Let $\vert\text{#}\rangle$ be the vector state of the cat, $\vert1\rangle$ the "alive" state, and $\vert0\rangle$ the "dead" state. Using the normalization condition $\langle \text{#}\vert\text{#}\rangle=1$:

\begin{equation} \vert \text{#}\rangle=a\vert1\rangle+b\vert0\rangle \end{equation}


\begin{equation} \vert a\vert^{2}+a^{*}b\langle 1\vert0\rangle+b^{*}a\langle 0\vert1\rangle+\vert b \vert^{2}=1 \end{equation}

where $\vert a\vert^{2}$ is the probability of the cat being at the state $\vert1\rangle$ (alive).

The equation leads to $a=b=\frac{1}{\sqrt{2}}$.

However, why is this? And how should $\langle 1\vert0\rangle$ and $\langle 0\vert1\rangle$ be interpreted?

  • $\begingroup$ In my interpretation with your notation, $\langle 1 \vert$ is the representation of the (probability) measurement captured the signal of alive. $\endgroup$
    – ChoMedit
    Commented Oct 19, 2018 at 9:50

4 Answers 4


Following John Rennie's analogy, let the cat be a spin with up (alive) or down (dead) state. Note that you've expanded $|\#\rangle$ in terms of the orthonormal basis vectors:

$$|\#\rangle = a|1\rangle + b|0\rangle$$

By definition of orthogonality (and the idea that "alive" and "dead" are orthogonal), $\langle 0|1\rangle = \langle 1|0\rangle = 0$.

Now you're computing the probability that the cat is alive (hopefully it always is) or dead:

$$ |{\langle 1 |\#\rangle }|^2 = |a|^2, \quad |\langle 0 |\#\rangle|^2 = |b|^2 $$

To determine the range of values $a$ and $b$ can take, compute:

$$ |\langle\#|\#\rangle|^2 = |a|^2 + |b|^2 = 1$$

This is $1$ because $|\#\rangle$ is assumed to be normalised. So the range of values for $a$ and $b$ are not strictly $1/\sqrt{2}$ for both (for example $a = 1/\sqrt{5}, b = 2i/\sqrt{5}$ also satisfy the equation, where their absolute squares indicate the probability of being 'alive' or 'dead'), but $(|a|,|b|) \in S^1, \forall\;a, b \in \mathbb{C}$, where $S^1$ is the unit circle.

  • 1
    $\begingroup$ It may be worth stating or showing explicitly that there’s no reason to assume a=b from the given information. $\endgroup$
    – alex_d
    Commented Oct 19, 2018 at 14:12

When you write:

$$ \vert \text{#}\rangle=a\vert1\rangle+b\vert0\rangle $$

you are assuming there exists an alive operator and that this operator has eigenstates $|1\rangle$ and $|0\rangle$ that you can use as a basis for writing the wavefunction of the cat. Neither of these assumptions seem reasonable so as it stands the question makes no sense.

However you could replace the cat by a spin that can be in a superposition of up and down states. In that case we are writing the $z$ component of the spin using the eigenstates of $\hat{L}_z$ as a basis, and we'd get the same equation:

$$ \vert \text{#}\rangle=a\vert1\rangle+b\vert0\rangle $$

where now our $|1\rangle$ and $|0\rangle$ states are well defined because they are the eigenstates of $\hat{L}_z$. And now it's clear that $\langle 0 | 1 \rangle$ and $\langle 1 | 0 \rangle$ are zero because eigenstates are orthogonal.

  • $\begingroup$ I was thinking about $\vert 1 \rangle$ and $\vert 0 \rangle$ as "alive" and "dead" states. Does that make sense? The cat sould be in a superposition of dead and alive states. $\endgroup$
    – IchVerlore
    Commented Oct 19, 2018 at 7:38
  • 2
    $\begingroup$ @IchVerloren a cat is a massively complex system, and being alive is also a complicated business. There isn't a precise distinction between alive and dead. So trying to define alive and dead states and use them as a basis doesn't make any sense. $\endgroup$ Commented Oct 19, 2018 at 8:52
  • $\begingroup$ In Schrodinger's cat the "alive" and "dead" states relate to the slightly simpler system of "has this atom decayed or not?" where if it has then the cat is dead. Is there no operator which might indicate whether an atom has decayed or not? $\endgroup$ Commented Oct 19, 2018 at 14:18

Your system has two mutually exclusive possible outcomes: dead cat $\vert 0\rangle$ or live cat $\vert 1\rangle$. In analogy with spin up/down states, the $\vert 0\rangle$ and $\vert 1\rangle$ states are orthogonal, in the sense that if the cat is found to be alive (in the state $\vert 1\rangle$) then it is not dead - or more precisely it has 0% probability of being found dead: that's the meaning of $\vert \langle 0\vert 1\rangle\vert^2=0$. Likewise, if the cat is found to be alive, it has 100% change of being alive: that's the meaning of $\vert \langle 1\vert 1\rangle\vert^2 =1$.

In this sense, a cat described by $$ \vert \#\rangle=a\vert 0\rangle + b\vert 1\rangle $$ has probability $\vert \langle \#\vert 0\rangle\vert^2 = \vert b\vert^2 $ of being found dead and $\vert \langle \#\vert 1\rangle\vert^2 = \vert a\vert^2 $ of being found alive. Note that the probabilities should sum to 1, i.e. $\vert a\vert^2+\vert b\vert^2=1$.

Please note also that $a$ and $b$ can, in general, be complex numbers although of course their magnitude squared, which is a probability, is a real number, i.e. one could, in principle, have $$ \vert \#\rangle =\frac{1}{\sqrt{3}}\vert 0\rangle + i\sqrt{\frac{2}{3}} \vert 1\rangle $$ which would lead to measuring a dead cat $1/3$ of the time and a living cat $2/3$ of the time.

Finally, despite the surely bizarre nature of the original proposition, there are serious people working on this stuff: this 2015 article published in The Guardian reports on attempts to place microbes in a superposition of states.


First of all, I do not think $a^2 + b^2 =1$ tells you that a = $\frac{1}{\sqrt 2}$. Basic trignometry.

And to tell the truth, no one knows why $a^2$ is the probability(Or more precisely, the square of the coefficients). It's Born's Postulate.

<0|1> can be viewed as what is the probability density to find state |1> in |0>. And those two happens to be orthogonal if $|0>$ and |1> are a set of basis.

By the way, I think your question can be found in any quantum physics textbooks. Hope this helps you!


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