmatrix}$, but then I would expect to see a bra vector like so: $\langle x \mid$.

So how should I read a notation like: $|0\rangle (|0\rangle+|0\rangle)$?


1 Answer 1


As a starting point, the state $\Psi = \frac{\vert 0 \rangle + \vert 1 \rangle}{\sqrt{2}}$ is a superposition of states $\vert 0 \rangle$ and $\vert 1 \rangle$, both with amplitude $\frac{1}{\sqrt{2}}$, indicating both states are equally likely to be measured. The critical point to understand is that $\vert 0 \rangle$ and $\vert 1 \rangle$ are orthogonal basis elements, so $\Psi$ doesn't reduce in the way you indicated (consider the analogous unit vector $\frac{x+y}{\sqrt{2}}$ in a Euclidean coordinate system).

Next, it's important to note that the sum of the square of the amplitudes always equals one because the magnitude of each amplitude squared represents the probability that we would observe that state if a measurement was taken. In other words, measurement will cause $\Psi$ to collapse to state $\vert 0 \rangle$ with probability $\left(\frac{1}{\sqrt{2}} \right)^2=\frac{1}{2}$, and the same for $\vert 1 \rangle$. So $\vert 0 \rangle + \vert 0 \rangle$ is not physical, since measuring $\vert 0 \rangle$ with probability $2$ is not physically realistic.

When you see two kets next to each other this implies a tensor product, i.e. $\vert i \rangle \vert j \rangle$ implies $\vert i \rangle \otimes \vert j \rangle$. So $\frac{1}{2}(|0\rangle+|1\rangle)(|0\rangle-|1\rangle)$ is a shorthand way of writing $$\frac{1}{2}\left(\vert 0 \rangle \otimes \vert 0 \rangle + \vert 1 \rangle \otimes \vert 0 \rangle - \vert 0 \rangle \otimes \vert 1 \rangle - \vert 1 \rangle \otimes \vert 1 \rangle \right).$$ You can see that this is properly normalized since $\sum \limits_i p_i = 4(\frac{1}{2})^2 =1$.

Reducing this to vector representations, if we take the vector representation $\vert 0 \rangle \equiv \begin{bmatrix} 1 \\ 0 \end{bmatrix}$ and $\vert 1 \rangle \equiv \begin{bmatrix}0 \\ 1\end{bmatrix}$, then $$\vert 0 \rangle \otimes \vert 0 \rangle = \begin{bmatrix} 1 \\0\\0\\0 \end{bmatrix}, \;\; \vert 1 \rangle \otimes \vert 0 \rangle = \begin{bmatrix} 0 \\0\\1\\0 \end{bmatrix}, \;\; \vert 0 \rangle \otimes \vert 1 \rangle = \begin{bmatrix} 0 \\1\\0\\0 \end{bmatrix}, \;\; \vert 1 \rangle \otimes \vert 1 \rangle = \begin{bmatrix} 0 \\0\\0\\1 \end{bmatrix}.$$

As a final point, since you mentioned outer products, in bra-ket notation outer products are indicated by a ket to the left of a bra, e.g. $\vert i \rangle \langle j \vert$.


Since the final paragraph opened the door to outer products, it might be helpful to clarify that notion as well.

Where a ket represents a basis vector, a bra represents a dual vector (sometimes referred to as a covector). In this context, for an arbitrary basis vector, the corresponding dual vector is the conjugate transpose of the basis vector. For $\vert 0 \rangle$ and $\vert 1 \rangle$, as defined above (conventionally known as the computational basis), this simply gives $$\langle 0 \vert = \begin{bmatrix} 1 & 0 \end{bmatrix}, \;\;\; \langle 1 \vert = \begin{bmatrix} 0 & 1 \end{bmatrix}.$$

For any orthonormal set of basis vectors, orthonormality is characterized by the inner product $$\langle j \vert k \rangle = \delta_{jk},$$ where $\delta_{jk}$ is the Kronecker delta. For any complete orthonormal set of basis vectors, completeness is characterized by the outer product $$\sum \limits_j \vert j \rangle \langle j \vert = I,$$ where $I$ is the identity matrix with appropriate dimensions. Using this, you can readily confirm that both the computational basis and the basis used in Deutsch's algorithm are both orthonormal and complete.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.