Can anyone explain to me the meaning of this picture? I Know that the argument is Quantum Physics and that cat is the Schrödinger's cat, but I don't know how to interpret the bra–ket notation and also the two numerical coefficients...

enter image description here

  • 2
    $\begingroup$ A ket (not related to the cat :D) is a mathematical object, which describes the state. Kets are denoted using special pair of brackets $| \phantom{a} \rangle$ and everything inside this brackets is just a label. The superposition of two states is represented as a linear combination of the corresponding kets. $\endgroup$ – Wildcat Jul 26 '13 at 13:19
  • $\begingroup$ So the combination of this two states $\frac{1}{\sqrt{2}}|\mathrm{Alive}\rangle +\frac{1}{\sqrt{2}}|\mathrm{Dead}\rangle$ result in a superposition that show that the cat is something between alive and dead, right? Thank you WildKET ;-) $\endgroup$ – Aurelius Jul 27 '13 at 15:45

The expression in the picture contains kets only. Kets represent states of a system. In this case, the "alive" state is the first one and the "dead" state the second.

The numerical factors are there for normalisation. It is assumed both states are equally likely, so they have the same numerical factor. If we call the expression in the picture $|\Psi\rangle$, the inner product$^1$ (analogous to the scalar product of two vectors)

$$\langle\Psi|\Psi\rangle = \int{\Psi^*(\vec{r})\Psi(\vec{r})d\vec{r}}$$

should be equal to $1$. This is because we interpret the wavefunction $\Psi(\vec{r})$ as a probability amplitude distribution and hence the scalar product of the wavefunction with itself as a probability distribution. Integrating a probability distribution over the entire space (I used the position space here, but you could just as well integrate over an abstract state space for example) should always give $1$, due to the nature of probability theory. This expresses the idea that the system (cat) must be in some state.

Now, there's a problem with this expression and that problem is also the resolution of the issue associated with Schrödinger's cat (the criticism that according to QM the cat would be both dead and alive at the same time). And that is that the cat is not a quantum system. A common mistake is that a conscious observer is needed for the "collapse" of the wavefunction, as it is described in the Copenhagen interpretation. This is not true. The cat will not be in a superposition of both dead and alive but in one particular state (see also decoherence).

$^1$ Some more information on the inner product $\langle\Psi|\Psi\rangle$.

Let's consider a general ket in some state space $\mathcal{V}$ (which is just a special kind of vector space and kets are special column vectors), say $|a\rangle$. Let's assume the dimension $d$ of this space is finite. It doesn't have to be, but it makes things conceptually easier to understand. Because then a finite basis $|e_i\rangle$ (with $i = 1\ldots d$) exists for this vector space, meaning we can write every ket in $\mathcal{V}$ as

$$|a\rangle = \sum_{i}^{d}{a_i|e_i\rangle}.\quad\quad(1)$$

For a ket space of infinite dimension (like position space) this sum becomes an integral.

Now we can construct a dual vector space $\mathcal{V^*}$ which contains all the linear functions $f(|a\rangle)$ mapping the kets in $\mathcal{V}$ to the complex numbers $\mathbb{C}$. We denote these functions as $\langle f|$ and call them bra's. You can think of these bra's as row vectors. The common notation therefore isn't $f(|a\rangle)$ but rather $\langle f|a\rangle$, a bra-ket, which can in turn be seen as a scalar (inner) product between a row vector and a column vector.

The elements of $\mathcal{V}^*$ are constructed in a special way. We choose the basis $\langle e_i|$ for this dual space such that for every $\langle e_i|$ it holds that $\langle e_i|a\rangle = a_i^*$ and therefore $\langle e_i|e_j\rangle = \delta_{ij}$ (orthonormality property). This allows us to associate a single bra $\langle a|$ with each ket $|a\rangle$ as follows

$$\langle a| = \sum_{i}^{d}{a_i^*\langle e_i|}.\quad\quad(2)$$

Again this sum becomes an integral if we consider a continuous label such as position instead of the discrete $i$'s.

Now the inner product of two kets in $\mathcal{V}$ is defined as

$$\left(|a\rangle\right) \cdot \left(|b\rangle\right) \equiv \langle a|b\rangle$$

which can be rewritten using $(1)$, $(2)$ and the orthonormality property, yielding:

$$\langle a|b\rangle = \sum_{i}^{d}{a_i^*b_i}$$

where the sum again becomes an integral in e.g. position space, recovering the expression I used above.

The difference between an expression of the form $\langle a|b\rangle$ and one of the form $c|b\rangle$ should now also be clear. In the first case we actually have a function $\langle a|$ with the ket $|b\rangle$ as its argument, the result of which is a complex number. In the second case we have a (possibly complex) number $c$ in front of a ket $|b\rangle$, the result of which is still a ket. And this ket still represents the same state as long as $c$ is a constant.

  • $\begingroup$ Why have you used the notation $\langle X|X\rangle$ and not only $X|X\rangle$ as in my picture? Because this $\langle X|X\rangle$ is a bra-ket and this $X|X\rangle$ is only a ket? What's the difference? $\endgroup$ – Aurelius Jul 27 '13 at 15:37
  • $\begingroup$ @FormlessCloud I've added a discussion of how the bra-ket is constructed with a paragraph pointing out the difference between expressions of the form $\langle X|X\rangle$ and ones of the form $x|X\rangle$. If there are still things that aren't clear, let me know. $\endgroup$ – Wouter Jul 27 '13 at 16:59

Ket(s) and Bra(s) represent state of a system. The simplest interpretation is to see kets and states as (complex) vectors. In your case, the state could be described by the vector :

$$\vec S = \frac{1}{\sqrt{2}} \vec{alive} + \frac{1}{\sqrt{2}} \vec{dead}$$

Here, $\vec{alive}$ and $\vec{dead}$ are a basis for the states, so they are orthogonal vectors : $\vec{alive}.\vec{dead} = 0$, and they are normed : $|\vec{alive}|=|\vec{dead}|=1$

A general state could be written: $$\vec S = a *~ \vec{alive} + b *~\vec{dead}$$ where $a$ and $b$ are $2$ complex numbers.

Generally, we want a normed state, that is $|a|^2 +|b|^2=1$, and this is the case in your particular state.

Now, if I take this general state, the probability that the cat is alive is $|a|^2$, and the probability that the cat is dead is $|b|^2$.

Said differently, the coefficients $a,b$ are probabilites amplitudes, there are not probabilities.


The easy part is the numerical coefficients which are there so the state is normalized, which means $\langle \psi | \psi \rangle=1$.

The easier part is what's written in the kets, one of them illustrates the state of a dead cat and the other a living one. The point is, due to the canonical form of quantum mechanics, before doing a measurement the cat's state is neither alive or dead; it's a linear superposition of them. When you do a measurement the cat's state will collapse in either one of them(but not both :-).


There are two states (in QM with ket's $\left|\phantom{H}\right>$ we mark possible states) of the Schrödinger's cat. 1st state means (the one on the left) the cat is alive and 2nd state (the one on the right) means it is dead. We could write this down like this:

$$\underbrace{\frac{1}{\sqrt{2}}}_{\llap{\text{amplitude for first state}}} \,\,\left|\text{alive}\right> + \underbrace{\frac{1}{\sqrt{2}}}_{\rlap{\text{amplitude for second state}}}\,\, \left|\text{dead}\right>$$

If you would deal with this problem in a classical way you would write probabilities instead of amplitudes like this (no square roots):

$$\underbrace{\frac{1}{2}}_{\llap{\text{probability for first state}}} \,\,\left|\text{alive}\right> + \underbrace{\frac{1}{2}}_{\rlap{\text{probability for second state}}}\,\, \left|\text{dead}\right> = \underbrace{1}_{\rlap{\text{total probability is allways 1}}}$$

For understanding the connection between probability and QM amplitude please read the accepted anwser to my own question which helped me to understand this: QM formalism is one big confusion - lack of geometrical explaination with images.

  • $\begingroup$ The second equal sign is merely suggestive. The object you've written down on the left hand side isn't a number, it's more of a list. Only the sum of the entries of that list is $1$. $\endgroup$ – Nikolaj-K Jul 26 '13 at 13:39
  • $\begingroup$ Thank you for the comment. I am still learning QM myself. $\endgroup$ – 71GA Jul 26 '13 at 13:48

I have provided an answer to this question in this video.

Your question was helpful for me to think about my understanding of QM notation since I am still learning myself.


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