<p>I just ran the first state you suggest (i.e. the GHZ state with negative phase in the Hadamard basis). Basically what I did was to write a circuit which creates that state, apply one of 5 stabilizer generators (XIXII,IXXII,IIXXI,IIXIX or -ZZZZZ) and then inverted the creation process. The circuit is shown below for the -ZZZZZ stabilizer.</p>

<p><a href="https://i.sstatic.net/K04RQ.png" rel="noreferrer"><img src="https://i.sstatic.net/K04RQ.png" alt="enter image description here"></a></p>

<p>There seems to be problem with the calibration or something, since I am seeing a bit flip occurring on the first qubit with roughly probability 2/3 that should not happen. The expected result should be 00000 in all cases, however 10000 seems to be twice as likely in almost all cases (prob ~0.1 vs ~0.2 over 1024 runs for each stabilizer generator).</p>

<p>Below are the probabilities of obtaining 00000 for each stabilizer generator over 1024 runs each:
    XIXII:0.10840
    IXXII:0.15625
    IIXXI:0.11621
    IIXIX:0.10645
    ZZZZZ:0.09375</p>

<p>And the probabilities of obtaining 10000 for each stabilizer generator:
    XIXII:0.21777
    IXXII:0.21875
    IIXXI:0.17871
    IIXIX:0.14648
    ZZZZZ:0.19726</p>

CC BY-SA 3.0
2016-06-20T19:14:30.097Z
@GilKalai: If you want the full statistics, just drop me an email. They are too long to post here.
Joe Fitzsimons
CC BY-SA 3.0
2018-01-14T03:42:39.440Z
you could post it in a pastebin and then link it.
Joshua Herman
CC BY-SA 3.0
2016-06-19T15:51:02.743Z
2016-06-19T15:51:02.743Z
Joe Fitzsimons
<p><strong>Update (January 2018)</strong> A new very interesting paper with various experiments on the IBM machine is <a href="https://arxiv.org/abs/1712.05642" rel="nofollow noreferrer">Five Experimental Tests on the 5-Qubit IBM Quantum Computer</a> by Diego García-Martín, Germán Sierra.</p>

<p><strong>Update (July 2017) Extended question:</strong> Of course, the question extends to the newly built IBM 16-qubits quantum computer, and the 50-qubits quantum computers that IBM, Google and various other players expect to demonstrate in the near future. A natural conjecture is that all these computers can only approximate "noise-stable" pure states.  namely pure states that are stable under noise operator with exponential decay of high order Pauli terms, (see e.g. <a href="https://arxiv.org/abs/0810.2435" rel="nofollow noreferrer">this paper</a> by Montanaro and Osborne).    </p>

<p><strong>Original question:</strong></p>

<p>IBM has recently built a 5-qubits quantum computers based on superconducting qubits. It is even possible to make <a href="http://www.research.ibm.com/quantum/" rel="nofollow noreferrer">experiments over the cloud</a>. </p>

<p>The space of pure states for 5 qubits is the unit sphere in the 32-dimensional space over the complex number. When you measure the state of the computer you get a probability distribution on strings of 5 bits. </p>

<h3>Give a "map" of the probability distributions and of the actual states that the IBM quantum computer can reach.</h3>

<p>Of course, the boundary between what can be reached and what is beyond reach is of special interest. And also the case of 3 and 4 qubits.</p>

<p>My question is motivated by <a href="https://rjlipton.wordpress.com/2016/04/22/quantum-supremacy-and-complexity/#comment-77178" rel="nofollow noreferrer">this comment</a> over Lipton abd Reagan's blog by Alexander Vlasov. </p>

<p>I would expect that distributions that can actually be reached are characterized by "noise stability" which is equivalent to "low degree Fourier-type" expansion. The relevant "noise" here is independent depolarizing noise. The relevant "Fourier expansion" might be the one described by Ashley Montanaro and Tobias J. Osborne in the paper <a href="http://arxiv.org/abs/0810.2435" rel="nofollow noreferrer">quantum Boolean function</a>. </p>

<p>Vlasov referred to the paper by Daniel Alsina and  José Ignacio Latorre <a href="http://arxiv.org/abs/1605.04220" rel="nofollow noreferrer">Experimental test of Mermin inequalities on a 5-qubit quantum computer</a>. A sentence from the abstract reads "The experimental results obtained using the quantum computer show violation of all Mermin inequalitites, with a clear degradation of the results in the 5 qubit case." </p>

<p>Update: Andrew Morgan suggests in the comment below that the proposal for the threshold might be too naive. A specific question inspired by Andrew's comment is:</p>

<h3>Can we well approximate with the IBM machine the state</h3>

<p>$$\frac 14 |11111\rangle+\frac 14 |11100\rangle+ \frac 14 |11010\rangle+\frac 14 |11001\rangle+
\frac 14 |10110\rangle+\frac 14 |10101\rangle+ \frac 14 |01110\rangle+\frac 14 |01101\rangle+
\frac 14 |10011\rangle+\frac 14 |01011\rangle+ \frac 14 |00111\rangle+\frac 14 |10000\rangle+
\frac 14 |01000\rangle+\frac 14 |00100\rangle+ \frac 14 |00010\rangle+\frac 14 |00001\rangle.$$
(Superposition of the computational basis elements with an odd numbers of 1's)</p>

<p>(Later update: In view of Peter Shor's comment below I suppose we need to look more carefully for "noise-sensitive" states that we would like to explore.) </p>

<p>Also of interest are the states $$\frac {1}{\sqrt {5}} |10000\rangle +
\frac {1}{\sqrt {5}} |01000\rangle +\frac {1}{\sqrt {5}} |00100\rangle + \frac {1}{\sqrt {5}} |00010\rangle +\frac {1}{\sqrt {5}} |00001\rangle $$ and the superposition of all basis elements with two 1's. (If somebody reading this question actually tried to program the IBM 5-qubit computers I'd love to hear about it.) </p>

CC BY-SA 3.0
2016-06-08T21:42:54.547Z
In what sense might qubit-efficient universal computation fit into this? I'm thinking of the following: Imagine for a moment that that any boolean function on $n$ bits can be computed using only the output qubit and no ancillary bits as $|x b\rangle \mapsto |x(b\oplus f(x))\rangle$. Then one might compute the uniform distribution over all even-weight strings by preparing a uniform superposition in the first $n-1$ bits and $|0\rangle$ as the last qubit, and then computing parity of the first $n-1$ bits into the last bit. Am I understanding right that this would disagree with your conjecture?
Andrew Morgan
CC BY-SA 3.0
2016-06-08T21:50:23.893Z
I am not sure I understand you. (Anyway this condition may well be naive.)
Also, I am not sure what kind of computations the IBM machine allows. Essentially I want to consider only probability distributions obtained by measuring a pure (as possible) state on the 5 qubits. (but maybe I miss something.)
Gil Kalai
CC BY-SA 3.0
2016-06-08T22:04:21.767Z
Sorry, I'm just trying to get my head around what all is going on. I'm assuming that you're trying to understand when, by starting from a simple proscribed initial state, applying some local unitary operations, and then sampling the state, a given distribution over bit strings can be reached. Qubit-efficient universal quantum computation is relevant since that helps understand what intermediate steps we can take. I just meant to give an example where (if I'm understanding your question and conjecture right) the output distribution might be more complicated than you suspect.
Andrew Morgan
CC BY-SA 3.0
2016-06-10T07:02:45.613Z
I think I get where I was coming from and why it was confusing: I was generalizing the problem to one where "the IBM computer" was replaced by "some fixed, idealized quantum computer with 5 qubits". I didn't realize that this could (ostensibly) drastically influence the answer, while your question is particular to the IBM computer.
Andrew Morgan
CC BY-SA 3.0
2016-06-10T15:35:09.303Z
The state you chose is a stabilizer state (stabilized by $-ZZZZZ$, $XXIII$, $IXXII$, $IIXXI$, $IIIXX$). Did you chose it on purpose ? Even if a quantum machine could produce and sample all stabilizer states, it could not prove any supremacy, because of the Gottesmann-Knill theorem.
Frédéric Grosshans
CC BY-SA 3.0
2016-06-11T19:36:01.840Z
I hope you realize that the state you wrote down, in a different basis, is $(|+++++\rangle - |-----\rangle)/\sqrt{2}$, where $|+\rangle = (|0\rangle +|1\rangle)/\sqrt{2}$ and $|-\rangle = (|0\rangle -|1\rangle)/\sqrt{2}$.
Peter Shor 
CC BY-SA 3.0
2016-06-13T05:19:38.900Z
Dear Peter, right! I suppose I have to think more carefully about good test cases for "noise sensitive" states for this case.
Gil Kalai
CC BY-SA 3.0
2016-06-13T18:16:39.727Z
Gil: ctually, in standard quantum theory, that is a relatively noise-sensitive state. However, maybe you think there are states which are much more noise-sensitive in your theory of decoherence. I'd like to see what they are and why they're more noise-sensitive.
Peter Shor 
CC BY-SA 3.0
2016-06-13T19:22:03.133Z
Peter, Thanks for the clarification. It will be interesting to see how well can the IBM 5-qubit machine achieve this state then. 

My theory of decoherence is based entirely  on the standard noise model (in fact I need to worry only about depolarizing noise). The non standard modeling of noise for larger systems emerges as a consequence of failure to lower the rate of noise to the level needed for quantum fault-tolerance for small systems. (There is nothing non-standard to start with.)
Gil Kalai
CC BY-SA 3.0
2016-06-19T14:13:11.610Z
In terms of density matrices, this is a 1023-dimensional space. I don't think stating it in terms of probability distributions makes it substantially easier. For arbitrarily many qubits, a generalization of this problem is probably unsolvable. This is not an easy question.
Peter Shor 
CC BY-SA 3.0
2016-06-19T16:06:42.417Z
@GilKalai: It is simply enough to run the stabilizer state, but the IBM system uses an approximately universal gate set where only T is non-Clifford. Also, programming is done via circuit diagrams. The result is that producing the W states you mention is extremely difficult in practice for normal users due to the need to approximate the required non-Clifford group gates.
Joe Fitzsimons
CC BY-SA 3.0
2017-09-15T13:22:10.263Z
@JoeFitzsimons Joe, does IBM provides an opportunity to experiment on the cloud for they larger than 5-qubits quantum computers (perhaps 16 or 17 qubits)? Is it possible to tell what kind of quantum states can be reached by this computers? Also, there were news that IBM used the quantum computer for simulation certain molecules. What kind of states were involved in thes simulations?
Gil Kalai
CC BY-SA 3.0
2018-04-04T00:45:45.130Z
You can ask this question at: https://quantumcomputing.stackexchange.com/ .
Rob
<cc.complexity-theory><quantum-computing><quantum-information>
States and Probability distributions that the 5-qubits IBM computer can produce
Gil Kalai
# States and Probability distributions that the 5-qubits IBM computer can produce

(Asked by: username_0 on 2016-06-08)

**Update (January 2018)** A new very interesting paper with various experiments on the IBM machine is [Five Experimental Tests on the 5-Qubit IBM Quantum Computer](https://arxiv.org/abs/1712.05642) by Diego García-Martín, Germán Sierra.

**Update (July 2017) Extended question:** Of course, the question extends to the newly built IBM 16-qubits quantum computer, and the 50-qubits quantum computers that IBM, Google and various other players expect to demonstrate in the near future. A natural conjecture is that all these computers can only approximate "noise-stable" pure states. namely pure states that are stable under noise operator with exponential decay of high order Pauli terms, (see e.g. [this paper](https://arxiv.org/abs/0810.2435) by Montanaro and Osborne).

**Original question:**

IBM has recently built a 5-qubits quantum computers based on superconducting qubits. It is even possible to make [experiments over the cloud](http://www.research.ibm.com/quantum/).

The space of pure states for 5 qubits is the unit sphere in the 32-dimensional space over the complex number. When you measure the state of the computer you get a probability distribution on strings of 5 bits.

### Give a "map" of the probability distributions and of the actual states that the IBM quantum computer can reach.

Of course, the boundary between what can be reached and what is beyond reach is of special interest. And also the case of 3 and 4 qubits.

My question is motivated by [this comment](https://rjlipton.wordpress.com/2016/04/22/quantum-supremacy-and-complexity/#comment-77178) over Lipton abd Reagan's blog by Alexander Vlasov.

I would expect that distributions that can actually be reached are characterized by "noise stability" which is equivalent to "low degree Fourier-type" expansion. The relevant "noise" here is independent depolarizing noise. The relevant "Fourier expansion" might be the one described by Ashley Montanaro and Tobias J. Osborne in the paper [quantum Boolean function](http://arxiv.org/abs/0810.2435).

Vlasov referred to the paper by Daniel Alsina and José Ignacio Latorre [Experimental test of Mermin inequalities on a 5-qubit quantum computer](http://arxiv.org/abs/1605.04220). A sentence from the abstract reads "The experimental results obtained using the quantum computer show violation of all Mermin inequalitites, with a clear degradation of the results in the 5 qubit case."

Update: username_1 suggests in the comment below that the proposal for the threshold might be too naive. A specific question inspired by Andrew's comment is:

### Can we well approximate with the IBM machine the state

$$\frac 14 |11111\rangle+\frac 14 |11100\rangle+ \frac 14 |11010\rangle+\frac 14 |11001\rangle+
\frac 14 |10110\rangle+\frac 14 |10101\rangle+ \frac 14 |01110\rangle+\frac 14 |01101\rangle+
\frac 14 |10011\rangle+\frac 14 |01011\rangle+ \frac 14 |00111\rangle+\frac 14 |10000\rangle+
\frac 14 |01000\rangle+\frac 14 |00100\rangle+ \frac 14 |00010\rangle+\frac 14 |00001\rangle.$$
(Superposition of the computational basis elements with an odd numbers of 1's)

(Later update: In view of Peter Shor's comment below I suppose we need to look more carefully for "noise-sensitive" states that we would like to explore.)

Also of interest are the states $$\frac {1}{\sqrt {5}} |10000\rangle +
\frac {1}{\sqrt {5}} |01000\rangle +\frac {1}{\sqrt {5}} |00100\rangle + \frac {1}{\sqrt {5}} |00010\rangle +\frac {1}{\sqrt {5}} |00001\rangle $$ and the superposition of all basis elements with two 1's. (If somebody reading this question actually tried to program the IBM 5-qubit computers I'd love to hear about it.)

**Tags:** cc.complexity-theory, quantum-computing, quantum-information

> **Comments:**
> **username_1**: In what sense might qubit-efficient universal computation fit into this? I'm thinking of the following: Imagine for a moment that that any boolean function on $n$ bits can be computed using only the output qubit and no ancillary bits as $|x b\rangle \mapsto |x(b\oplus f(x))\rangle$. Then one might compute the uniform distribution over all even-weight strings by preparing a uniform superposition in the first $n-1$ bits and $|0\rangle$ as the last qubit, and then computing parity of the first $n-1$ bits into the last bit. Am I understanding right that this would disagree with your conjecture?  
> **username_0**: I am not sure I understand you. (Anyway this condition may well be naive.) Also, I am not sure what kind of computations the IBM machine allows. Essentially I want to consider only probability distributions obtained by measuring a pure (as possible) state on the 5 qubits. (but maybe I miss something.)  
> **username_1**: Sorry, I'm just trying to get my head around what all is going on. I'm assuming that you're trying to understand when, by starting from a simple proscribed initial state, applying some local unitary operations, and then sampling the state, a given distribution over bit strings can be reached. Qubit-efficient universal quantum computation is relevant since that helps understand what intermediate steps we can take. I just meant to give an example where (if I'm understanding your question and conjecture right) the output distribution might be more complicated than you suspect.  
> **username_1**: I think I get where I was coming from and why it was confusing: I was generalizing the problem to one where "the IBM computer" was replaced by "some fixed, idealized quantum computer with 5 qubits". I didn't realize that this could (ostensibly) drastically influence the answer, while your question is particular to the IBM computer.  
> **username_2**: The state you chose is a stabilizer state (stabilized by $-ZZZZZ$, $XXIII$, $IXXII$, $IIXXI$, $IIIXX$). Did you chose it on purpose ? Even if a quantum machine could produce and sample all stabilizer states, it could not prove any supremacy, because of the Gottesmann-Knill theorem.  
> **username_3**: I hope you realize that the state you wrote down, in a different basis, is $(|+++++\rangle - |-----\rangle)/\sqrt{2}$, where $|+\rangle = (|0\rangle +|1\rangle)/\sqrt{2}$ and $|-\rangle = (|0\rangle -|1\rangle)/\sqrt{2}$.  
> **username_0**: Dear Peter, right! I suppose I have to think more carefully about good test cases for "noise sensitive" states for this case.  
> **username_3**: Gil: ctually, in standard quantum theory, that is a relatively noise-sensitive state. However, maybe you think there are states which are much more noise-sensitive in your theory of decoherence. I'd like to see what they are and why they're more noise-sensitive.  
> **username_0**: Peter, Thanks for the clarification. It will be interesting to see how well can the IBM 5-qubit machine achieve this state then.   My theory of decoherence is based entirely  on the standard noise model (in fact I need to worry only about depolarizing noise). The non standard modeling of noise for larger systems emerges as a consequence of failure to lower the rate of noise to the level needed for quantum fault-tolerance for small systems. (There is nothing non-standard to start with.)  
> **username_3**: In terms of density matrices, this is a 1023-dimensional space. I don't think stating it in terms of probability distributions makes it substantially easier. For arbitrarily many qubits, a generalization of this problem is probably unsolvable. This is not an easy question.  
> **username_4**: @GilKalai: It is simply enough to run the stabilizer state, but the IBM system uses an approximately universal gate set where only T is non-Clifford. Also, programming is done via circuit diagrams. The result is that producing the W states you mention is extremely difficult in practice for normal users due to the need to approximate the required non-Clifford group gates.  
> **username_0**: @JoeFitzsimons Joe, does IBM provides an opportunity to experiment on the cloud for they larger than 5-qubits quantum computers (perhaps 16 or 17 qubits)? Is it possible to tell what kind of quantum states can be reached by this computers? Also, there were news that IBM used the quantum computer for simulation certain molecules. What kind of states were involved in thes simulations?  
> **username_5**: You can ask this question at: https://quantumcomputing.stackexchange.com/ .  

---

## 1 answers

### Answer by username_4 on 2016-06-19. Score: 7

I just ran the first state you suggest (i.e. the GHZ state with negative phase in the Hadamard basis). Basically what I did was to write a circuit which creates that state, apply one of 5 stabilizer generators (XIXII,IXXII,IIXXI,IIXIX or -ZZZZZ) and then inverted the creation process. The circuit is shown below for the -ZZZZZ stabilizer.

[![enter image description here](https://i.sstatic.net/K04RQ.png)](https://i.sstatic.net/K04RQ.png)

There seems to be problem with the calibration or something, since I am seeing a bit flip occurring on the first qubit with roughly probability 2/3 that should not happen. The expected result should be 00000 in all cases, however 10000 seems to be twice as likely in almost all cases (prob ~0.1 vs ~0.2 over 1024 runs for each stabilizer generator).

Below are the probabilities of obtaining 00000 for each stabilizer generator over 1024 runs each:
XIXII:0.10840
IXXII:0.15625
IIXXI:0.11621
IIXIX:0.10645
ZZZZZ:0.09375

And the probabilities of obtaining 10000 for each stabilizer generator:
XIXII:0.21777
IXXII:0.21875
IIXXI:0.17871
IIXIX:0.14648
ZZZZZ:0.19726

> **Comments:**
> **username_4**: @GilKalai: If you want the full statistics, just drop me an email. They are too long to post here.  
> **username_6**: you could post it in a pastebin and then link it.  

---

